Splitting Fields
Definition.
Let $F$ be a field, and let $p(x) \in F[x] \setminus F$ be a non-constant polynomial. A splitting field of $p(x)$ over $F$ is a minimal field extension $E/F$ such that $p(x)$ splits completely in $E$. That is, there exist $\alpha_i \in E$ such that
\[p(x) = \prod_{i=1}^k (x - \alpha_i)\]
More generally, if $\mathcal{P} \subset F[x] \setminus F$ is a set of non-constant polynomials, then a splitting field of $\mathcal{P}$ over $F$ is a minimal field extension $E/F$ such that every polynomial in $\mathcal{P}$ splits completely in $E$.
Theorem (Existence and Uniqueness of Splitting Fields).
Let $F$ be a field and let $p(x) \in F[x] \setminus F$ be a non-constant polynomial. If we fix an algebraic closure $\overline{F}$ of $F$, then there exists a unique splitting field of $p(x)$ over $F$ inside $\overline{F}$.
Later, we will also prove (using the isomorphism extension theorem) that the splitting field is unique up to isomorphism, even without reference to a fixed $\overline{F}$.
Proof.
Let $\alpha_1, \ldots, \alpha_k \in \overline{F}$ be the roots of $p(x)$. We claim that
\[F(\alpha_1, \ldots, \alpha_k)\]
is the splitting field of $p(x)$ over $F$.
First, since all the roots $\alpha_i$ lie in $F(\alpha_1, \ldots, \alpha_k)$, the polynomial $p(x)$ splits completely in this field.
Now suppose $E$ is another field extension of $F$ in which $p(x)$ splits completely. Then all the roots $\alpha_i$ lie in $E$, so the field $F(\alpha_1, \ldots, \alpha_k)$ is contained in $E$. Therefore, $F(\alpha_1, \ldots, \alpha_k)$ is the minimal such field — i.e., a splitting field.
Uniqueness inside $\overline{F}$ follows similarly: if $E’ \subset \overline{F}$ is another splitting field of $p(x)$, then it must contain all roots $\alpha_i$, and hence also contain $F(\alpha_1, \ldots, \alpha_k)$. Conversely, since $E’$ is minimal, it must equal $F(\alpha_1, \ldots, \alpha_k)$.
Remark (Splitting field of a set of polynomials).
Let
\[\mathcal{P} = \{ p_1(x), \ldots, p_n(x) \} \subset F[x] \setminus F\]
be a finite set of non-constant polynomials. Then the splitting field of $\mathcal{P}$ over $F$ is the same as the
splitting field of the single polynomial
$p(x) = \prod_{i=1}^n p_i(x)$, since all the roots of the $p_i$ are also roots of $p$, and vice versa.
If $\mathcal{P}$ is an infinite set of non-constant polynomials, we proceed as follows. Choose an increasing chain
of finite subsets
\[\mathcal{P}_1 \subset \mathcal{P}_2 \subset \cdots \subset \mathcal{P} \;\;\;\;\;\;\;\;\;\;\;\;\bigcup_{i \in I} \mathcal{P}_i = \mathcal{P}\]
Let $E_i \subset \overline{F}$ be the splitting field of $\mathcal{P}_i$ over $F$. By the previous theorem, each $E_i$ exists and is uniquely determined inside $\overline{F}$. Moreover, we have an increasing chain of subfields:
\[E_1 \subset E_2 \subset \cdots\]
We define:
\[E = \bigcup_{i \in I} E_i\]
Then $E$ is a field extension of $F$ in which every $p(x) \in \mathcal{P}$ splits completely, and it is minimal with
respect to this property. Therefore, $E$ is the unique splitting field of $\mathcal{P}$ over $F$ inside $\overline{F}$.
Theorem.
Let $E$ be the splitting field of a set of polynomials $\mathcal{P} \subset F[x]$ over a field $F$, inside a fixed algebraic closure $\overline{F}$. Then for every $\sigma \in \operatorname{Aut}(\overline{F}/F)$, we have:
\[\sigma(E) = E\]
That is, the splitting field $E$ is stable under the action of all automorphisms of $\overline{F}$ fixing $F$.
