3.8 Galois Group of a Polynomial

Definition

Let $f\in K[T]$. Define the Galois group of $f$ over $K$ as ${\mathrm{Aut}}_{K}E$ where $E$ is a splitting field of $f$ over $K$. (Note that $E$ is normal but not necessary separable.)

Example. Let $f=x^3-2$, and let $F=\mathbb{Q}(\sqrt[3]{2},e^{2\pi i/3})$. Then $|{\mathrm{Aut}}_{K}F|=[F:K]=6$. Note that ${\mathrm{Aut}}_{K}F\in \{S_3,{\mathbb{Z}}_6\}$. On the other hand, ${\mathrm{Aut}}_{K}F$ induces a permutation of $\{\sqrt[3]{2},\sqrt[3]{2}{e}^{2\pi i/3},\sqrt[3]{2}{e}^{4\pi i/3}\}$ and there is an embedding ${\mathrm{Aut}}_{K}F\hookrightarrow S_3$. Thus ${\mathrm{Aut}}_{K}F\simeq S_3$. In fact, for any $\rho \in {\mathrm{Aut}}_{K}F$, $\rho (\sqrt[3]{2})$ is a root of $x^3-2$ and $\rho (e^{2\pi i/3})$ is a root of $x^2+x+1$, and so we can write all elements in ${\mathrm{Aut}}_{K}F$ explicitly.

Remark. Algebra - 1974 - Hungerford.pdf gives computations of Galois group when $\mathrm{deg}f\in \{2,3\}$.

Remark. There are several exercises about computing the Galois group of some polynomials, and here is a very nice example. In fact, Let $f(x)\in \mathbb{Q}(x)$ be an irreducible polynomial of degree $5$ with exactly three real roots and let 𝐾 be the splitting field of 𝑓, then $\mathrm{Gal}(K/\mathbb{Q})\simeq S_5$.

Moreover, the basic method for finding the Galois group of a field extension is:

• First, verify, the field extension is a Galois extension.
• Then, find the degree of the extension. This will give you the size of the Galois group. (the trickiest part)
• Then, find a table of all the finite groups of that order, and apply some sort of logical reasoning to decide which group you have. (E.g., if you know the Galois group is abelian, this will help narrow things down.) Some list is helpful.

Theorem

Let $f\in K(x)$ with Galois group $G$.

• $G$ is isomorphic to some subgroup of $S_m$ for some $m⩽\mathrm{deg}f$.
• If $f(x)$ is irreducible and separable of degree $n$, then $n\mid |G|$ and $G$ is isomorphic to a transitive subgroup of $S_n$.

\begin{proof} i) Suppose $f(x)={\prod }_{i=1}^n(x-u_i)$ for some $u_i\in E$, where $E$ is a splitting field of $f$. WLOG suppose $u_1,\cdots ,u_m$ are all the distinct roots. Then $G\curvearrowright \{u_1,\cdots ,u_m\}$ induces a group homomorphism $G\to S_m$, which is injective.

ii) In this case, $E/K$ is Galois. Let $u$ be a root of $f$. Then $[K(u):K]=n$ and so ${\mathrm{Aut}}_{K(u)}E⩽G$ is a subgroup of $G$ of index $n$. Hence, $n\mid |G|$. For any $i\ne j$, there exists an isomorphism $\sigma :K(u_i)\stackrel{\sim }{\to }K(u_j)$. By ^28f933, $\sigma$ extends an isomorphism $\tilde{\sigma }:E\to E$. Hence $\tilde{\sigma }\in \mathrm{Gal}(E/K)$ and $\tilde{\sigma }(u_i)=u_j$. By the arbitrary of $i,j$, $G$ is transitive. \end{proof}

For simplicity, assume that $\mathrm{char}K=0$.

Corollary

Let $f(x)\in K[x]$ with $\mathrm{char}K=0$ be a monic, irreducible polynomial of degree $2$, and let $G$ be its Galois group. Then $G\simeq S_2\simeq \mathbb{Z}/2\mathbb{Z}$.

Remark. If $\mathrm{char}K=2$ and $f$ is NOT separable, then $G\simeq \{1\}$. For example, $f=x^2-y$ has only one root for any $y\in K$.

Definition

Let $f(x)\in K[x]$ be a separable polynomial of degree $n$ with $f(x)={\prod }_{i=1}^n(x-u_i)$ in a splitting field $F$. Let $\Delta ={\prod }_{i<j}(u_i-u_j)\in F$, and let $D={\Delta }^2\in F$. $D$ is called the discriminant of $f(x)$.

For example, when $f=a{x}^2+bx+c$, $\Delta =b^2-4ac$.

Proposition

• $D={\Delta }^2\in K$.
• Let $\sigma \in G<S_n$. Then $\sigma$ is even iff $\sigma (\Delta )=\Delta$, and $\sigma$ is odd iff $\sigma (\Delta )=-\Delta$.

\begin{proof} i) Since $D={\Delta }^2$, we have $\sigma (D)=D$ for all $\sigma \in G$ and so $D\in F^G=K$.

ii) Consider Vandermonde matrix $M$ with $\det M=(-1{)}^{\left(\genfrac{}{}{0ex}{}{n}{2}\right)}\Delta$. Then $(ts)\Delta =\det (M^{(ts)})=-M$ and we finish the proof. \end{proof}

Corollary

Suppose $f(x)$ is separable with degree $n$, and suppose $G$ is a Galois group of $f$. Then the intermediate set $K\subseteq K(\Delta )\subseteq F$ corresponds to $G\cap A_n$, via the fundamental theorem of the Galois theory.

criterion of Galois group of polynomial of degree $3$

Suppose $\mathrm{char}K=0$ and $f(x)\in K[x]$ is an irreducible and separable polynomial of degree $3$. Then $G\simeq \{\begin{array}{l}{S}_3,\Delta \notin K,\\ A_3,\Delta \in K.\end{array}$

\begin{proof} By ^e42a92, $\Delta \in K$ iff $G\cap A_n=G$ iff $G=A_3$. \end{proof}

how to compute $\Delta$

Suppose $f(x)=x^3+b{x}^2+cd+d\in K[x]$. Let $g(x)=f(x-b/3)$. Then $g(x)$ has the form $x^3+px+q$, and ${\Delta }_{f(x)}^2={\Delta }_{g(x)}^2=-4p^3-27q^2$.

\begin{proof} Suppose that $u_1,u_2,u_3$ are roots of $f$, then $u_1+b/3,u_2+b/3,u_3+b/3$ are roots of $g$. Hence ${\Delta }_g={\Delta }_f$. To compute ${\Delta }_g$, expand $g={\prod }_{i=1}^3(x-v_i)$ and use Vita’s theorem. \end{proof}

Examples. Let $f_1=x^3-3$ and $f_2=x^3-3x+1$. Then the Galois groups of $f_1$ and $f_2$ are $S_3$ and $A_3$, respectively.