Galois Group of x^4+px+p (14.6.15)

Dummit and Foote Abstract Algebra, section 14.6, exercise 15:

Prove that the polynomial $f(x)=x^4+px+p∈ℚ[x]$ is irreducible for every prime $p$ and for $p \neq 3,5$ has Galois group $S_4$. Prove the Galois group for $p=3$ is dihedral of order 8 and for $p=5$ is cyclic of order $4$.

Proof: We check for rational roots to see if it has a linear factor; $f(-p)=p(p^3-p+1)≠0$, $f(-1)=1$, and $f(1),f(p) > 0$, so we check to see if $f(x)$ splits into two irreducible quadratics. $$f(x)=(x^2+ax+b)(x^2+cx+d)$$ Assume without loss that $d= \pm 1$ and $b= \pm p$. Then assuming $d=1$ and $b=p$ we have by the coefficient of $x$ the relation $a-ap=p$ so $a(1-p)=p$ and testing $a=-1,1,p,-p$ all yield immediate contradictions except where $a=-p$ implying $p=2$, but then $c=2$ and we evidently have $x^4+2x+2≠(x^2-2x+2)(x^2+2x+1)$. The case when $d=-1$ and $b=-p$ is parallel.

Hence $f(x)$ is irreducible over $ℚ$. Now, we note that the resolvent cubic $g(x)$ of $f(x)$ is $x^3-4px+p^2$. To test irreducibility it suffices to check for rational roots. We see $g(p)=p^2(p-3)$ so $p=3$ is a special case here. We see $g(1)=1-4p+p^2≡1~\text{mod p}$ so $g(1)≠0$. We see $g(-1)=-1+4p+p^2≠0$. Finally, $g(-p)=-p^2(p-5)$ so $p=5$ is also a special case. Otherwise, the the resolvent cubic is irreducible. Now, we examine the determinant $D=p^3(256-27p)$. When $p≠2$ we see $p \not \mid 256-27p$ so $D$ has a nonsquare term in its prime factorization. When $p=2$ we have $D=1616=2^4·101$ which is also not a square. Hence for $p≠3,5$ the Galois group is $S_4$.

We further examine $x^4+3x+3$ when $p=3$. As we saw $g(x)=x^3-12x+9=(x-3)(x^2+3x-3)$. Since $3^2+4·3=21$ is not a square we see the quadratic is irreducible. We see $D=4725$ so we examine if $x^4+3x+3$ factors over $ℚ(\sqrt{4725})$. We see the latter is contained in $ℝ$, so it suffices to show $x^4+3x+3$ has no real roots by showing it evaluates positive for all real values. We calculate its derivative $4x^3+3$ is zero when $x=-\sqrt[3]{\dfrac{3}{4}}$, which must be the local minimum as the second derivative $12x^2$ is positive. But assuming $f(-\sqrt[3]{\dfrac{3}{4}})≤0$ implies $\dfrac{3}{4}≥\sqrt[3]{\dfrac{3}{4}}$, a contradiction. Hence $x^4+3x+3$ remains irreducible and the Galois group is dihedral.

We further examine $x^4+5x+5$ when $p=5$. As we saw $g(x)=x^3-20x+25=(x+5)(x^2-5x+5)$. Since $5^2-5·4=5$ is not a square we see the quadratic is irreducible. We see $D=15125=5^3·11^2$ so $ℚ(\sqrt{D})=ℚ(\sqrt{5})$.

Galois Groups of a Particular Quartic Polynomial (14.6.13)

Dummit and Foote Abstract Algebra, section 14.6, exercise 13:

(a) Let $±α$, $±β$ denote the roots of the polynomial $f(x)=x^4+ax^2+b∈\mathbb{Z}[x]$. Prove that $f(x)$ is irreducible if and only if $α^2$ and $α±β$ are not elements of $\mathbb{Q}$.
(b) Suppose $f(x)$ is irreducible and let $G$ be the Galois group of $f(x)$. Prove that
(i) $G≅V$, the Klein 4-group, if and only if $\sqrt{b}∈\mathbb{Q}$ if and only if $αβ∈ℚ$.
(ii) $G≅C$, the cyclic group of order $4$, if and only if $\sqrt{b(a^2-4b)}∈\mathbb{Q}$ if and only if $\mathbb{Q}(αβ)=\mathbb{Q}(α^2)$.
(iii) $G≅D_8$, the dihedral group of order $8$, if and only if $\sqrt{b},\sqrt{b(a^2-4b)}∉ℚ$ if and only if $αβ∉ℚ(α^2)$.

Proof: (a) Note that if $f(α)=0$ then $f(-α)=0$ so it makes sense to refer to the roots as above. ($⇒$) Note that $x^2-α^2~|~f(x)$ so $α^2∉ℚ$. As well, the Galois group is transitive on roots so let $φ(α)=-α$. Then we have $φ(β)=±β$ and thus $φ(α±β)=-α±β$ so either $-α+β=α+β$ in which case $α=0$, or $-α-β=α+β$ in which case $α=-β$ so $f(x)$ is inseparable and hence not irreducible, or $-α+β=α-β$ in which case $α=β$ a contradiction for the same reason, or $-α-β=α-β$ implying $α=0$. Hence $α±β∉ℚ$. ($⇐$) If $b=0$ then either $α=0$ or $α^2∈ℚ$. Note that $α^2=b/β^2$ so also $β^2∉ℚ$. Since neither of $α$ or $β$ are rational we must have $f(x)$ is either irreducible or a product of irreducible quadratics. In the latter case, write $x^4+ax^2+b=(x^2+px+q)(x^2+rx+s)$. Since $p=0$ implies $α^2$ or $β^2$ is rational, simple algebra will show $r=-p$ and $s=q$. Thus the four roots are $\dfrac{±p±\sqrt{p^2-4q}}{2}$ and $α±β$ is rational.

(b)(i) We find that the resolvent cubic is $x^3-2ax^2+(a^2-4b)x=x(x-a-2\sqrt{b})(x-a+2\sqrt{b})$ hence the equivalence is clear after seeing $αβ=\sqrt{b}$.

(ii) ($1⇒2$) Note now that $D=16b(a^2-4b)^2$. By the hypothesis and the procedure for Galois groups $f(x)$ is reducible in $ℚ(\sqrt{D})=ℚ(\sqrt{b})$. Since $f(x)$ is irreducible over $ℚ[x]$, any root generates a fourth degree extension over $ℚ$ and hence is not contained in $ℚ(\sqrt{b})$ so $f(x)$ splits into two irreducible quadratics over $ℚ(\sqrt{b})[x]$. Writing this factoring generally as a system of algebraic equations, we discover either $$x^4+ax^2+b=(x^2+px+\dfrac{p^2+a}{2})(x^2-px+\dfrac{p^2+a}{2}),~~~0≠p∈ℚ(\sqrt{b})$$ or $$x^4+ax^2+b=(x^2+p)(x^2+q)$$ In the latter case, we thus have $x^2+ax+b$ factoring in $ℚ(\sqrt{b})$ so since $x^2+ax+b$ doesn't factor in $ℚ[x]$ hence $\sqrt{a^2-4b}∉ℚ$ we observe $ℚ(\sqrt{b})=ℚ(\sqrt{a^2-4b})$. Therefore $b$ and $a^2-4b$ differ in a square, i.e. $bz^2=a^2-4b$ and $b(a^2-4b)$ is a square.

In the former case, we have some solution $(\dfrac{x^2+a}{2})^2=b$ in $ℚ(\sqrt{b})[x]$ so $x^2+a±2\sqrt{b}$ splits in $ℚ(\sqrt{b})[x]$. Writing the solution $x=v_1+v_2\sqrt{b}$ we see $$v_1^2+bv_2^2=-a$$ $$2v_1v_2=±2$$ so $v_2=±1/v_1$ and substituting into the first equation we see $f(v_1)=0$, so $f(x)$ has a linear factor, a contradiction.

($1⇐2$) Suppose $z=\sqrt{b(a^2-4b)}∈ℚ$. Then $z/\sqrt{b}=\sqrt{a^2-4b}$ so $ℚ(z)=ℚ(\sqrt{b})$. We see $x^2+ax+b$ is reducible in $ℚ(\sqrt{a^2-4b})$, so $x^4+ax^2+b$ is reducible in $ℚ(\sqrt{b})$. Since $ℚ(\sqrt{D})=ℚ(\sqrt{b})$ and $\sqrt{b}∉ℚ$ (else $\sqrt{a^2-4b}∈ℚ$ and $f(x)∈ℚ[x]$ is reducible), we have the resolvent cubic (above) splits into a linear factor and an irreducible quadratic, and $f(x)$ is reducible in $ℚ(\sqrt{D})$, so by the procedure the Galois group is $C$.

($1,2⇒3$) Since $α^2$ is a root of irreducible $x^2+ax+b$ and $αβ=\sqrt{b}$ is a root of irreducible (above) $x^2-b$, we have $α^2$ generates $\sqrt{a^2-4b}$ over $ℚ$ and $αβ$ generates $\sqrt{b}$ over $ℚ$, by (2) the implication follows.

($1,2⇐3$) Since $α^2∉ℚ$ by (a), and $ℚ(α^2)=ℚ(αβ)=ℚ(\sqrt{b})$, we have $b$ is not a square hence the resolvent cubic splits into a linear factor and irreducible quadratic, and since $ℚ(\sqrt{D})=ℚ(\sqrt{b})=ℚ(αβ)=ℚ(α^2)$ as well as $f(x)=(x^2-α^2)(x^2-β^2)$ we observe $f(x)$ is reducible in $ℚ(\sqrt{D})[x]$, so by the procedure the Galois group is $C$.

(iii) This is evident by exhaustion of the previously characterized Galois groups.$~~\square$

Dihedral Galois Closures and Quadratic Extensions (14.6.11-12)

Dummit and Foote Abstract Algebra, section 14.6, exercises 11-12:

11. Let $F$ be a non-Galois extension of degree $4$ over $ℚ$. Show that the Galois closure $L$ of $F$ has Galois group either $S_4$, $A_4$, or $D_8$. Further, show that the Galois group is dihedral if and only if $F$ contains a quadratic extension of $ℚ$.

12. Let $F$ be an extension of degree $4$ over $ℚ$. Show that $F$ can be generated over $ℚ$ by an element with minimal polynomial of the form $x^4+ax^2+b$ if and only if $F$ contains a quadratic extension.

Proof: (11) Let $F=ℚ(θ)$. We observe its minimal polynomial $m_θ(x)$ of degree $4$. It is irreducible, and since simultaneously we have $F⊆L$ and $F$ is not Galois, by the procedure for quartics only the groups of order greater than $4$ remain, which are those above.

($⇒$) By the fundamental theorem, since every subgroup of order $2$ in $D_8$ is contained in a subgroup of order $4$, we have $F$ contains a subfield of index $4$ in $L$, i.e. degree $2$ over $ℚ$.

($⇐$) Let $ℚ(\sqrt{D})⊆F$ be quadratic. Then we have $F=ℚ(\sqrt{D})(\sqrt{a+b\sqrt{D}})=ℚ(\sqrt{a+b\sqrt{D}})$ for any nonsquare $a+b\sqrt{D}$, which exists since quadratic extensions are generally extensions via square roots. Let $α=\sqrt{a+b\sqrt{D}}$. We observe $x^4-2ax^2+a^2-b^2D$ is the minimal polynomial for $α$. The Galois closure is the splitting field for this polynomial. We observe the resolvent cubic for this quartic is $x^3+4ax^2+4b^2Dx$ which clearly has a factor of $x$. By the procedure the Galois group cannot be either $S_4$ or $A_4$, hence must be $D_8$.

(12) ($⇒$) Let $α$ have such a minimal polynomial. Factor $x^2+ax+b=(x-c)(x-d)$ so that $x^4+ax^2+b=(x^2-c)(x^2-d)$. Necessarily $c$ and $d$ are not contained in $ℚ$ as the minimal polynomial must be irreducible, so they must generate a quadratic extension. Hence $α^2$ generates a quadratic extension in $F$.

($⇐$) As we saw previously, without assuming $F$ is non-Galois, we can arrive at a minimal polynomial $x^4-2ax^2+a^2-b^2D$ for a generator of $F$.$~\square$

Maximal Real Subfields (14.5.7,9)

Dummit and Foote Abstract Algebra, section 14.5, exercises 7 and 9:

7. Show that complex conjugation restricts to the automorphism $σ_{-1}∈\text{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$. Show that the field $K^+=\mathbb{Q}(\zeta_n+\zeta_n^{-1})$ is the subfield of real elements in $K=\mathbb{Q}(\zeta_n)$, called the maximal real subfield of $K$.

9. Let $K_n=\mathbb{Q}(\zeta_{2^{n+2}})$ for $n ≥ 0$.
(a) Prove that $K_n^+/\mathbb{Q}$ is cyclic of degree $2^n$.
(b) Prove $K_n/K_{n-1}^+$ is biquadratic and that two of the three intermediate subfields are $K_{n-1}$ and $K^+$. Prove the remaining intermediate field is a cyclic extension of $\mathbb{Q}$ of degree $2^n$.

Proof: (7) Note that complex conjugation is an automorphism of $\mathbb{C}$, and also restricts to an automorphism of $K$ since $\overline{\zeta_n}=\zeta_n^{-1}$ ($|\zeta_n^n|=|\zeta_n|^n=1$ implies for $\zeta_n=a+bi$ that $|\zeta_n|=a^2+b^2=\zeta_n\overline{\zeta_n}=1$). Now, when $n > 2$ we note $\zeta_n∉\mathbb{R}$ so $K_n/K^+$ is of degree greater than or equal to $2$. As well, $\dfrac{1}{2}(\zeta_n+\zeta_n^{-1})=a$ and $1-a^2=b^2$, so when $α$ is a root of the equation $x^2+b^2$ we observe $K^+(α)=K_n$ and hence $K_n/K^+$ is of degree exactly $2$. Since the maximal real subfield is necessarily properly contained in $K_n$, we must have $K^+$ is the maximal real subfield.

(9)(a) The degree of the extension follows from above. Here we may observe the isomorphism $\text{Gal}(K/ℚ)≅(ℤ/2^{n+2}ℤ)^×≅ℤ/2ℤ×ℤ/2^nℤ$ shown in Exercises 2.3.22-23. We show $(ℤ/2^{n+2}ℤ)^×$ is generated by $5$ and $-1$. This is clearly the case when $n=0$ so assume $n ≥ 1$. Now $5$ is of order $2^n$ by the mentioned exercises and it suffices to show $5$ doesn't generate $-1$; we observe this by demonstrating $k_n$ such that $5^{k_n}≡2^{n+1}+1~\text{mod }2^{n+2}$, a different element of order $2$. First, we see $k_1=1$. Now, given $k_n$, we have $k_{n+1}=2k_n$ since $(5^{k_1})^2=(2^{n+1}+1+m2^{n+2})^2≡2^{(n+1)+1}+1~\text{mod }2^{(n+1)+2}$ when $n ≥ 1$. Thus $-1$ and $5$ generate $(ℤ/2^{n+2}ℤ)^×$, and since we notice the quotient group of $(ℤ/2^{n+2}ℤ)^×$ over the subgroup generated by $-1$ is the cyclic group generated by the image of $5$, we conclude $\text{Gal}(K^+/ℚ)≅\text{Gal}(K/ℚ)/\text{Gal}(K/K^+)$ is cyclic.

