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 $F$.$~\square$