Proof.
First suppose $\mathcal{P}$ is a finite set of polynomials. Then, by a previous remark, $E$ is also the splitting field of a single polynomial
$p(x) = \prod_{f \in \mathcal{P}} f(x)$, and we can write
\[E = F(\alpha_1, \ldots, \alpha_k)\]
where the $\alpha_i \in \overline{F}$ are all roots of $p(x)$. Let $\sigma \in \operatorname{Aut}(\overline{F}/F)$. Since $\sigma$ fixes $F$ pointwise and $p(x) \in F[x]$, we have:
\[p(\sigma(\alpha_i)) = \sigma(p(\alpha_i)) = \sigma(0) = 0\]
so $\sigma(\alpha_i)$ is also a root of $p(x)$ and hence belongs to ${ \alpha_1, \ldots, \alpha_k } \subset E$. Therefore,
\[\sigma(E) = \sigma(F(\alpha_1, \ldots, \alpha_k)) = F(\sigma(\alpha_1), \ldots, \sigma(\alpha_k)) \subset E\]
and since $\sigma$ is invertible, we must have equality: $\sigma(E) = E$.
Now suppose $\mathcal{P}$ is an infinite set. Then, by a previous remark, we may write
\[E = \bigcup_{i \in I} E_i\]
where each $E_i$ is the splitting field of a finite subset $\mathcal{P}_i \subset \mathcal{P}$.
By the finite case, we know that $\sigma(E_i) = E_i$ for all $i$. Now for any $e \in E$, there exists some $i$ such that $e \in E_i$, and hence $\sigma(e) \in E_i \subset E$. Therefore, $\sigma(E) \subset E$. Applying the same argument to $\sigma^{-1}$ gives the reverse inclusion, so $\sigma(E) = E$.
Theorem (Galois Action on Roots).
Let $E$ be the splitting field of a nonconstant polynomial $p(x) \in F[x]\setminus F$ inside a fixed algebraic closure $\overline{F}$,
and let $X = \{ \alpha_1, \ldots, \alpha_n \} \subset \overline{F}$ be the set of roots of $p(x)$. Then $\operatorname{Aut}(E/F)$ acts on $X$ via:
\[\sigma \cdot \alpha := \sigma(\alpha)\]
This action induces an injective group homomorphism:
\[\operatorname{Aut}(E/F) \hookrightarrow S_X \cong S_n\]
As a consequence, we have the bound:
\[|\operatorname{Aut}(E/F)| \leq n!\]
Proof.
Since $E$ is the splitting field of $p(x)$ over $F$, it contains all the roots of $p(x)$, so $X \subset E \subset \overline{F}$.
Let $\sigma \in \operatorname{Aut}(E/F)$ and $\alpha \in X$. Because $p(x) \in F[x]$ and $\sigma$ fixes $F$, we have:
\[p(\sigma(\alpha)) = \sigma(p(\alpha)) = \sigma(0) = 0\]
So $\sigma(\alpha) \in X$, and the action $\sigma \cdot \alpha := \sigma(\alpha)$ preserves the set $X$.
The identity map acts trivially, and composition behaves as expected:
\[(\sigma\tau)(\alpha) = \sigma(\tau(\alpha)) = \sigma \cdot (\tau \cdot \alpha)\]
Hence, this defines a group action of $\operatorname{Aut}(E/F)$ on $X$, inducing a group homomorphism:
\[\operatorname{Aut}(E/F) \to S_X\]
We now show that this map is injective. Note that $E = F(\alpha_1, \ldots, \alpha_n)$, so any $\sigma \in \operatorname{Aut}(E/F)$ is uniquely determined by its values on the $\alpha_i$.
If $\sigma|_X$ is the identity map, then $\sigma(\alpha_i) = \alpha_i$ for all $i$. Therefore, $\sigma$ fixes a generating set of $E$ over $F$, and hence $\sigma = \mathrm{id}$.
Thus, the kernel is trivial, and the homomorphism is injective.