(b) We note that since $K_n/ℚ$ is Galois necessarily $K_n/K_{n-1}^+$ is Galois. This Galois extension by degree must be either cyclic or biquadratic, and since it cannot be cyclic by the presence of two distinct subfields of degree two, it must be biquadratic. We see that $K_{n-1}$, $K_n^+$, and the third subfield manifest as the fixed fields of the (isomorphic pullbacks) of the elements of order $2$ in $(ℤ/2^{n+2}ℤ)^×$, and viewing their Galois groups as the quotient of this group with the subgroups generated by those elements, we see as a result of only one of them intersecting nontrivialy with the order-$2^n$ subgroup generated by $5$ (that is, $2^{n+1}+1$ as shown above, which corresponds to $K_{n-1}/ℚ$ which we see is not cyclic) that the third subfield's Galois group's representation as a quotient is cyclic.$~\square$

Galois Traces and the Möbius Function (14.5.6,11)

Dummit and Foote Abstract Algebra, section 14.5, exercise 6:

(6) Show $$\text{Tr}_{\mathbb{Q}(\zeta_n)/\mathbb{Q}}(\zeta_n)=\mu(n)$$ (11) Show the primitive $n^{th}$ roots of unity form a basis of $\mathbb{Q}(\zeta_n)$ if and only if $n$ is squarefree.

Proof: (6) Let $n=p_1^{α_1}...p_k^{α_k}$. Since for any choice of primitives we have $\zeta_n=(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^z$ for some $z$, by distributing powers we may simply assume $\zeta_n=\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}}$. Since $\mu(ab)=\mu(a)\mu(b)$ for $(a,b)=1$, it suffices to show $\text{Tr}(\zeta_a\zeta_b)=\text{Tr}(\zeta_a)\text{Tr}(\zeta_b)$ for $(a,b)=1$ and that the proposition holds for $n$ a prime power.

Note that for $1≤m≤n$ relatively prime to $n$, we have $(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^m=\zeta_{p_1^{α_1}}^{r_1}...\zeta_{p_k^{α_k}}^{r_k}$ for $0 ≤ r_i < p_i^{α_i}$ and $(r_i,p_i^{α_i})=1$, and as well for such a selection of $r_i$ we have $\zeta_{p_1^{α_1}}^{r_1}...\zeta_{p_k^{α_k}}^{r_k}=(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^m$ for some $(m,n)=1$ by the Chinese Remainder Theorem is the two-sided inverse of this operation. Since $$\text{Tr}(\zeta_n)=\sum_{(m,n)=1, 1≤m < n} \zeta_n^m$$ we therefore have $\text{Tr}(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})=\text{Tr}(\zeta_{p_1^{α_1}})...\text{Tr}(\zeta_{p_k^{α_k}})$.

Now, we may observe $\text{Tr}(\zeta_p)=\zeta_p+\zeta_p^2+...+\zeta_p^{p-1}=-1$. As well, assume $q=p^a$ for $a≥2$. Then $$\text{Tr}(\zeta_q)=\zeta_q+\zeta_q^2+...+\zeta_q^{p-1}+\zeta_q^{p+1}+...+\zeta_q^{2p-1}+\zeta_q^{2p+1}+...+\zeta_q^{p^a-1}=$$ $$(\zeta_q+...+\zeta_q^{p-1})+\zeta_q^p(\zeta_q+...+\zeta_q^{p-1})+\zeta_q^{2p}(\zeta_q+...+\zeta_q^{p-1})+...+$$ $$\zeta_q^{p(p^{a-1}-1)}(\zeta_q+...+\zeta_q^{p-1})=$$ $$(1+\zeta_q^p+\zeta_q^{2p}+...+\zeta_q^{p(p^{a-1}-1)})(\zeta_q+\zeta_q^2+...+\zeta_q^{p-1})=f(\zeta_q^p)(\zeta_q+\zeta_q^2+...+\zeta_q^{p-1})$$ where $f(x)=\dfrac{x^{p^{a-1}}-1}{x-1}$ which is valid since $\zeta_q^p≠1$. Thus we note $f(\zeta_q^p)=0$ as $(\zeta_q^p)^{p^{a-1}}-1=\zeta_q^{p^a}-1=0$ and so $\text{Tr}(\zeta_q)=0$ and $\text{Tr}(\zeta_q)$ agrees with $\mu(q)$ on prime powers, and since both are relatively multiplicative the proposition is complete.

(11) ($⇒$) By the above, we see $\text{Tr}(\zeta_n)=0$ if $n$ is not squarefree. ($⇐$) Let $n=p_1p_2...p_k$. Then we see $\mathbb{Q}(\zeta_n)=\mathbb{Q}(\zeta_{p_1}\zeta_{p_2}...\zeta_{p_k})$ is the composite of $\mathbb{Q}(p_k)$ and $\mathbb{Q}(\zeta_{p_1...p_{k-1}})$. By induction on prime width, we see the latter field has for basis over $\mathbb{Q}$ the elements $\zeta_{p_1}^{α_1}...\zeta_{p_{k-1}}^{α_{k-1}}$ for $1≤α_i < p_i$. Since the fields have relatively prime degree, we see the basis $\zeta_{p_k}^{α_k}$ for $1≤α_k < p_k$ for $\mathbb{Q}(\zeta_{p_k})$ over $\mathbb{Q}$ remains linearly independent over $\mathbb{Q}(\zeta_{p_1...p_{k-1}})$ (Corollary 13.2.22). Hence a basis for $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$ is $\zeta_{p_1}^{α_1}...\zeta_{p_k}^{α_k}$ for $1≤α_i < p_i$, which by the Chinese Remainder Theorem are all the primitive $n^{th}$ roots.$~\square$

p-Extensions and Galois Closures (14.4.5)

Dummit and Foote Abstract Algebra, section 14.4, exercise 5:

Let $p$ be a prime and let $F$ be a field. Let $K$ be a Galois extension of $F$ whose Galois group is a $p$-group. Such an extension is called a $p$-extension.

Let $L/K$ be a $p$-extension. Prove the Galois closure of $L$ over $F$ is a $p$-extension of $F$.

Proof: Since $L/K$ and $K/F$ are Galois of $p$-power degree, we have $L/F$ is separable and finite (and of $p$-power degree), thus simple. Let $L=F(α)$. We note that the Galois closure of $L$ must contain all the roots of $m_α(x)$ and also that the splitting field of $m_α(x)$ over $F$ is Galois, hence the latter is precisely the closure. Since $K/F$ is Galois, by 14.4.4 we see that in particular $m_α(x)∈F[x]$ splits over $K$ into a product of $n$ irreducibles of same degree $d$, and since $dn=p^a$ (the degree of $L/F$ and $m_α(x)$) we must have $d$ is a $p$-power thus $K(β)/K$ is of $p$-power degree for any root $β$ of $m_α(x)$.

Since the isomorphism $φ~:~K(α)→K(β)$ from mapping $α↦β$ induces an isomorphism $\text{Aut}(K(α)/K)≅\text{Aut}(K(β)/K)$ given by $σ↦φσφ^{-1}$, we see $K(β)/K$ is Galois. By observing degrees in Proposition 14.4.21 we see that composites of $p$-extensions are themselves $p$-extensions ($\text{Gal}(K_1/F)×\text{Gal}(K_2/F)$ is a $p$-group, and so too are its subgroups), hence it follows that the composite of the extensions $K(β_i)$ for roots $β_i$ of $m_α(x)$ (i.e. the splitting field of $m_α(x)$ over $F$, the Galois closure of $L$ over $F$) is in fact of $p$-power degree over $K$ and a $p$-extension of $F$.$~\square$

Automorphism Computation (Research)

Compute $|\text{Aut}(Z_2^4 × D_8)|$.

Let $A=Z_2^4$, $B=D_8$, and $G=\text{Aut}(A×B)$. We note $A$ is abelian, so we consider $F=\text{Aut}_A(A × B)≤G$. We also note $Z(A×B)=Z_2^4 × \langle r^2 \rangle ≅Z_2^5$. Since $A×1 ≤ Z(A×B) \text{ char } A×B$ we observe $[G~:~F]=[\text{Aut}^{A×B}(Z(A×B))~:~\text{Aut}_A^{A×B}(Z(A×B))]$.

Let $Φ∈G$. Then we note $Φ(r^2)=Φ(r)^2$ so since $r^2$ and also $Φ(r^2)$ are in $Z(A×B)$ we have $Φ(r^2)=r^2$ ($r^2$ is the only nonidentity square in $Z(A×B)$). Conversely, let $σ∈\text{Aut}(Z(A×B))$ be such that $σ(r^2)=r^2$. We observe the relations of $A×B$ generated in $\langle a,b,c,d,s,r \rangle$: $$a^2=b^2=c^2=d^2=s^2=r^4=1$$ $$aba^{-1}b^{-1}=aca^{-1}c^{-1}=ada^{-1}d^{-1}=asa^{-1}s^{-1}=ara^{-1}r^{-1}=...=1$$ $$(rs)^2=1$$ where the second line lists all the commutation relations (all commute but $r$, $s$). Some extending map $σ'$ defined by its action on $a,...,r,s$ is an automorphism iff $σ'(a)$, $...$ , $σ'(r),σ'(s)$ generate $A×B$ and $σ'$ is one on these relations generators in the free group. If we let $σ'$ act on $a$, $b$, $c$, $d$ as $σ$ does and set $σ'(s)=s$ and $σ'(r)=r$ we observe $σ'(r^2)=σ(r^2)=r^2$ and it follows that $σ'$ extends $σ$, and is an automorphism since the orders line is fulfilled ($Z(A×B)$ has exponent $2$, so $σ'(a)^2=σ(a)^2=1$ and similar for $b,c,d$ and clearly $σ'(s)^2=σ'(r)^4=1$), the commutation line is fulfilled (all of these relations involve $a$, $b$, $c$, or $d$, and $σ'$ preserves their centricity), the last relation is fulfilled ($σ'$ fixes $D_8$), and the generation is established as $a,...,d$ is in the image of $σ$ on $Z(A×B)$, hence $σ'$ on $A×B$, plus $s=σ'(s)$ and $r=σ'(r)$. Thus $\text{Aut}^{A×B}(Z(A×B))$ is isomorphic to the group of isomorphisms fixing the last coordinate of $Z_2^5$, whose order we compute $(2^5-2^1)(2^5-2^2)(2^5-2^3)(2^5-2^4)=322,560$.

Utilizing the previous reasoning we note that $σ∈\text{Aut}_A(Z(A×B))$ extends iff $σ(r^2)=r^2$, so $|\text{Aut}_A^{A×B}(Z(A×B))|$ is the number of automorphisms of $Z_2^5$ fixing the last coordinate and restricting to an automorphism of $Z_2^4 × 1$, i.e. the number of automorphims of $Z_2^4$ which is computed to be $(2^4-2^0)(2^4-2^1)(2^4-2^2)(2^4-2^3)=20,160$. Thus we calculate $[\text{Aut}^{A×B}(Z(A×B))~:~\text{Aut}_A^{A×B}(Z(A×B))]=16$ distinct cosets of $\text{Aut}_A(A × B)$ in $\text{Aut}(A×B)$.

We have seen $|\text{Aut}_A(A × B)|=|\text{Aut}(A)|·|\text{Hom}(B,A)|·|\text{Aut}(B)|$. We have already computed $|\text{Aut}(Z_2^4)|=20,160$. As well, utilizing the relations for $D_8$ yields $|\text{Aut}(D_8)|=8$. We see $\text{Hom}(B,A)$ is the set of all maps defined freely on $r$ and $s$ in the free group factoring through the relations of $D_8$, i.e. mapping one on the relations of $D_8$, and since $Z_2^4$ has exponent $2$ we see every such mapping is a homomorphism, so $|\text{Hom}(B,A)|=16^2=256$. $$|\text{Aut}(A×B)|=|G|=[G~:~F]·|F|=$$ $$[G~:~F]·|\text{Aut}(A)|·|\text{Hom}(B,A)|·|\text{Aut}(B)|=$$ $$16·20,160·256·8=660,602,880$$ ~~~~~
Alternatively, we calculate the number of automorphisms of the characteristic center that extend $|\text{Aut}^{A×B}(Z_2^4×\langle r^2 \rangle)|=322,560$ (see above) which represent the cosets of the subgroup of automorphisms that fix the center. We see that sending $r$ to either $r$ or $r^3$ with any coordinates for $a$, $b$, $c$, $d$ together with any mapping of $s$ to $s$, $rs$, $r^2s$, or $r^3s$ and any coordinates of $a$, $b$, $c$, $d$ are all automorphisms that fix the center, so $2,048·322,560=660,602,880$.

Reductions in Finite Fields (14.3.6-7,11)

Dummit and Foote Abstract Algebra, section 14.3, exercises 6-7, 11:

6. Suppose $K=\mathbb{Q}(θ)=\mathbb{Q}(\sqrt{D_1},\sqrt{D_2})$ with $D_1,D_2∈\mathbb{Z}$ and that $θ=a+b\sqrt{D_1}+c\sqrt{D_2}+d\sqrt{D_1D_2}$ for integers $a$, $b$, $c$, $d$. Prove $m_θ(x)$ is reducible modulo every prime $p$. In particular show the polynomial $x^4-10x^2+1$ is irreducible over $\mathbb{Z}[x]$ but reducible modulo every prime.
7. Prove that one of $2$, $3$, or $6$ is a square in $\mathbb{F}_p$. Conclude $$x^6-11x^4+36x^2-36=(x^2-2)(x^2-3)(x^2-6)$$ has a root modulo $p$ for every prime $p$ but has no root in $\mathbb{Z}$.
11. Prove that $x^{p^n}-x+1$ is irreducible over $\mathbb{F}_p$ only when $n=1$ or $n=p=2$.

Proof: (6) Lemma 1: Let $f(x),g_i(x)∈\mathbb{Z}[x]$ be irreducible with finite indexing set $I$. If $θ∈\mathbb{Q}(α_1,...)$ for some roots $g_i(α_i)=0$ and the coefficients of $θ$ are given in integers, then if $Φ$ is reduction modulo $p$ and $β_i$ is a root of $Φ(g_i(x))$ we have $θ'$ (the corresponding integers having been reduced mod $p$, and $α_i$ replaced with $β_i$) is a root of $Φ(f)$. Proof: $Φ$ is an additive homomorphism $\mathbb{Z}(α_i)→\mathbb{F}_p(β_i)$ by its construction on a basis, and further it is multiplicative between two basis elements $Φ(\prod α_i)Φ(\prod α_j)=Φ(\prod α_i \prod α_j)$. Thus it is a ring homomorphism and we see $Φ(f)(θ')=Φ(f(θ))=Φ(0)=0$.$~\square$

Now, since a biquadratic extension of $\mathbb{F}_p$ would entail only the four automorphisms negating $\sqrt{D_1}$ and $\sqrt{D_2}$, none of which having order $4$ and thus none representing Frobenius $σ_p$, we must have $\mathbb{F}_p(\sqrt{D_1},\sqrt{D_2})$ is contained in $\mathbb{F}_{p^2}$ and now $m_θ(x)$ of degree $4$ cannot be irreducible mod $p$. With some work we find $\pm \sqrt{2} \pm \sqrt{3}$ are the solutions to the irreducible $x^4-10x^2+1$ and generate $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Hence $m_{\sqrt{2}+\sqrt{3}}(x)=x^4-10x^2+1$ generates only a second degree extension of $\mathbb{F}_p$ and is not irreducible.

