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 $2*3=6$.