(7) Lemma 2: One of $a,b,ab∈\mathbb{F}_{p^n}$ is a square in $\mathbb{F}_{p^n}$. Proof: Since $\_^2$ is a multiplicative endomorphism of $\mathbb{F}_{p^n}^×$ and for each square $α^2$ we have only two possible elements in the fiber by $x^2-α^2=(x-α)(x+α)$, we must have $[\mathbb{F}_{p^n}^×~:~\text{img }\_^2]≤2$. Now if $a$, $b$ are not squares then $a,b \neq 1$ in $\mathbb{F}_{p^n}^×/\text{img }\_^2$. Hence $ab=1$, i.e. $ab$ is a square.$~\square$

Thus we have proved either $2$, $3$, or $6$ is a square in $\text{F}_p$ and thus the polynomial above has a root in $\text{F}_p$.

(11) We note that if $α$ is a root, then so is $α+a$ for any $a∈\mathbb{F}_{p^n}$. Thus $\mathbb{F}_{p^n}⊆\mathbb{F}_p(α)$ as else the latter would contain all the roots of $x^{p^n}-x+1$ and then a nontrivial extension would yet more roots, a contradiction. Now we observe any automorphism of $\text{Gal}(\mathbb{F}_{p}(α)/\mathbb{F}_{p^n})$ must be defined by $α↦α+a$ for some $a∈\mathbb{F}_{p^n}$ and since these automorphisms fix $\mathbb{F}_{p^n}$ they all have order $p$. Since a Galois group of degree $k$ is always cyclic over $\mathbb{F}_p$ with generator $σ_p$ and hence always cyclic over $\mathbb{F}_{p^n}$, so $[\mathbb{F}_p(α)~:~\mathbb{F}_{p^n}]=p$.

Thus we must have $pn=p^n$. We write $n=p^k$ for some $k≥0$. If $k > 1$ then we notice $k+1=p^k$ and since $p≥2$ implies $k+1 < 2^k ≤ p^k$ when $k=2$ and inductively $(k+1)+1 < 2^k+1 ≤ 2^k+(2^{k+1}-2^k) = 2^{k+1} ≤ p^{k+1}$ we must have $k=0$ (and $n=1$) or $k=1$ (in which case $n=p$ and $p^2=p^n$ and $n=p=2$). We have previously verified $x^p-x+1$ is irreducible ($γ$ implies $γ+1$ a root, thus all roots are in the same extension, thus there is the same degree $d$ among $k$ irreducibles implying $kd=p$ and $d=1$, but there are no roots in $\mathbb{F}_p$) and one may check $x^4-x+1∈\mathbb{F}_2[x]$ has no roots and is not the square of the only irreducible quadratic $(x^2+x+1)^2=x^4+x^2+1$.$~\square$

Scratch Ideas (Research)

Let $R$ be a ring, $I$ an indexing set for which each $M_i$ is an $R$-module. We set $M=\bigoplus_{i∈I}M_i$. We say an element $s∈M$ is condensed if all but one of the coordinates of $s$ is zero (including $s=0$). We say a submodule $N⊆M$ is condensing if for every $m∈M$ we have $\overline{m}=\overline{s}∈M/N$ for some condensed $s∈M$. $N$ is further said to be halt-condensing if for condensed $s_1 \neq s_2$ we have $\overline{s_1} \neq \overline{s_2}$.

Let $N$ be a condensing submodule. We have a homomorphism of the modules $M_i$ into $M/N$, and the union of their images is all of $M/N$. Conversely, if there is such a homomorphism of the $M_i$ into a module $K$ with the union property, then we have a surjective homomorphism $φ : M → K$ given by evaluation of the sum of the nonzero components in $K$, giving rise to a condensing submodule as the kernel. The equivalence regarding halt-condensing submodules is modified by requiring the mapped modules to inject and be disjoint at nonzero values.
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Let $R$ be a UFD with group of units $V$. Choose a list of primes $p_i$ of cardinality $|S|=c$ to represent the associativity classes of prime elements. Letting $A=\text{Aut}(V)$ (group automorphisms), $S_c$ be the symmetric group on $c$ objects, and $V^*$ be the group of maps $S→V$ where the operation is multiplication $φ_1φ_2(s)=φ_1(s)φ_2(s)$, we claim $A × S_c × V^*$ contains (a copy of) $\text{Aut}(R)$ (ring automorphisms) where the former is under a custom operation elucidated below.

Let $φ∈\text{Aut}(R)$. Since automorphisms map units to units, we must have $φ|_V∈\text{Aut}(V)$. Now automorphisms also map primes to primes, and more generally associativity classes of primes to other associativity classes, so $φ$'s action on $p_i$ might be completely represented by an element $σ$ from $S_c$ together with a map $ψ:S→V$ given by $ψ(s)=φ(p_s)/p_{σ(s)}$ (i.e. so we may construct $φ$ from $ψ$ and $σ$ by $φ(p_s)=ψ(s)p_{σ(s)}$). Now, $φ$ is uniquely determined by this action on the units and primes, so we have an injection $\text{Aut}(R)$ into $A × S_c × V^*$. In order to claim an algebraic embedding the structure imposed on the latter is not simply componentwise multiplication but is rather modified to imitate composition of automorphisms of $R$. The operation is defined as such:

Let $x_1,x_2=(φ_1,σ_1,ψ_1),(φ_2,σ_2,ψ_2)∈A × S_c × V^*$. Then $x_1x_2=(φ_1∘φ_2,σ_1∘σ_2,(φ_1∘ψ_2)·(ψ_1∘σ_2))$, where multiplication and composition of maps are here distinct.
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Let $A,B$ be groups. Consider $\text{Aut}_A(A \times B)$, the set of automorphisms of $A \times B$ mapping $A \times 1$ to $A \times 1$. This is seen to be a subgroup of $\text{Aut}(A \times B)$ which we shall now classify. Let $\Phi \in \text{Aut}_A(A \times B)$. By definition $\Phi$ restricts to an automorphism $φ$ on $A \times 1$, and by simultaneously defining $(\psi(b), \sigma(b))=\Phi(1,b)$ we obtain homomorphisms $\psi : B \rightarrow A$ and $\sigma : B \rightarrow B$. Further, we see $\sigma$ is injective because if $\sigma(b)=1$ then $\Phi(1,b) \in A \times 1$ and the automorphism $φ$ leads to an element $\Phi(a,1)=\Phi(1,b)$ implying $a,b=1$. Further, $σ$ is surjective seeing as $Φ$ is surjective. Assume $\text{img }ψ \not ∈ Z(A)$; then $aψ(b) \neq ψ(b)a$ for some $a∈A$, $b∈B$. Then we have $$Φ((1,b)(φ^{-1}(a),1))=Φ(1,b)Φ(φ^{-1}(a),1)=$$ $$(ψ(b),σ(b))(a,1) \neq (a,1)(ψ(b),σ(b)) = Φ((φ^{-1}(a),1)(1,b))$$ a contradiction. Since $φ,ψ,σ$ uniquely determine $Φ$, we have an injection of sets $\text{Aut}_A(A × B) → \text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$.

Conversely, let $φ∈\text{Aut}(A)$, $ψ∈\text{Hom}(B,Z(A))$, and $σ∈\text{Aut}(B)$. Define $Φ: A×B → A×B$ by $Φ(a,b)=(φ(a)ψ(b),σ(b))$. We observe $Φ((a_1,b_1)(a_2,b_2))=Φ(a_1,b_1)Φ(a_2,b_2)$, so it is homomorphic. As well, it is injective as $Φ(a,b)=(1,1)$ implies $σ(b)=1$ so $b=ψ(b)=1$ and consequently $a=1$. Finally, we observe surjectivity $(a,b)=Φ(φ^{-1}(aψ(b)^{-1}),σ^{-1}(b))$, so $Φ$ is an automorphism. Thus we have a bijection of sets $\text{Aut}_A(A × B) → \text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$.

Suppose $Φ_1$ associates to $(φ_1,ψ_1,σ_1)$ and $Φ_2$ associates to $(φ_2,ψ_2,σ_2)$. Then we observe $Φ_1Φ_2$ associates to $(φ_1φ_2,φ_1ψ_2+ψ_1σ_2,σ_1σ_2)$. Thus, defining such a binary operation on $\text{Aut}(A) × \text{Hom}(B,Z(A)) × \text{Aut}(B)$ induces a group isomorphism between the two.

Footnotes: This machinery works especially well when $Z(B)$ is manageable, and in particular when $Z(B)=1$ and $A$ is abelian we have $\text{Aut}_A(A × B) = \text{Aut}(A × B)$, as then $Z(A × B) = A × 1$ and so $A × 1$ is characteristic. Otherwise, one may possibly discover $A × 1~\text{char}~A × Z(B)~\text{char}~A × B$ and thus $A × 1~\text{char}~A × B$ to the same effect.
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For groups $A≤B≤G$ let $\text{Aut}^G(A)$ be the subgroup of automorphisms of $A$ extending to automorphisms of $G$. Let $\text{Aut}_A^{G}(B)$ be the subgroup of automorphisms of $B$ mapping $A$ to $A$ and extending to automorphisms of $G$.

Let $H ≤ N \text{ char } G$. Then we have a bijection $ψ$ of coset representatives $[\text{Aut}(G)~:~\text{Aut}_H(G)]=[\text{Aut}^G(N)~:~\text{Aut}_H^G(N)]$ given by restriction $ψ(σ)=σ|_N$. This is injective as $σ_1|_N\text{Aut}_H^G(N)=σ_2|_N\text{Aut}_H^G(N)$ implies $σ_2^{-1}|_Nσ_1|_N∈\text{Aut}_H^G(N)$ so $σ_2^{-1}σ_1∈\text{Aut}_H(G)$, and surjective seeing as the representative $σ|_N$ by definition has an extension $σ$ and letting $σ'$ represent $σ$ we see $σ'|_N\text{Aut}_H^G(N)=σ|_N\text{Aut}_H^G(N)$.
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Let $(X,d)$ be a metric space, and let $\mathcal{C}$ be the set of all nonempty compact subspaces of $X$. Let $A∈\mathcal{C}$. Then given $x∈X$, define $d(x,A)=d(A,x)=\min_{a∈A}d(x,a)$. Note that since $A$ is compact this is well defined. Note also that fixing $A$ this function is continuous $X→ℝ$. Hence given $A,B∈\mathcal{C}$ $$D(A,B)=\max \{\max_{a∈A} d(a,B), \max_{b∈B} d(b,A)\}$$ is well defined. It is clear $D(A,A)=0$, and $D(A,B)=0$ quickly implies $A⊆B$ and $B⊆A$. As well, $D(A,B)=D(B,A)$. Now, we show it satisfies the triangle inequality.

Let $A,B,C∈\mathcal{C}$ and assume $D(A,C) > D(A,B) + D(B,C)$. We may assume $\max_{a∈A} d(a,C) ≥ \max_{c∈C} d(c,A)$, for otherwise interchanging $A,C$ would result in such a situation.

Let $a∈A$ and $β∈B$ be such that $d(a,β)=d(a,B) ≥ d(x,B)$ for all $x∈A$, let $b∈B$ and $γ∈C$ be such that $d(b,γ)=d(b,C)≥d(x,C)$ for all $x∈B$, and let $a'∈A$ and $γ'∈C$ be such that $d(a',γ')=d(a',C)≥d(x,C)$ for all $x∈A$. Since $d(a',B)≤d(a,B)$, let $x∈B$ be such that $d(a',x)≤d(a,β)$. Since $d(x,C)≤d(b,C)$, let $y∈C$ be such that $d(x,y)≤d(b,γ)$. Now we exhibit $$D(A,C) = d(a',γ') ≤ d(a',y) ≤ d(a',x)+d(x,y)$$ $$≤ d(a,β)+d(b,γ) ≤ D(A,B)+D(B,C)$$ Hence $(\mathcal{C},D)$ is a metric space. We garner some facts about this metric space.

Let $A⊆X$, and for each $a∈A$ let $f_a$ be a path from $f_a(0)=a$ to $f_a(1)$. The family $\{f_a\}$ is said to be uniform if for each $ε > 0$ and $r∈[0,1]$ there exists a neighborhood $U$ of $r$ in $[0,1]$ such that $f_a(U)⊆B_d(f_a(r),ε)$ for each $a∈A$. In other words, when $X^A$ is given the uniform topology, if the map $f : [0,1]→X^A$ given by $f(r)=\prod f_a(r)$ is continuous. When $B=∪f_a(1)$, we say $\{f_a\}$ is a uniform path from $A$ to $B$.
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Let $σ : ℕ→I^2$ be such that $\overline{σ(ℕ)}=I^2$. Given $ε > 0$ and a function $g : ℕ→\{0,1\}$, we may define $$h_g : I^2→I$$ $$h_g(x)=\lim_{N→∞}\sum_{σ(i)∈B(x,ε)}^N \dfrac{g(i)}{N}$$ when the above limit exists for all $x∈I^2$ (also: consider weighting the sum for guaranteed convergence). When $σ$ and $ε$ are fixed, which continuous functions $f:I^2→I$ are constructible as $h_g$ for some $g∈ℕ^{\{0,1\}}$?

Let $p : I→I^2$ be a path with only finitely many self-intersection points, i.e. $|p^{-1}(x)|=1$ for all but finitely many $x∈I^2$. Choose a countable dense subset $Q⊆[0,1]$ ordered by $φ : ℕ→Q$ with the property that, for all open $U⊆[0,1]$, the sequence $$\dfrac{|\{φ(1),φ(2),...,φ(N)\}∩U|}{N}$$ converges as $N→∞$ (the dyadic rationals under the usual ordering suffice). Let $V⊆[0,1]$ be a countable dense subset disjoint from $p(I)$ with ordering $ψ : ℕ→V$. When $σ : ℕ→I^2$ is defined $σ(1)=p∘φ(1)$, $σ(2)=ψ(1)$, $σ(3)=ψ(2)$, $σ(4)=p∘φ(2)$, $...$ (alternating between primes and non-primes), then $σ$ is an ordering of the countable dense subset $Q∪V$ (with the exception of finitely many points; modify without loss). Define $g : ℕ→\{0,1\}$ by $g(i)=1$ if $σ(i)∈Q$, and $g(i)=0$ otherwise. Then $h_g(x)$ denotes a well-defined function, which when the dyadic rationals as above are chosen and the limit of $$\dfrac{|\{φ(1),φ(2),...,φ(N)\}∩U|}{N}$$ as $N→∞$ thus coincides with the Lebesgue measure, and measures the "length" of the segment contained the $ε$-ball about $x$. For example, when $ε=1/2$ and $p(x)=0×x$, we observe $$h_g(x,y)=\text{min }\{1,y+\sqrt{1-4x^2}/2\}-\text{max }\{0,y-\sqrt{1-4x^2}/2\}$$ for $x≤1/2$ and $h_g(x,y)=0$ elsewhere.

Quaternion Galois Group (14.2.27)

Dummit and Foote Abstract Algebra, section 14.2, exercise 27:

Let $α=\sqrt{(2+\sqrt{2})(3+\sqrt{3})}$ and consider the extension $E=\mathbb{Q}(α)$.
(a) Show that $a=(2+\sqrt{2})(3+\sqrt{3})$ is not a square in $F=\mathbb{Q}(\sqrt{2},\sqrt{3})$.
(b) Conclude from (a) that $[E~:~\mathbb{Q}]=8$. Prove that the roots of the minimal polynomial over $\mathbb{Q}$ for $α$ are the 8 elements $\pm \sqrt{(2 \pm \sqrt{2})(3 \pm \sqrt{3})}$
...
(f) Conclude $\text{Gal}(E/\mathbb{Q})≅Q_8$.

Proof: (a) We observe that if $a=c^2$ then $aφa=c^2φc^2=(cφc)^2$ where $φ$ is the automorphism of $F$ fixing $\sqrt{2}$ and negating $\sqrt{3}$. Since $cφc$ is actually the image of $c$ under $N_{F/\mathbb{Q}(\sqrt{2})}$ we have $cφc=\sqrt{aφa}=\sqrt{6(3+\sqrt{3})^2}=\pm (3\sqrt{2}+3\sqrt{6})∈\mathbb{Q}(\sqrt{2})$ and now $\sqrt{6}∈\mathbb{Q}(\sqrt{2})$, a contradiction.

(b-f) (We take a slightly different path from the authors, especially for parts c and d) We have shown $[\mathbb{Q}(\sqrt{2},\sqrt{3},α)~:~\mathbb{Q}]=8$. Since $\dfrac{α^2}{2+\sqrt{2}}-3=\sqrt{3}$, we have $\mathbb{Q}(\sqrt{2},\sqrt{3},α)=\mathbb{Q}(\sqrt{2},α)=E(\sqrt{2})$. Assume $[E(\sqrt{2})~:~E]=2$; then $E(\sqrt{2})$ is Galois of degree $2$ over $E$ (splitting $x^2-2$), and the map $φ:\sqrt{2}↦-\sqrt{2}$ is an automorphism of $E(\sqrt{2})$ fixing $E$. But $φ$ is in particular an isomorphism of $F$ allowing us to observe $φ(α^2)=φ((2+\sqrt{2})(3+\sqrt{3}))=(2-\sqrt{2})(3+\sqrt{3}) \neq α^2$, a contradiction. So $E=E(\sqrt{2})$ and $[E~:~\mathbb{Q}]=8$.

Now as we saw above, $E$ is the splitting field for $x^2-α^2$ over $F$. Therefore every automorphism of $F$ extends to an automorphism of $E$. This is an order $4$ subgroup of $\text{Aut}(E/\mathbb{Q})$. Since $E/F$ is Galois we also have the automorphism $ψ:α↦-α$ fixing $F$. Letting $n=|\text{Aut}(E/\mathbb{Q})|$ by Lagrange we see $4~|~n$, by Galois we see $n≤8$, and by counting we see $n > 4$, so that $n=8$ and $E/\mathbb{Q}$ is Galois.

We see that the 8 elements mentioned above are distinct by observing squares, and since $φ(x^2)=φ(x)^2$ for general automorphisms we have $φ(x)=\pm \sqrt{φ(x^2)}$. Letting $H=\text{Aut}(E/F)$ we see $1,ψ$ form a set of right coset representatives for $H$ in $\text{Aut}(E/\mathbb{Q})$. By letting $λ∈H$ fix $\sqrt{3}$ and negate $\sqrt{2}$, for example, we see $λψ(α^2)=(2-\sqrt{2})(3+\sqrt{3})$ and thus $λψ(α)=\pm \sqrt{(2-\sqrt{2})(3+\sqrt{3})}$. Similarly, we can see any automorphism maps $α$ to one of the 8 forms above, and thus must map to all of them, and these are the 8 distinct roots of the minimal polynomial for $α$ over $\mathbb{Q}$.

Let $σ$ map $α$ to $β=\sqrt{(2-\sqrt{2})(3+\sqrt{3})}$. We see $σ(α^2)=β^2$ so that $σ(\sqrt{2})=-\sqrt{2}$ and $σ(\sqrt{3})=\sqrt{3}$, and together with $αβ=\sqrt{2}(3+\sqrt{3})$ we have $σ(αβ)=-αβ$ and thus $σ(β)=-α$. Now $σ$ can be seen to be of order $4$ and together with $\tau$ mapping $α$ to $γ=\sqrt{(2+\sqrt{2})(3-\sqrt{3})}$ we similarly find the relations $σ^4=\tau^4=1$, $σ^2=\tau^2$, and $σ\tau = \tau σ^3$ (keeping in mind $β=\dfrac{\sqrt{2}(3+\sqrt{3})}{α}$), so that $\text{Gal}(E/\mathbb{Q})≅Q_8$.$~\square$

Calculation of Galois Groups (14.2.10,12)

Dummit and Foote Abstract Algebra, section 14.2, exercises 10, 12:

10. Determine the Galois group of the splitting field over $\mathbb{Q}$ of $x^8-3$.
12. Determine the Galois group of the splitting field over $\mathbb{Q}$ of $x^4-14x^2+9$.

Proof: (10) Letting $θ=\zeta_8=\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}i$, we have the $8$ roots of this polynomial are $θ^a\sqrt[8]{3}$ for $a=0,1,...,7$. Therefore the splitting field for this polynomial is $\mathbb{Q}(\sqrt[8]{3})(\sqrt{2})(i)$. We note $x^8-3$ is irreducible over $\mathbb{Q}$ by Eisenstein, so the first extension is degree $8$, and assuming $x^2-2$ isn't irreducible over $\mathbb{Q}(\sqrt[8]{3})$ leads to a solution $$(a_0+a_1\sqrt[8]{3}+...+a_7\sqrt[8]{3}^7)^2=2$$ However, we notice the coefficient of the basis element $1$ here is $$a_0^2+6a_1a_7+6a_2a_6+6a_3a_5+3a_4^2=2$$ We notice that the integral domain of elements of the form $b_0+b_1\sqrt[8]{3}+...+b_7\sqrt[8]{3}^7$ for integers $b_i$ has for field of fractions $\mathbb{Q}(\sqrt[8]{3})$, because the latter contains the former and the former contains the latter by writing fractional coefficients under a common denominator. Thus by Gauss's lemma we may assume the $a_i$ are integers, and reducing modulo $3$ the equality is impossible. Therefore $\mathbb{Q}(\sqrt[8]{3})(\sqrt{2})$ is of degree $16$, and since this field is contained in $\mathbb{R}$ we see $K=\mathbb{Q}(\sqrt[8]{3})(\sqrt{2})(i)$ is of degree $32$ over $\mathbb{Q}$.

We see there are $8*2*2=32$ permutations of the roots, therefore these are all automorphisms, so $\text{Gal}(K/\mathbb{Q})$ is a group generated by the automorphisms $$α~:~\sqrt[8]{3}↦θ\sqrt[8]{3}$$ $$β~:~\sqrt{2}↦-\sqrt{2}$$ $$γ~:~i↦-i$$ We see these elements satisfy $α^8=β^2=γ^2=1$, $βγ=γβ$, $βα=α^5β$, and $γα=α^3γ$, and also these relations on a free group of three generators is sufficient to write any element in the form $α^aβ^bγ^c$, of which there are $32$ combinations, so this is precisely the set of relations. $$\text{Gal}(K/\mathbb{Q})=<α,β,γ~|~α^8=β^2=γ^2=1,~βγ=γβ,~βα=α^5β,~γα=α^3γ>$$ Now, since $3^2≡5^2≡7^2≡1~\text{mod }8$ we observe $\text{Aut}(Z_8)=Z_2^2$, so letting $φ$ be the isomorphism between these two groups, letting $a$ generate $Z_8$, and $b,c∈Z_2^2$ be such that $φ(b)(a)=a^5$ and $φ(c)(a)=a^3$, we observe the same relations between these elements in $Z_2^2 \rtimes_φ Z_8$ also of order $32$, so that finally we may say $$\text{Gal}(K/\mathbb{Q})=Z_2^2 \rtimes_φ Z_8$$ (12) Finding the roots of the polynomial in $x^2$, we obtain the solutions $α=\sqrt{7+2\sqrt{10}}$, $β=\sqrt{7-2\sqrt{10}}$, $-α$, and $-β$. We note $αβ=\sqrt{7^2-(2\sqrt{10})^2}=3$, so that $β=3/α$ and the splitting field is merely $\mathbb{Q}(α)$. We see that $\mathbb{Q}(α)$ is of degree $4$ since the polynomial points to $≤4$ and $\mathbb{Q}(\sqrt{10}) ⊂ \mathbb{Q}(α)$ (for the proper inclusion, consider a solution to $x^2-α^2$ over $\mathbb{Z}(\sqrt{10})$ by Gauss's lemma). The automorphisms of this extension must be the permutations of $α$ about the roots of its minimal polynomial, so we observe $φ~:~α ↦ -α$, $ψ~:~α↦β=3/α$, and $φψ$ are all of order $2$, so $$\text{Gal}(\mathbb{Q}(α)/\mathbb{Q})=Z_2^2~\square$$

Computations with Galois Theory (14.2.1-3)

Dummit and Foote Abstract Algebra, section , exercise :

1. Determine the minimal polynomial over $\mathbb{Q}$ for the element $\sqrt{2}+\sqrt{5}$.
2. Determine the minimal polynomial over $\mathbb{Q}$ for the element $1+\sqrt[3]{2}+\sqrt[3]{4}$.
3. Determine the Galois group of $(x^2-2)(x^2-3)(x^2-5)$.

Proof: Lemma: Let $K/F$ be Galois over a perfect field and let $α∈K$. Then the minimal polynomial of $α$ over $F$ is the squarefree part of the polynomial $$\prod_{σ∈\text{Gal}(K/F)}(x-σα)$$ Proof: We show $\{σα~|~σ∈\text{Gal}(K/F)\}$ is the full set of zeros for the minimal polynomial $p(x)$. They are all zeros as the automorphisms fix the coefficients of $p(x)$, and as well for any other root $β$ of $p(x)$ we have the isomorphism $F(α)→F(β)$ and extending automorphism $K$ preserving this map, since being Galois $K$ is a mutual splitting field for $F$, $F(α)$, and $F(β)$.

As well, $F$ being perfect, $p(x)$ is separable and thus has no repeated roots, and we may now say $p(x)~|~\prod_{σ∈\text{Gal}(K/F)}(x-σα)$. Since the zeros of the latter are precisely the (nonrepeated) zeros of the former, we have the squarefree part is indeed $p(x)$.$~\square$ This solves the problem of determining minimal polynomials (over perfect fields) when the Galois group of a containment Galois extension is known (and still provides much information when $F$ isn't perfect).

(1) We see $\sqrt{2}+\sqrt{5}∈\mathbb{Q}(\sqrt{2},\sqrt{5})=\mathbb{Q}(\sqrt{2})(\sqrt{5})$, where the latter is computed to be Galois (splitting field of $(x^2-2)(x^2-5)$) of degree $4$. The four automorphisms must be the four uniquely defined by the identity, $α~:~\sqrt{2}↦-\sqrt{2}$, $β~:~\sqrt{5}↦-\sqrt{5}$, and the composite $αβ$. Thus the four roots of the minimal polynomial are $\pm \sqrt{2} \pm \sqrt{5}$ and the minimal polynomial is calculated to be $x^4-14x^2+9$.

(2) We put the element in the Galois extension $\mathbb{Q}(ρ,\sqrt[3]{2})$ and observe the effect of the elements of the Galois group previously determined. The result is the polynomial $$(x-1-\sqrt[3]{2}-\sqrt[3]{4})(x-1-ρ\sqrt[3]{2}-ρ^2\sqrt[3]{4})(x-1-ρ\sqrt[3]{4}-ρ^2\sqrt[3]{2})=$$ $$x^3-3x^2-3x-1$$ (3) With some simple algebra we may see $x^2-5$ is irreducible over the field $\mathbb{Q}(\sqrt{2},\sqrt{3})$ of degree $4$ over $\mathbb{Q}$, so the splitting field in question $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ is Galois of degree $8$. We see the automorphisms are precisely the automorphisms generated by the maps $α~:~\sqrt{2}↦-\sqrt{2}$, $β~:~\sqrt{3}↦-\sqrt{3}$, $γ~:~\sqrt{5}↦-\sqrt{5}$. By noting that these automorphisms all commute and are all of order $2$, we conclude the Galois group is $Z_2^3$.$~\square$

Automorphisms of Polynomial Rings over Fields (14.1.6, 8-9)

Dummit and Foote Abstract Algebra, section 14.1, exercises 6, 8-9:

6. Let $k$ be a field. Show that the automorphisms of $k[t]$ fixing $k$ are precisely the linear transformations defined by $t ↦ at+b$ for $a \neq 0$.
8. Show that the automorphism of $k(t)$ fixing $k$ are precisely the fractional linear transformations defined by $t ↦ \dfrac{at+b}{ct+d}$ for $ad-bc \neq 0$.
9. Determine the fixed field of the automorphism $t ↦ t+1$ of $k(t)$.

Proof: (6) Let $φ$ be such a mapping. It is seen to be a ring homomorphism as evaluation at any polynomial is seen to be a ring homomorphism. Moreover, letting $k[x]_i$ denote the subspace spanned by $1,t,...,t^i$ over $k$, when $a \neq 0$ we see the basis elements of $k[x]_i$ are mapped to linear combinations of its preimage basis, showing $φ$ is bijective on $k[x]_i$ and by extension on $k[x]$ since $∪k[x]_i=k[x]$.

Conversely, any automorphism fixing $k$ is uniquely defined by its action on $t$, so observe the polynomial $f(t)=φ(t)$. We have $φ(g(t))=g(f(t))=t$ for some $g(t)∈k[x]$ by surjectivity, and since $\text{deg }g(f(t)) = \text{deg }f(t)~\text{deg }g(t)$, necessarily $f(t)=at+b$ for some $a \neq 0$.

(8) As before, evaluation is an endomorphism. When $ad-bc \neq 0$, $at+b$ and $ct+d$ are relatively prime (one of them may be in $k$) and by 13.2.18 $[k(t)~:~\text{img }φ]=1$ so $φ$ is surjective. Now, assume $f(\dfrac{at+b}{ct+d})=0$ for $f(t) = ∑a_kt^k$ ($f(t)$ being fractional implies the existence of a nonfractional polynomial satisfying such). Letting $n=\text{deg }f(t)$ and observing $(ct+d)^nf(\dfrac{at+b}{ct+d})= a_n(at+b)^n + (ct+d)g(t) = 0$ for some polynomial $g(t)$, we see by the fact that $k[t]$ is a UFD that $ct+d$ is a unit and injectivity follows by (6).

As before, all endomorphisms are evaluations, and writing them in the form $φ~:~t↦p(t)/q(t)$ for relatively prime $p(t),q(t)$ by 13.2.18 for it to be surjective necessarily the greatest degree is $1$. Clearly $ad-bc \neq 0$ as otherwise $\text{img }φ=k$.

(9) By the previous this is indeed an automorphism. Now, let $f(t)=\dfrac{pt)}{q(t)}$ for relatively prime $p(t),q(t)$ and monic $p(t)$ be a typical element of the fixed field, i.e. $f(t)=f(t+1)$. Then $\dfrac{p(t)}{q(t)}=\dfrac{p(t+1)}{q(t+1)}$ and $p(t)q(t+1)=p(t+1)q(t)$. Assuming $p(t) \neq p(t+1)$ implies $p(t) \not \mid p(t+1)$ since they are monic of the same degree, so there is some irreducible factor on the left not present on the right, a contradiction. Now $p(t)q(t+1)=p(t)q(t)$ so $q(t)=q(t+1)$ and it suffices to find the collection of polynomials in $k[t]$ fixed by $t ↦ t+1$.

If $k$ is of characteristic $0$ then $f(t)=f(t+1)$ implies $f(α)=0$ implies $f(α+1)=0$, so $f(t)$ has no zeros in any field and $f(t)∈k$. In this case the fixed field is merely $k$. Now consider the polynomial $λ(t)=t(t-1)...(t-(p-1))$ in $k[t]$, where $p$ is the characteristic of $k$. Clearly $λ(t)=λ(t+1)$, so all polynomials generated as a ring by $λ(t)$ and $k$ are fixed by $t↦t+1$. Conversely, if $f(t)=f(t+1)$ and $f(0)=a_0$, then for the polynomial $F(t)=f(t)-f(0)$ we have $F(t)=F(t+1)$ and $F(0)=0$ so also $F(1),...,F(p-1)=0$ and $λ(t)~|~F(t)$. By induction on degree $F(t)/λ(t)$ is in the ring generated by $λ(t)$ and $k$ and so too is $f(t)$, so this ring in $k[t]$ (also known as the image of $φ~:~f(t)↦f(λ(t))$ on $k[t]$, to provide a way of efficiently determining whether a polynomial is fixed by $t↦t+1$) extended to a field in $k(t)$ is precisely the field fixed by $t↦t+1$.$~\square$

Canonical forms of the Fröbenius Endomorphism (13.6.11-12)

Dummit and Foote Abstract Algebra, section 13.6, exercises 11-12:

11-12. Let $φ$ denote the Fröbenius map $$φ:\mathbb{F}_{p^n}→\mathbb{F}_{p^n}$$ $$x↦x^p$$ Find the rational and Jordan (when it exists) canonical form of $φ$.

Proof: (11) We saw $φ^n=1$, so since $[\mathbb{F}_{p^n}~:~\mathbb{F}_p]=n$ we note $φ$ is an $n \times n$ matrix and the invariant factors of $φ$ all divide $x^n-1$. Assume there are $m > 1$ invariant factors, so $h_1(x)~|~h_2(x)$ are invariant factors and write $h_2(x)=f(x)h_1(x)$. Observe elements of the form $\bigoplus_{k=1}^m a_k$ where $f(x)~|~a_k$ when $k=2$, and $a_k=0$ when $k > 2$. These are all in the kernel of $h_1(x)$, and yet there are more than $p^a$ of them, implying there are more than $p^a$ solutions to the polynomial of degree $p^a$ represented by the linear transformation $h_1(x)$, contradiction. Therefore the sole invariant factor is one of degree $n$ dividing $x^n-1$, necessarily $x^n-1$ itself and the Jordan canonical form is the matrix with $1$s along the subdiagonal and a $1$ in $1,n$.

(12) Let $n=p^km$ for $p \not | m$. We see $x^n-1=(x^m-1)^{p^k}$ in $\mathbb{F}_{p^n}[x]$, so that by previous investigations the invariant factors are some powers of $(x-\alpha)$ where $\alpha$ is a power of an $n^{th}$ primitive root of unity, and moreover there are $m$ such distinct roots. By the same reason as above, and since the degree of the product of the polynomials must be $n$, the Jordan canonical form (when it exists, i.e. when there is an $m^{th}$ primitive root in $\mathbb{F}_{p^n}$, iff $m~|~p^n-1$) of $φ$ is the matrix with Jordan blocks $(x-\zeta_{m})^{p^k}$.

Finite Extensions of Q and Roots of Unity (13.6.5)

Dummit and Foote Abstract Algebra, section 13.6, exercise 5:

Prove there are only a finite number of roots of unity in any finite extension $K$ of $\mathbb{Q}$.

Proof: Let $ψ(k)=p_1p_2...p_k$, where $p_i$ is the $i^{th}$ prime.

Lemma 1 (Local Extrema of Totient Ratio): If $n≤ψ(k)$ then $φ(n)/n ≥ \dfrac{(p_1-1)(p_2-1)...(p_k-1)}{p_1p_2...p_k}$. Proof: Collect $n$ such that $φ(n)/n$ is minimal, and then choose $n=q_1^{α_1}q_2^{α_2}...q_m^{α_m}$ minimal from this collection. Assume $α_i > 1$ for some $i$; then $$φ(n)/n=\dfrac{q_1^{α_1-1}(q_1-1)q_2^{α_2-1}(q_2-1)...q_m^{α_m-1}(q_m-1)}{q_1^{α_1}q_2^{α_2}...q_m^{α_m}}=$$ $$\dfrac{(q_1-1)(q_2-1)...(q_m-1)}{q_1q_2...q_m}=φ(n/q_i)/(n/q_i)$$ and $n/q_i < n$, violating minimality. So $n$ is squarefree. Let $p_j$ be the smallest prime not dividing $n$, which must be $≤p_k$ else $n=ψ(k)$ and $φ(n)/n$ equals the bound given above. If there is no prime larger than $p_k$ dividing $n$ then set $m=p_jn ≤ ψ(k)$, and otherwise let $q_v$ be this prime and set $m=p_jn/q_v < n ≤ ψ(k)$. In the first case we see $φ(m)/m=\dfrac{p_j-1}{p_j}φ(n)/n < φ(n)/n$ and in the second $φ(m)/m = \dfrac{q_v(p_j-1)}{(q_v-1)p_j}φ(n)/n < φ(n)/n$, invariably violating minimality. Thus the bound holds.$\square$

Lemma 2: $φ(n)→∞$. Proof: Choose finite positive $z$, and let $k$ be such that $(p_1-1)(p_3-1)...(p_k-1) > z$ (index $2$ is missing). We show when $n > ψ(k)$ that $φ(n) > z$. Let $k'$ be such that $ψ(k'-1) < n ≤ ψ(k')$ so that $k' > k ≥ 3$. We observe $$φ(n) = (φ(n)/n)n ≥ \dfrac{(p_1-1)(p_2-1)...(p_{k'}-1)}{p_1p_2...p_{k'}}p_1p_2...p_{k'-1} =$$ $$\dfrac{(p_1-1)(p_2-1)...(p_{k'}-1)}{p_{k'}} ≥ (p_1-1)(p_3-1)...(p_{k'-1}-1) ≥$$ $$(p_1-1)(p_3-1)...(p_k-1) > z~~\square$$ Now, since there are only a finite number of primitive roots for any $n$, $K$ must contain $n^{th}$ primitive roots for $n$ arbitrarily large. Since the degree $φ(n)$ of the cyclotomic minimal polynomial for these primitive roots also becomes arbitrarily large, we must have $K/\mathbb{Q}$ is not finite.$~\square$

Facts About General Roots of Unity (13.6.1-4)

Dummit and Foote Abstract Algebra, section 13.6, exercises 1-4:

1. Suppose $m$ and $n$ are relatively prime integers. Prove $\zeta_m \zeta_n$ is a primitive $mn^{th}$ root of unity.

2. Let $d~|~n$. Prove $\zeta_n^d$ is a primitive $(n/d)^{th}$ root of unity.

3. Prove that if a field contains the $n^{th}$ roots of unity for odd $n$ then it also contains the $2n^{th}$ roots of unity.

4. Prove that if $n=p^km$ for $p~\not \mid~m$ then there are precisely $m$ distinct $n^{th}$ roots of unity over a field of characteristic $p$.

Proof: (1) Let the field in question be of characteristic $p$ or $0$, and assume without loss that $p~\not \mid~n$. Further assume that $p~\not \mid m$ or that the characteristic is $0$. Then by (4) we may assume $\zeta_m,\zeta_n$ are of orders $m,n$ respectively. We see $(\zeta_m\zeta_n)^{mn}=1$ so the order of $\zeta_m\zeta_n$ is $≤mn$. As well, $(\zeta_m\zeta_n)^m=\zeta_n^m$ is of order $n$, so $n$ divides the order of $\zeta_m\zeta_n$. Similarly $m$ divides this order and $\zeta_m\zeta_n$ is of order $mn$ and thus a primitive $mn^{th}$ root of unity (still keeping mind of the case of characteristic $p$, as the $mn$ zeros of $x^{mn}-1$ are all distinct).

Now assume exclusively characteristic $p$ and $m=p^km'$ for $p~\not \mid~m'$. Then $x^m-1=(x^{m'}-1)^{p^k}$ and so $\zeta_m$ is of order $m'$ and $\zeta_m\zeta_n$ is of order $m'n$. Finally, we see $x^{mn}-1=(x^{m'n}-1)^{p^k}$ so there are exactly $m'n$ distinct solutions and thus $\zeta_m\zeta_n$ is a primitive $mn^{th}$ root.

(2) Suppose characteristic $0$. Then clearly $\zeta_n^d$ is of order $≤n/d$ and at least of this order, so is a primitive root. Suppose characteristic $p$ and write $d=p^jd'$ and $n=p^kn'$ for maximal $j,k$. Then $x^{n/d}-1=(x^{n'/d'}-1)^{p^{k-j}}$ has $n'/d'$ solutions, so since $\zeta_n$ is of order $n'$ by (4) and thus $\zeta_n^d$ of order $n'/d'$, we thus have $\zeta_n^d$ is primitive.

(3) Since $-1$ is the $2^{nd}$ root of unity, by (1) $-\zeta_n$ are the $2n^{th}$ roots of unity.

(4) We have $x^n-1=(x^m-1)^{p^k}$ where $D_x x^m-1 = mx \neq 0$ has no zeros in common with $x^m-1$, so there are $m$ distinct roots of $x^m-1$ and thus of $x^n-1$.$\square$

Characterization of Finite Subfield Structure (13.5.3-4)

Dummit and Foote Abstract Algebra, section 13.5, exercises 3-4:

3. Prove $d~|~n⇔x^d-1~|~x^n-1$ in $\mathbb{Z}[x]$.

4. Let $1 < a ∈ \mathbb{Z}$. Prove for positive $n,d∈\mathbb{Z}$ that $d~|~n⇔a^d-1~|~a^n-1$. Conclude in particular that $\mathbb{F}_{p^d}⊆\mathbb{F}_{p^n}⇔d~|~n$.

Proof: (3) ($⇒$) Evidently $(x^d-1)(x^{n-d}+x^{n-2d}+...+x^d+1)=x^n-1$. ($⇐$) Write $f(x)(x^d-1)=x^n-1$ and assume $d~\not \mid~n$ so $n=qd+r$ for some $0 < r < d$. We thus have every solution of $x^d-1$ is a solution of $x^n-1$ in any field containing $\mathbb{Z}$. Let $\zeta$ be a primitive $d^{th}$ root of unity. Then $\zeta^n-1=\zeta^{qd+r}-1=\zeta^r-1 \neq 0$, a contradiction.

(4a) First we note that $a^n-1 ≡ a^r-1~\text{mod }a^d-1$ if $n≡r~\text{mod }d$. This is because, after writing $n=qd+r$ for $0 ≤ r < d$, we have $1≡(a^d)^q ≡ a^{qd} ≡ a^{n-r}$ and now $a^n ≡ a^r$. Since $0 ≤ a^r - 1 < a^d - 1$ we have $a^n - 1 ≡ 0$ if and only if $n ≡ 0~\text{mod }d$.

(4b) Assume $d~|~n$; then $p^d-1~|~p^n-1$ so $x^{p^d-1}-1~|~x^{p^n-1}-1$ and the splitting field of the latter contains the former, i.e. $\mathbb{F}_{p^d}⊆\mathbb{F}_{p^n}$. Assume $\mathbb{F}_{p^d}⊆\mathbb{F}_{p^n}$; then we see these are exactly the splitting fields of $x^{p^d-1}-1$ and $x^{p^n-1}-1$ respectively, and $$x^{p^d-1}-1=\prod_{α∈\mathbb{F}_{p^d}} (x-α)~|~\prod_{α∈\mathbb{F}_{p^n}} (x-α) = x^{p^n-1}-1$$ implying $p^d-1~|~p^n-1$ implying $d~|~n$.$~\square$

Splitting Field Computations (13.4.1-4)

Dummit and Foote Abstract Algebra, section 13.4, exercises 1-14:

Determine the degree of the splitting field over $\mathbb{Q}$ for the following polynomials:
$x^4-2$
$x^4+2$
$x^4+x^2+1$
$x^6-4$

Proof: $x^4-2$: Letting $α$ be a solution to this polynomial, we see the elements $α$,$iα$,$-α$, and $-iα$ are the four distinct solutions. Therefore we have the splitting field contains $α$ and $i$ and also a field containing $α$ and $i$ contains the splitting field, so the splitting field is precisely $\mathbb{Q}(α,i)$. We shall show $x^4-2$ is irreducible over $\mathbb{Q}[x]$ by first observing it has no roots in $\mathbb{Z}$ and also does not decompose into two quadratics: $$x^4-2=(x^2+ax+b)(x^2+cx+d)$$ $$bd=-2⇒d=-2/b$$ $$ad+bc=0⇒c=2a/b^2$$ $$b+d+ac=0⇒a^2=-b(b^2-2)/2$$ $$a+c=0⇒a=2a/b^2⇒a=0,b^2-2=0⇒b \not ∈ \mathbb{Z}$$ Thus $[\mathbb{Q}(α)~:~\mathbb{Q}]=4$, and since $\mathbb{Q}(α)⊂\mathbb{R}$ we have $x^2+1$ irreducible over $\mathbb{Q}(α)$, so the computed degree is $8$.

$x^4+2$: Again letting $α$ be a root we have the splitting field is $\mathbb{Q}(α,i)$. We shall show $x^4+2$ is irreducible over $\mathbb{Q}(i)$ by observing it doesn't have a root (else the splitting field would be $\mathbb{Q}(i)$ despite the fact that $α^2=\pm \sqrt{2}i \not ∈ \mathbb{Q}(i)$) and by computations similar as above it doesn't decompose into quadratics unless there exists $b$ such that $b^2-2=0$, despite $\pm \sqrt{2} \not ∈ \mathbb{Q}(i)$. Thus the degree is $8$.

$x^4+x^2+1$: After treating this as a quadratic in $x^2$ and applying some algebra, we come to the factorization in $\mathbb{C}[x]$ $$x^4+x^2+1=(x-1/2-\sqrt{3}/2i)(x-1/2+\sqrt{3}/2i)$$ $$(x+1/2-\sqrt{3}/2i)(x+1/2+\sqrt{3}/2i)$$ Therefore the splitting field is precisely $\mathbb{Q}(\sqrt{-3})$ and the degree is $2$.

$x^6-4$: We observe $x^6-4=(x^3-2)(x^3+2)$. Letting $α$ be a solution to $x^3-2$ and $\zeta$ being a primitive third root of unity, we see $α$, $\zeta α$, and $\zeta^2 α$ are the three solutions. As well, $-α$, $-\zeta α$, and $-\zeta^2 α$ are the solutions to $x^3+2$. Therefore the splitting field is $\mathbb{Q}(α,\zeta)$. Letting $α$ be the positive real solution, since $\zeta$ is not real and of degree two we must have the degree over $\mathbb{Q}$ is $2*3=6$.

Tensor Products and Field Extensions (13.3.22)

Dummit and Foote Abstract Algebra, section 12.3, exercise 22:

Let $K/F$ be a finite field extension, and let $K_1,K_2⊆K$ be field extensions of $F$. Show the $F$-algebra $K_1 ⊗_F K_2$ is a field if and only if $[K_1K_2~:~F] = [K_1~:~F][K_2~:~F]$.

Proof: Let $A$ be the set of finite sums of elements of the form $k_1k_2$ for $k_1∈K_1,k_2∈K_2$, let $φ : K_1 × K_2 → A$ be the bilinear map defined by $φ(k_1,k_2)=k_1k_2$, and let $Φ : K_1 ⊗ K_2 → A$ be the corresponding $F$-linear transformation. We observe $$Φ(k_1⊗k_2)Φ(k_1'⊗k_2') = k_1k_1'k_2k_2' = Φ((k_1⊗k_2)(k_1'⊗k_2'))$$ allowing us to show $$Φ(\sum_i k_{i1} ⊗ k_{i2})Φ(\sum_j k_{j1}'⊗k_{j2}')=\sum_{i,j} Φ(k_{i1} ⊗ k_{i2})Φ(k_{j1}' ⊗ k_{j2}') =$$ $$\sum_{i,j} Φ((k_{i1} ⊗ k_{i2})(k_{j1}' ⊗ k_{j2}')) = Φ((\sum_i k_{i1} ⊗ k_{i2})(\sum_j k_{j1}' ⊗ k_{j2}'))$$ so that $Φ$ is an $F$-algebra homomorphism. Note $A = \text{img Φ}$. As well, let $K_1$ have for basis over $F$ $\{n_i\}$ and let $K_2$ have $\{m_j\}$.

($⇒$) We thus have $Φ$ is a nonzero field homomorphism, and is thus an isomorphism. Now $A$ is a field and by definition we have $K_1K_2 ⊆ A$ and by construction we observe $A ⊆ K_1K_2$ so that $K_1K_2 ≅ K_1 ⊗ K_2$ as $F$-algebras, the latter of which has for basis $\{n_i ⊗ m_j\}$ of order $[K_1~:~F][K_2~:~F]$. ($⇐$) We still have $A ⊆ K_1K_2$, and since we observe $[K_1~:~F][K_2~:~F]$ elements $n_im_j$ of $A$ linearly independent over $F$ by Proposition 13.2.21, we must have this is a basis for $K_1K_2$ and thus again $A = K_1K_2$. The $F$-algebra homomorphism $Φ$ above sends basis to basis and is thus an isomorphism, and now $K_1 ⊗ K_2$ is a field.$~\square$

Applications of Algebraic Extensions (13.2.16-17)

Dummit and Foote Abstract Algebra, section 13.2, exercises 16-17:

16. Let $K/F$ be an algebraic extension and let $R$ be a ring where $F ⊆ R ⊆ K$. Show $R$ is a field.

17. Let $f(x)∈F[x]$ be irreducible of degree $n$, and let $g(x)∈F[x]$. Prove that every irreducible factor of $f(g(x))$ has degree divisible by $n$.

Proof: (16) It suffices to show that every element $r∈R$ has a multiplicative inverse. Since $r∈K$ is algebraic over $F$, we observe $r^{-1}∈F(r) ⊆ R$.

(17) Let $h(x) \mid f(g(x))$ be irreducible of degree $k$, and let $α$ be a solution to $h(x)$, so that $h(α)=0$ and thus $f(g(α))=0$. Since $f(x)$ is irreducible, we must have $g(α)$ is of degree $n$. We thus have $$\text{deg }h(x) = k = [F(α)~:~F] =$$ $$[F(α)~:~F(g(α))] \cdot [F(g(α))~:~F] = [F(α)~:~F(g(α))] \cdot n$$ so that $n \mid k$.$~\square$

Convergence of Matrices (12.3.40-45)

Dummit and Foote Abstract Algebra, section 12.3, exercises 40-45:

40. Letting $K$ be the real or complex field, prove that for $A,B∈M_n(K)$ and $α∈K$:
(a) $||A+B|| ≤ ||A|| + ||B||$
(b) $||AB|| ≤ ||A|| \cdot ||B||$
(c) $||αA|| = |α| \cdot ||A||$

41. Let $R$ be the radius of convergence of the real or complex power series $G(x)$.
(a) Prove that if $||A|| < R$ then $G(A)$ converges.
(b) Deduce that for all matrices $A$ the following power series converge: $$\text{sin}(A)=\sum_{k=0}^∞ (-1)^k \dfrac{A^{2k+1}}{(2k+1)!}$$ $$\text{cos}(A)=\sum_{k=0}^∞(-1)^k \dfrac{A^{2k}}{(2k)!}$$ $$\text{exp(A)}=\sum_{k=0}^∞\dfrac{A^k}{k!}$$ 42. Let $P$ be a nonsingular $n \times n$ matrix, and denote the variable $t$ by the matrix $tI$ (in light of the theory of differential equations).
(a) Prove $PG(At)P^{-1}=G(PAtP^{-1})=G(PAP^{-1}t)$, so it suffices to consider power series for matrices in canonical form.
(b) Prove that if $A$ is the direct sum of matrices $A_1,...,A_m$, then $G(At)$ is the direct sum of the matrices $G(A_1t),...,G(A_mt)$.
(c) Show that if $Z$ is the diagonal matrix with entries $z_1,...,z_n$ then $G(Zt)$ is the diagonal matrix with entries $G(z_1t),...,G(z_nt)$.

43. Letting $A$ and $B$ be commuting matrices, show $\text{exp}(A+B)=\text{exp}(A)\text{exp}(B)$.

44. Letting $λ ∈ K$, show $$\text{exp}(λIt+M)=e^{λt}\text{exp}(M)$$ 45. Let $N$ be the $r \times r$ matrix with $1$s on the first superdiagonal and zeros elsewhere. Show $$\text{exp}(Nt)=\begin{bmatrix}1 & t & \dfrac{t^2}{2!} & \cdots & \cdots & \dfrac{t^{r-1}}{(r-1)!} \\ ~ & 1 & t & \dfrac{t^2}{2!} &~ & \vdots \\ ~ & ~ & \ddots & \ddots & \ddots & \vdots \\ ~ & ~ & ~ & \ddots & t & \dfrac{t^2}{2!} \\ ~ & ~ & ~ & ~ & 1 & t \\ ~ & ~ & ~ & ~ & ~ & 1 \end{bmatrix}$$ Deduce that if $J$ is the $r \times r$ elementary Jordan matrix with eigenvalue $λ$ then $$\text{exp}(Jt)=\begin{bmatrix}e^{λt} & te^{λt} & \dfrac{t^2}{2!}e^{λt} & \cdots & \cdots & \dfrac{t^{r-1}}{(r-1)!}e^{λt} \\ ~ & e^{λt} & te^{λt} & \dfrac{t^2}{2!}e^{λt} &~ & \vdots \\ ~ & ~ & \ddots & \ddots & \ddots & \vdots \\ ~ & ~ & ~ & \ddots & te^{λt} & \dfrac{t^2}{2!}e^{λt} \\ ~ & ~ & ~ & ~ & e^{λt} & te^{λt} \\ ~ & ~ & ~ & ~ & ~ & e^{λt} \end{bmatrix}$$ Proof: (40)(a) We have $$||A+B||=\sum_{i,j}|a_{ij}+b_{ij}| ≤ (\sum_{i,j}|a_{ij}|)+(\sum_{i,j}|b_{ij}|)=||A||+||B||$$ (b) We have $$||AB||=\sum_{i,j}|\sum_{k=0}^n a_{ik}b_{kj}| ≤ \sum_{i,j} \sum_{k=0}^n |a_{ik}| \cdot |b_{kj}| ≤$$ $$\sum_{i,j,i',j'} |a_{ij}| \cdot |b_{i'j'}| = (\sum_{i,j}|a_{ij}|)(\sum_{i,j}|b_{ij}|) = ||A|| \cdot ||B||$$ (c) We have $$||αA|| = \sum_{i,j}|αa_{ij}| = |α| \sum_{i,j} |a_{ij}| = |α| \cdot ||A||$$ (41)(a)(Method due to Project Crazy Project) For each entry $i,j$ entry $a_{(k)ij}$ of $A^k$, we note that $|a_(k){ij}| ≤ ||A^k|| ≤ ||A||^k < R^k$. Therefore, we note $\sum_{k=0}^N |α_ka_{(k)ij}| ≤ \sum_{k=0}^N |α_k| \cdot r^k$, where $r$ has been chosen $||A|| < r < R$. By the Cauchy-Hadamard theorem this last power series displays the same radius of convergence $R$ and thus converges for $r$ as $N$ approaches infinity. Thus the partial power series for $G(A)$ in the $i,j$ coordinate converges absolutely, and thus converges.

(b) Since these functions are shown to have radius of convergence $R = ∞$ over $K$, by the above they converge for matrices of arbitrary absolute value, i.e. for all matrices.

(42)(a) Lemma 1: Let $x_n → x$ and $y_n → y$ be convergent sequences of $m × m$ matrices over $K$. Then $x_ny_n → xy$. Proof: We see $x → x_n$ if and only if for any $ε > 0$ we have $||x-x_N|| < ε$ for sufficiently large $N$, since the forward is evident when the entries all converge within range of $ε/m^2$, and the converse forces all entries to converge within $ε$. For sufficiently large $N$ we have - for some asymptotically small matrices $||ε_{(N)1}||,||ε_{(N)2}||$ - the result $$||xy-x_Ny_N|| = ||xy-(x+ε_{(N)1})(y+ε_{(N)2})|| =$$ $$||xε_{(N)2}+ε_{(N)1}y+ε_{(N)1}ε_{(N)2}|| ≤ ||x|| \cdot ||ε_{(N)2}|| + ||y|| \cdot ||ε_{(N)1}|| + ||ε_{(N)1}|| \cdot ||ε_{(N)2}||$$ which vanishes to zero.$~\square$

Now we see $$PG(At)P^{-1}=(\text{lim }P)(\text{lim }G_N(At))(\text{lim }P)=$$ $$\text{lim }PG_N(At)P^{-1} = \text{lim }G_N(PAtP^{-1}) = G(PAtP^{-1}) = G(PAP^{-1}t)$$ (b) By considering lemma 2 of 12.3.38, we see that the power series' summands may be computed in blocks, leading to independent convergences as blocks.

(c) This is simply a special case of (b).

(43) Lemma 2: $\lim_{n → ∞}\dfrac{x^n}{\lfloor n/2 \rfloor !}=0$ for any real $x$. Proof: When $n > x^2$ $$\lim_{n → ∞}\dfrac{x^n}{\lfloor n/2 \rfloor !} ≤ \lim_{n → ∞}x\prod_{k=1}^{\lfloor n/2 \rfloor} \dfrac{x^2}{k} = \lim_{n → ∞}x \prod_{k = 0}^{\lfloor x^2 \rfloor}\dfrac{x^2}{k} \prod_{k = \lceil x^2 \rceil}^{\lfloor n/2 \rfloor}\dfrac{x^2}{k} =$$ $$x \prod_{k = 0}^{\lfloor x^2 \rfloor}\dfrac{x^2}{k} \lim_{n → ∞} \prod_{k = \lceil x^2 \rceil}^{\lfloor n/2 \rfloor}\dfrac{x^2}{k} = 0~~\square$$ Lemma 3: Let $x_n → x$ and $y_n → y$ be convergent sequences of $m × m$ matrices over $K$ such that $|x_n-y_n| → 0$. Then $x=y$. Proof: Assume $x \neq y$ so $|x-y| > 0$. Let $|x-x_n| < |x-y|/2 - ε$ for $n > n_1$ and some chosen $0 < ε < |x-y|/2$, let $|y-y_n| < |x-y|/2$ for $n > n_2$, let $|x_n-y_n| < ε$ for $n > n_3$, and choose $N > \text{max}(n_1,n_2,n_3)$. We have $$|x-y| = |(x-x_N)+(x_N-y_N)+(y_N-y)| ≤$$ $$|x-x_N|+|x_N-y_N|+|y-y_N| < |x-y|$$ a contradiction.$~\square$

Utilizing lemma 1, we have $\lim_{N → ∞}\text{exp}_n(A)\lim_{N → ∞}\text{exp}_n(B)=\lim_{N → ∞}\text{exp}_n(A)\text{exp}_n(B)$, so we may compare terms of the two sequences $$\text{exp}_n(A)\text{exp}_n(B)=(\sum_{k=0}^n\dfrac{A^k}{k!})(\sum_{k=0}^n\dfrac{B^k}{k!})=\sum_{j=0}^n\sum_{k=0}^n\dfrac{A^jB^k}{j!~k!}=\sum_{j,k ≤ n} \dfrac{A^jB^k}{j!~k!}$$ $$\text{exp}_n(A+B)=\sum_{k=0}^n \dfrac{(A+B)^k}{k!} = \sum_{k=0}^n \dfrac{\sum_{j=0}^k \dfrac{k!}{j!(k-j)!}A^jB^{k-j}}{k!} =$$ $$\sum_{k=0}^n \sum_{j=0}^k \dfrac{A^jB^{k-j}}{j!(k-j)!} = \sum_{j+k ≤ n}\dfrac{A^jB^k}{j!~k!}$$ and then compare their differences (without loss of generality assume $||A|| ≤ ||B||$) $$|\text{exp}_n(A+B)-\text{exp}_n(A)\text{exp}_n(B)| = |\sum_{j,k ≤ n < j+k}\dfrac{A^jB^k}{j!~k!}| ≤ \sum_{j,k ≤ n < j+k} \dfrac{||A||^j||B||^k}{j!~k!} ≤$$ $$\dfrac{n^2||B||^{2n}}{\lfloor n/2 \rfloor !} = \dfrac{||B||^2 \prod_{k=2}^n\dfrac{k^2||B||^{2k}}{(k-1)^2||B||^{2k-2}}}{\lfloor n/2 \rfloor !} ≤ ||B||^2\dfrac{(4||B||^2)^{n-1}}{\lfloor n/2 \rfloor !}$$ When $4||B||^2 < 1$ we clearly have the above vanishing to zero as $n$ approaches infinity, so assume $4||B||^2 ≥ 1$ $$||B||^2\dfrac{(4||B||^2)^{n-1}}{\lfloor n/2 \rfloor !} ≤ ||B||^2\dfrac{(4||B||^2)^n}{\lfloor n/2 \rfloor !}$$ and by lemma 2, the term still tends toward $0$. Thus by lemma 3 we conclude $\text{exp}(A+B)=\text{exp}(A)\text{exp}(B)$.

(44) This is clear from the fact $\text{exp}(λt) = e^{λt}$, the ring isomorphism between $K$ and matrices $KI$, and the preceding exercise.

(45) The first part is clear from the fact that $N^k$ is the matrix with $1$s along the $k^{th}$ superdiagonal (observable most easily by induction from the linear transformation form of $N$), and since $N^r = 0$, we have $\text{exp}(Nt)=\text{exp}_r(Nt)$ is the matrix described above. The second part is clear from the observation in (44) with $M=Nt$ coupled with the first part.$~\square$

Matrix Roots (12.3.37-39)

Dummit and Foote Abstract Algebra, section 12.3, exercises 37-39:

37. Let $J$ be a Jordon block of size $n$ with eigenvalue $λ∈\mathbb{C}$.
(a) If $λ \neq 0$, prove the Jordan canonical form of $J^2$ is the Jordan block of size $m$ with eigenvalue $λ^2$.
(b) If $λ = 0$, prove the Jordan canonical form of $J^2$ is two Jordan blocks of sizes $\dfrac{m}{2}$, $\dfrac{m}{2}$ if $m$ is even and of sizes $\dfrac{m-1}{2}$, $\dfrac{m+1}{2}$ if $m$ is odd.
38. Determine the necessary and sufficient conditions for a matrix over $\mathbb{C}$ to have a square root.
39. Let $J$ be a Jordan block of size $m$ with eigenvalue $λ$ over a field $F$ with characteristic 2. Determine the Jordan canonical for for the matrix $J^2$. Determine the necessary conditions for a matrix over $F$ to have a square root.

Proof: We consider a more general case - determining when a matrix over $\mathbb{F}$ has an $n^{th}$ root for any natural $n$, first when $n \neq 0$ in $F$, and when $n = 0$ in $F$ and $n$ is prime in $\mathbb{Z}$.

(37)(a) Let $T$ be the linear transformation of $J$ with regard to some basis $e_1,...,e_m$. We shall show $$J^n=\begin{bmatrix}λ^n{n \choose 0} & λ^{n-1}{n \choose 1} & \cdots & {n \choose n}λ^0 & 0 & \cdots \\ 0 & λ^n{n \choose 0} & λ^{n-1}{n \choose 1} & \ddots & 0 & \cdots \\ 0 & 0 & λ^n{n \choose 0} & \ddots & 0 & \cdots \\ \ddots & & & \cdots & \cdots & \cdots \\ & \ddots & & & \cdots & \cdots \end{bmatrix}$$ In the language of linear transformations, this is equivalent to showing $T^n$ is the linear transformation acting on the basis by $$T^n(e_k)=\sum_{j=0}^{k-1}λ^{n-j}{n \choose j}e_{k-j}$$ This is evident for $n=1$, so we prove the proposition by induction: $$T^n(e_k)=T \circ T^{n-1}(e_k) = T(\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}e_{k-j}) =$$ $$\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}T(e_{k-j})$$ We note at this point that $T$ acts on the basis by $T(e_k)=λe_k+e_{k-1}$ if $k > 1$ and $T(e_1)=λe_1$, so we may continue $$\sum_{j=0}^{k-1}λ^{n-1-j}{n-1 \choose j}T(e_{k-j})=$$ $$[\sum_{j=0}^{k-2}{n-1 \choose j}(λ^{n-j}e_{k-j}+λ^{n-j-1}e_{k-j-1})]+{n-1 \choose k-1}λ^{(n-1)-(k-1)+1}e_1=$$ $$λ^ne_k+[\sum_{j=1}^{k-2}({n-1 \choose j} λ^{n-j}+{n-1 \choose j-1}λ^{n-1-(j-1)})e_{k-j}]+$$ $$({n-1 \choose k-1}λ^{n-k+1}+{n-1 \choose k-2}λ^{n-1-(k-2)})e_1=$$ $$λ^ne_k+[\sum_{j=1}^{k-2}λ^{n-j}{n \choose j}e_{k-j}]+λ^{n-(k-1)}e_1=\sum_{j=0}^{k-1}λ^{n-j}{n \choose j}e_{k-j}$$ And so our claim is proven. Now, since $J^n-λ^n$ is an upper triangular matrix, it is nilpotent and thus $m_J(x) | (x-λ^n)^m$. As well, note the nullity of $T^n-λ^n$ on $V$; we have $(T^n-λ^n)V$ is generated by $λ^{n-j}{n \choose j}e_{m-j}$ for $1≤j≤m-1$, which arises as a linear combination of the linearly independent elements $e_1,...,e_{m-1}$ when $λ \neq 0$, ${n \choose 1} = n \neq 0$ and so $(T^n-λ^n)V$ is of rank $m-1$ and $T^n-λ^n$ has nullity of $1$. Thus there is one Jordan block in the Jordan canonical form of $J$ with eigenvalue $λ^n$ and by the minimal monomial this must be the only block, and hence the Jordan canonical form of $J^n$ for $J$ a Jordan block of size $m$ with eigenvalue $λ \neq 0$ is a Jordan block of size $m$ with eigenvalue $λ^n$.

(b) When $λ=0$, by the above we have $J^n$ is the matrix where the $i,j$ entry is $1$ if $j=i+n$ and all other entries are $0$. Thus $T^n$ acts on the basis by $T(e_k)=0$ if $k ≤ n$ and $T(e_k)=e_{k-n}$ when $k > n$. Write $m = an+b$ where $0 ≤ b < n$. Assume $m < n$ so $a = 0$: Then $J^n=0$ and there are $m=b$ blocks of size $1=a+1$ so the form is of the same pattern as follows when $a ≥ 1$: We may see $r_k = m-nk$ when $k ≤ a$ and $r_k = 0$ when $k > a$, where $r_k = \text{dim}(T^n)^kV$. We count the Jordan blocks by 12.3.30: $$r_{a-1}-2r_a+r_{a+1}=m-n(a-1)-2(m-na)=$$ $$m-na-2(m-na)+n=n-b$$ $$r_a-2r_{a+1}+r_{a+2}=m-na=b$$ Thus there are $n-b$ blocks of size $a$ and $b$ blocks of size $a+1$ (all with eigenvalue $0$, remember). We count their combined dimension over $F$ by $(n-b)a+b(a+1)=na+b=m$ so this describes the Jordan form entirely.

38. We prove a matrix $A$ over $F$ wherein $n \neq 0$ and which contains all the eigenvalues of its matrices (e.g. $\mathbb{C}$) has an $n^{th}$ root if and only if:

1) Each of its nonzero eigenvalues has an $n^{th}$ root in $F$, and
2) It is possible to divide its list of exponents of its invariant factors $x^k$ into regions with generation rules:
     i) Size $m$ containing only $1$s for $m < n$, and
     ii) Size $n$ containing elements which all differ from each other by $1$ or $0$.

Lemma 1: Let $G$ be a ring. An element $g∈G$ has a multplicative $n^{th}$ root in $G$ if and only if each of the elements in its similarity class has an $n^{th}$ root. Proof: ($⇐$) is clear, and ($⇒$) follows by $f^n = g ⇒ (hfh^{-1})^n = hf^nh^{-1}=hgh^{-1}$.$~\square$

Lemma 2: Let $A$ be the block matrix with blocks $A_1,...,A_n$ and let $B$ be the block matrix with blocks $B_1,...,B_n$ where $A_i$ is the same size as $B_i$. Then $AB$ is the block matrix with blocks $A_1B_1,...,A_nB_n$. Proof: Expanding $AB=(A_1'+...+A_n')(B_1'+...+B_n')$ where $A_i'$ is the block matrix in position of a matrix of size $A$, and similar for $B_i'$, by viewing these individual blocks as collapses onto disjoint subspaces followed by a particular linear transformation, we see $A_i'B_j'=0$ for $i \neq j$ and $A_i'B_i'=(A_iB_i)'$. Therefore $(A_1'+...+A_n')(B_1'+...+B_n')=(A_1B_1)'+...+(A_nB_n)'$ is the block matrix described above.$~\square$

By the above lemmas and our knowledge of the Jordan canonical forms of powers of Jordan blocks, the logical equivalence now holds. $A$ has an $n^{th}$ root if and only if there is a matrix for which $B^n$ is of the same Jordan canonical form as $A$ if and only if there is such $B$ that is in Jordan canonical form itself (since $(P^{-1}BP)^n=P^{-1}B^nP$ and $B^n$ are in the same similarity class), and we see $B^n$ has blocks each of which can be separately conjugated into Jordan canonical form giving rise to a block matrix $P$ of conjugating blocks conjugating $B^n$ into its Jordan canonical form as the block sum of the Jordan canonical forms of its component blocks. By the nature of how these canonical forms arise in these blocks by (37), this completes the proof.

39. The second requirement is the same as above but applies to each class of eigenvalues as well as $0$, as when $A_λ$ is the matrix in Jordan canonical form with eigenvalue $λ$ of some fixed size, then $A_λ^n-λ^n=A_0^n$, since when $n$ is prime $n | {n \choose k}$ for all $0 < k < n$.$~\square$

General Canonical Forms (12.3.25-28)

Dummit and Foote Abstract Algebra, section 12.3, exercises 25-28:

25. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{Q}$ whose entries are all equal to $1$.
26. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{F}_p$ whose entries are all equal to $1$.
27. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{Q}$ whose diagonal entries are all equal to $0$ and other entries are all $1$.
28. Determine the Jordan canonical form of the $n \times n$ matrix over $\mathbb{F}_p$ whose diagonal entries are all equal to $0$ and other entries are all $1$.

Proof: When $n=1$ all these matrices are already in Jordan canonical form, so assume $n > 1$.

(25) Let $B=A-n$ and $C=AB$. We see $$c_{ij}=\sum_{k=1}^n a_{ik}b_{kj}=\sum_{k=1}^nb_{kj}=b_{jj}+\sum_{k=1,k \neq j}b_{kj}=1-n+n-1=0$$ so now $m_A(x) \mid x(x-n)$ and since $A,B \neq 0$ we have $m_A(x)=x(x-n)$ and $A$ is diagonalizable with $n$s and $0$s.

Assume there are multiple $n$s down the diagonal of the Jordan form of $A$, so there are multiple $x-n$ factors in the invariant decomposition. Letting $1_a,1_b∈\mathbb{Q}^n$ be the $\mathbb{Q}[x]$ generators of these summands we see $x(1_a)$ and $x(1_b)$ are linearly independent over $\mathbb{Q}$. But we see $A \begin{bmatrix}f_1 \\ ... \\ f_n \end{bmatrix}=\begin{bmatrix}\sum f_i \\ ... \\ \sum f_i \end{bmatrix}$ so that all elements under the image of $A$ are either zero or associate to $\begin{bmatrix}1 \\ ... \\ 1 \end{bmatrix}$, a contradiction. Therefore the Jordan canonical form of this matrix is the $n \times n$ matrix with $a_{1,1}=n$ and all other entries $0$.

(26) Assume $p \not \mid n$. Then as before we can conclude $m_A(x)=x(x-n)$ and that there is only one $x-n$ invariant factor and the Jordan canonical form is as above. Assume $p \mid n$. Then $x=x-n$ and since $A \neq 0$ we have $m_A(x)=x^2$. Once again we see that assuming there is more than invariant factor not dividing $x$ leads to linearly independent elements in the image of $A$, so that we may conclude the invariant decomposition is $x,...,x,x^2$ and the Jordan canonical form is simply the matrix with $a_{1,2}=1$ and all other entries $0$.

(27) Letting $A$ be this matrix and $A'$ the matrix of (25), we see $A+1=A'$ and thus $((x+1)-n)(x+1)=(x-n+1)(x+1)=0$. As before we note there is one associativity class in the image of $A'=A+1$ so that there is exactly one invariant factor of $x-n+1$ in the decomposition, and so the Jordan canonical form is the matrix with $a_{1,1}=n-1$, the other diagonal entries $-1$ and the other entries $0$.

(28) When $p \not \mid n$ we have $n \not ≡ 0$ and so $x-n+1 \neq x+1$ and the Jordan canonical form is as above. When $p \mid n$ we again notice there is one invariant factor of $(x+1)^2$ and thus the Jordan canonical form is the matrix with $a_{1,2}=1$, the diagonal entries all $-1$ and the other entries $0$.$~\square$

Diagonalization of Special Matrices (12.3.21-22)

Dummit and Foote Abstract Algebra, section 12.3, exercise 21-22:

21. Let $A$ be a matrix such that $A^2=A$. Show that $A$ can be diagonalized with $1$s and $0$s down the diagonal.

22. Let $A$ be a matrix such that $A^3=A$. Show that $A$ can be diagonalized over $\mathbb{C}$. Is this true over any field $F$?

Proof: (21) Let $f_i(x)^{α_i}$ be the $i^th$ invariant factor in the variant factor decomposition of $V$ over $F[x]$. Let $1$ be viewed as the $F[x]$ generator of this direct summand. Since $x^2(1)=x(1)$ we have $x(x-1)=0$. Since $-1 \neq 0$ and $f_i(x)^{α_i}$ is a power of a single prime power dividing $x(x-1)$ we must have $f_i(x)^{α_i}∈\{x,x-1\}$. Thus in the Jordan form of $A$ it is diagonal with $1$s and $0$s down the diagonal.

(22) Restarting as above we can see $f_i(x)^{α_i}$ divides $x^3-x=x(x-1)(x+1)$. When $F$ has characteristic greater than $2$ these are all distinct prime factors and thus $A$ is diagonalizable with $-1$s, $0$s, and $1$s down the diagonal. When $F$ has characteristic $2$ we note $x-1=x+1$ so $x^3-x=x(x-1)^2$ and thus by choosing the Jordan canonical form of the matrix with invariant factors $(x-1)^2$ we obtain a matrix that is not diagonalizable and is seen to satisfy $$\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}^3=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}~\square$$

Prime Cyclotomic Polynomials Modulo p (12.2.20)

Dummit and Foote Abstract Algebra, section 12.2, exercise 20:

Let $n$ be prime, and let $Φ_n(x) = \dfrac{x^n-1}{x-1} = x^{n-1}+...+1$ be the $n^{th}$ cyclotomic polynomial, which is irreducible over $\mathbb{Q}$. Fix a prime $p$. This exercise determines when $Φ_n(x)∈\mathbb{F}_p[x]$ is irreducible and thus provides information about the conjugacy classes of matrices of prime order over prime fields.

(a) Let $p = n$. Show $x-1$ divides $Φ_n(x)$.
(b) Let $p \neq n$. Let $f$ be the multiplicative order of $p$ in $\mathbb{F}_n^\times$. Show that $m=f$ is the smallest integer for which $\text{GL}_m(\mathbb{F}_p)$ contains a matrix $A$ of order $n$.
(c) Show that no polynomial in $\mathbb{F}_p[x]$ of degree less than $f$ divides $Φ_n(x)$, and that $m_A(x)$ is an irreducible polynomial of degree $f$ dividing $Φ_n(x)$.
(d) In particular, prove $Φ_n(x)$ is irreducible over $\mathbb{F}_p$ if and only if $p$ is a primitive root modulo $n$, i.e. $p$ generates $\mathbb{F}_n$.

Proof: (a) Since $Φ_n(1)≡0$, $x-1$ divides $Φ_n(x)$.

(b) We have $|GL_m(\mathbb{F}_p)|=(p^m-1)...(p^m-p^{m-1})$. By Lagrange's and Cauchy's theorems, $GL_m(\mathbb{F}_p)$ has a matrix of order $p$ if and only if its order is divisible by $p$, so - since $\mathbb{F}_p$ is an integral domain - if and only if $n \mid p^m-p^k=p^k(p^{m-k}-1)$ for some $0≤k≤m$. Since this first occurs when $m=f$ and $k=0$, the claim is proven.

(c) Assuming $g(x)~|~Φ_n(x)~|~x^n-1$ implies $m_B(x)~|~c_B(x) = g(x)~|~x^n-1$ where $B$ is the companion matrix of $g(x)$. Thus $B^n=1$ and by (b) $g(x)$ is of degree $≥f$. Letting $A$ be as in (b) we see $m_A(x)~|~x^n-1=(x-1)Φ_n(x)$ and by the restriction of polynomials dividing $Φ_n(x)$ and the size of $A$ we have $m_A(x)$ is of degree $f$. Since $m_A(x)$ is a monomial of minimal degree dividing $Φ_n(x)$ it must be irreducible, and since $A \neq 1$ we have $m_A(x) \neq x-1$ and now $m_A(x)~|~Φ_n(x)$.

(d) The case is clear when $p=n$. Otherwise, let $p$ be a primitive root. Then $f=n-1$ and no polynomial of degree smaller than $n-1$ divides $Φ_n(x)$, and thus $Φ_n(x)$ is irreducible. If $Φ_n(x)$ is irreducible, then a polynomial of degree smaller than $n-1$ divides $Φ_n(x)$ and thus the order of $p$ in $\mathbb{F}_n$ is smaller than $n-1$.$~\square$

Matrix Similarity Classes Over Extension Fields of Q (12.2.13)

Dummit and Foote Abstract Algebra, section 12.2, exercise 13:

Show that there are the same number of similarity classes of $3 \times 3$ matrices over $\mathbb{Q}$ for a given characteristic polynomial over $\mathbb{Q}[x]$ as there are for when the entries are over any extension field of $\mathbb{Q}$. Give an example to show this is not true in general for $4 \times 4$ matrices.

Proof: Consider all the cases of the decomposition of $c_A(x)∈\mathbb{Q}[x]$. Let $a \neq b \neq c$.

$(x+a)(x+b)(x+c)$: The only choice for minimal polynomial is $c_A(x)$, and there is one similarity class over $\mathbb{Q}$ and $F$.

$(x+a)^2(x+b)$: Whether viewed over $\mathbb{Q}$ or $F$, there are two similarity classes.

$(x+a)^3$: As before, there are invariably three similarity classes. These are all a result of $x+a$ being irreducible in both $\mathbb{Q}[x]$ and $F[x]$.

$(x^2+ax+b)(x+c)$: There is only one similarity class over $\mathbb{Q}$. If $x^2+ax+b=(x+v_1)(x+v_2)$ for $v_1 \neq v_2$, then there is again only one similarity class. If $x^2+ax+b=(x+v_1)^2=x+2v_1x+v_1^2$, then $2v_1 ∈ \mathbb{Q}$ so $v_1 ∈ \mathbb{Q}$, even though $x^2+ax+b$ doesn't factor and thus doesn't have zeros in $\mathbb{Q}$.

$x^3+ax^2+bx+c$: If this polynomial decomposes in $F[x]$ to three distinct linear factors, or an irreducible quadratic and a linear factor, or doesn't decompose further, then the minimal polynomial remains the same. If it decomposes into $(x+v)^3$, then comparing coefficients of $x^2$ we obtain $3v∈\mathbb{Q}$ so $v∈\mathbb{Q}$. Therefore $x^3+ax^2+bx+c = (x+v_1)^2(x+v_2)$ and we obtain the relations $$2v_1+v_2 = a$$ $$v_1^2+2v_1v_2 = b$$ $$v_1^2v_2 = c$$ We first observe $v_2 = -2v_1 + a$ to manipulate the second equation $$v_1^2 = \dfrac{2}{3}av_1-\dfrac{1}{3}b$$ Substituting these both into the third equation yields a rational quadratic expression over $v_1$, so employing the first observation again and rearranging yields $$(6b-2a^2)v_1 = 9c-ab$$ Since the original polynomial can't have zeros in $\mathbb{Q}$, $-v_1 \not ∈ \mathbb{Q}$ so $v_1 \not ∈ \mathbb{Q}$ implying $6b-2a^2=0$ implying $b = \dfrac{1}{3}a^2$ and $c = \dfrac{1}{27}a^3$. Now $x^3+ax^2+bx+c=(x+\dfrac{1}{3}a)^3$ decomposes in $\mathbb{Q}[x]$, a contradiction.

Observe the polynomial $x^4+2x^2+1=(x^2+1)^2$ in $\mathbb{Q}[x]$. There are two lists of invariant factors for matrices with this characteristic polynomial, and thus there are two similarity classes over $\mathbb{Q}$. In $\mathbb{C}$ this polynomial decomposes to $(x-i)^2(x+i)^2$, and there are seen to be four lists of invariant factors and thus four similarity classes over $\mathbb{C}$.$~\square$

Complex Computations and Inequalities (1.13-15)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercises 13-15:

13. For complex $x,y$ show $$||x|-|y||≤|x-y|$$ 14. If $z$ is complex and $|z|=1$, compute $$|1+z|^2+|1-z|^2$$ 15. Under what conditions does equality hold in the Schwarz inequality? I.e., $$|\sum a_j\overline{b_j}|^2 = \sum |a_j|^2 \sum |b_j|^2$$ for complex $a_j,b_j$.

Proof: (13) Assume $|x|≥|y|$. Then we must show $|x|-|y| > |x-y|$ is impossible. Multiply both sides by (positive) $|x|+|y|$ to obtain $|x|^2-|y|^2 > |x-y|(|x|+|y|) ≥ |x-y||x+y| = |x^2-y^2|$ so $|x^2| > |x^2-y^2|+|y^2| ≥ |x^2|$, a contradiction. The case is parallel when $|y|≥|x|$.

(14) Then $z=a+bi$ such that $a^2+b^2=1$. We see $|1-z|=|z-1|$ and now $|z+1|^2+|z-1|^2=(a+1)^2+b^2+(a-1)^2+b^2=2(a^2+b^2)+2=4$.

(15) Let $v_a,v_b∈\mathbb{C}^k$ be the the vectors of the $a_j$ and $b_j$. Let $A = \sum |a_j|^2$,$B = \sum |b_j|^2$, and $C = \sum a_j\overline{b_j}$. We claim $|C|^2=AB$ if and only if $v_b$ is associate to $v_a$ (i.e. $v_b=cv_a$ for some $c∈\mathbb{C}$) or at least one of $v_a$ or $v_b$ is zero. ($⇒$) Assume $|C|^2=AB$ and $v_a,v_b \neq 0$ so $A,B > 0$. As we have seen, $∑|Ba_j-Cb_j|^2 = B(AB-|C|^2)$. When $|C|^2=AB$ we then thus have $a_j=\dfrac{C}{B}b_j$ for all $j$, and thus $v_b = \dfrac{C}{B}v_a$. ($⇐$) When either one of $v_a$ or $v_b$ is zero the equality clearly holds, so assume $v_a,v_b \neq 0$ and $v_b = cv_a$ for some complex $c$. We can thus manipulate $$B=\sum |b_j|^2 = \sum |ca_j|^2 = |c|^2\sum |a_j|^2 = |c|^2A$$ and $$\overline{c}A=\overline{c}\sum a_j\overline{a_j} = \sum a_j\overline{b_j} = C$$ so for all $j$ we have $$Ba_j = |c|^2Aa_j = \dfrac{|c|^2a_j\overline{a_j}A}{\overline{a_j}} = \dfrac{|ca_j|^2}{\overline{a_j}}A = b_j \dfrac{\overline{b_j}}{\overline{a_j}} A = b_j \overline{c} A = Cb_j$$ so that $∑ |Ba_j - Cb_j|^2 = B(|C|^2 - AB) = 0$. Since $v_b \neq 0$ and thus $B > 0$, we have $|C|^2 = AB$.$~\square$

Real Logs (1.7)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercise 7:

Fix $b > 1, y > 0$ and prove that there is a unique real $x$ such that $b^x = y$ by completing the following outline. Say $x = \text{log}_b y$.
(a) For natural $n$, $b^n - 1 ≥ n(b-1)$.
(b) Hence $b - 1 ≥ n(b^{1/n}-1)$.
(c) If $t > 1$ and $n > (b-1)/(t-1)$ then $b^{1/n} < t$.
(d) If $w$ is such that $b^w < y$ then $b^{w+1/n} < y$ for sufficiently large $n$.
(e) If $b^w > y$ then $b^{w-1/n} > y$ for sufficiently large $n$.
(f) Let $A = \{w~|~w < y\}$ and show $b^{\text{sup }A} = y$.
(g) Show $x$ is unique.

Proof: (a) We show $b^n ≥ n(b-1)+1$. This is clear when $n=1$, so by induction $$b^n = bb^{n-1} ≥ b((n-1)(b-1)+1)) = (n-1)b(b-1)+b$$ and we must show $(n-1)b(b-1)+b ≥ n(b-1)+1$. Collecting terms on the left and manipulating we must thus show $(n-1)(b-1)^2 ≥ 0$, which is clear as all these terms are nonnegative.
(b) This is clear from the previous, as $b^{1/n} > 1$ by noting $c^n ≤ 1$ for $c ≤ 1$ by induction.
(c) Assume $b^{1/n} ≥ t$. Then $b - 1 < n(t-1) ≤ n(b^{1/n}-1)$, a contradiction by (b).
(d) We have $1 < \dfrac{y}{b^{w}}$. Observe $b^{1/n} ≤ \dfrac{b-1}{n}+1$ by (b) so we may choose $n$ such that $\dfrac{b-1}{n}+1 < \dfrac{y}{b^{w}}$ by conditioning $\dfrac{b-1}{n} < \dfrac{y}{b^w}-1 > 0$ and now $n > \dfrac{b-1}{\dfrac{y}{b^w}-1}$ so that $b^{w+1/n} < y$.
(e) As before, since we showed $b^{1/n}$ tends toward 1 we can choose $n$ such that $b^{1/n} < \dfrac{b^w}{y}$ to fulfill.
(f) By (a) we have $b^w$ is divergent so $A$ has $x = \text{sup }A$. Suppose $b^x < y$; then by (d) we have some $b^{x+1/n} < y$ so $x$ is not an upper bound. Suppose $b^x > y$; then by (e) choose $n$ for $b^{x-1/n} > y$ and $x$ is not a minimal upper bound. Therefore $b^x = y$.
(g) If $x' \neq x$ then $b^x - b^{x'} = b^{x'}(b^{x-x'}-1) = 0$ is a contradiction.$~\square$

Real Exponents (1.6)

Walter Rudin Principles of Mathematical Analysis, chapter 1, exercise 6:

Fix $b>1$.
(a) Let $m,n,p,q$ be integers, $n,q > 0$ with $m/n=p/q$. Prove $$(b^m)^{1/n}=(b^p)^{1/q}$$ Hence it makes sense to define $b^{m/n}=(b^{m})^{1/n}$.
(b) Prove $b^{r+s}=b^rb^s$ for $r,s∈\mathbb{Q}$.
(c) For $x∈\mathbb{R}$, define $B(x) = \{b^t~|~t∈\mathbb{Q},~t≤x\}$. When $r∈\mathbb{Q}$, prove $$b^r=\text{sup }B(r)$$ Hence it makes sense to define $$b^x=\text{sup }B(x)$$ (d) For $x,y∈\mathbb{R}$, prove $b^{x+y}=b^xb^y$.

Proof: (a) Recall $(z_1z_2)^{1/n}=z_1^{1/n}z_2^{1/n}$ for natural $n$. For natural $n_1,n_2$ it is also clear that $(z^{n_1})^{n_2}=z^{n_1n_2}$, and when $n_1,n_2$ may be negative the equality holds by case analysis together with the definition $z^{-n}=(z^n)^{-1}$. Therefore, we derive $$(b^m)^{1/n}=(b^{1/n})^m=(((b^{1/n})^m)^q)^{1/q}=((b^{1/n})^{mq})^{1/q}=$$ $$((b^{1/n})^{pn})^{1/q}=(((b^{1/n})^n)^p)^{1/q}=(b^p)^{1/q}$$ (b) The case is clear for integer $r,s$. As well, we note $z^{1/n_1n_2}=(z^{1/n_1})^{1/n_2}$ as $((z^{1/n_1})^{1/n_2})^{n_1n_2}=(((z^{1/n_1})^{1/n_2})^{n_2})^{n_1}=z$. Otherwise let $r=r_1/r_2$ and $s=s_1/s_2$. We have $$b^{r+s}=b^{r_1/r_2+s_1/s_2}=b^{(r_1s_2+r_2s_1)/(r_2s_2)}=(b^{r_1s_2+r_2s_1})^{1/r_2s_2}=$$ $$(b^{r_1s_2}b^{r_2s_1})^{1/r_2s_2}=(((b^{r_1})^{s_2})^{1/s_2})^{1/r_2}(((b^{s_1})^{r_2})^{1/r_2})^{1/s_2}=b^rb^s$$ (c) Lemma 1: When $a = a_1/a_2 > 0$ is rational we have $b^a > 1$. This implies $b^a > b^c$ for rationals $a > c$. Proof: Rewrite $a_2 > 0$ so that $a_1 > 0$ so by induction $b^{a_1} > 1$ and it suffices to prove $b^{1/a_2} > 1$ for $b > 1$. Here it suffices to prove for $c ≤ 1$ then $c^n ≤ 1$ for natural $n$, which is clear by induction $c^n = cc^{n-1} ≤ c ≤ 1$.

Now, for $t≤r$ we show $b^r ≥ b^t ⇔ b^t(b^{r-t}-1) ≥ 0$, so since $b^t$ is nonnegative and by the lemma $b^{r-t}-1$ is nonnegative we have $b^r$ is an upper bound of $B(r)$. As well, if $z < b^r$ then since $b^r∈B(r)$ we have $z$ is not an upper bound of $B(x)$ and thus $b^r = \text{sup }B(r)$.

(d) Lemma 2: Let $X⊆\mathbb{R}$ be bounded above such that when $x∈X$ and $y≤x$, then $y∈X$. We see $\text{sup }X = \text{sup }(X \setminus \{\text{sup }X\})$. Proof: $z = \text{sup }X$ bounds the latter, so assume $y < z$ also bounds. But $y$ doesn't bound $X$ and thus $y < x$ for some $x∈X$ and we may choose some $y < x_1 < x ≤ z$ so $x_1 ∈ X \setminus \{\text{sup }X\}$ yet $y < x_1$ and $z$ is the smallest bound.

Let $B^*(x) = \{v ≤ b^t~|~t∈\mathbb{Q},~t≤x\}$ so that $\text{sup }B^*(x) = \text{sup }B(x)$ and it serves as an equivalent definition for $b^x$. Now if $b^x∈B^*(x)$ then by the above lemma we may remove it, and as well $b^t$ for $t < x$ to define $B^{**}(x)$ with an equivalent supremum. Now we must show that when $z < \text{sup }B^{**}(x)$ then $z < b^v$ for some rational $v < x$. We already see this is the case when for some rational $v ≤ x$, so we must construct rational $v' < v$ such that $z < b^{v'} < b^v$; this may be done by observing lemma 1 and parts (a),(b), and (g) to show $b^{1/n}$ tends to 1 and setting $v' = v - 1/n$ for $1 < b^{1/n} < b^v/z$.

Now, suppose $b^xb^y < b^{x+y}$; then $b^xb^y < b^v$ for some rational $v < x+y$ by the above discussion. Set rational $v - y < v_1 < x$ and $v - v_1 < v_2 < y$ so $v_1+v_2 > v$ and $b^xb^y ≥ b^{v_1}b^{v_2} = b^{v_1+v_2} > b^{v}$.

Suppose $b^{x+y} < b^xb^y$. Then $(b^x)^{-1}b^{x+y} < b^{v_2}$ for some rational $v_2 < y$. Choose rational $v_2 < v_2 < y$ and set $v_4 = v_3 - v_2 > 0$, and then choose $x-v_4 < v_1 < x$. We have $b^{v_2} > (b^x)^{-1}b^{x+y} ≥ (b^x)^{-1}b^{v_1+v_2+v_4}$ implying $1 > (b^x)^{-1}b^{v_1}b^{v_4}$ then $b^x > b^{v_1+v_4}$. But $v_1+v_4 > x$ and choosing rational $x < x' < v_1+v_4$ we see $b^{x'}$ is an upper bound of $b^x$ and $b^{v_1+v_4} > b^{x'} ≥ b^x$ is a contradiction.$~\square$