Notes - Galois Theory HT25, Galois extensions

$K \supseteq F$ is said to be a splitting field of $f$ if:

• $f$ splits completely in $K[t]$, and
• $K$ is generated as a field by $F$ together with the roots of $f$.

• There exists a non-constant polynomial $f \in F[x]$ such that $K$ is the splitting field of $f$.
• Equivalently (? @todo), there is a separable polynomial $f \in F[x]$ such that $K/F$ is the splitting field of $F$.

If $f \in F[t]$ and $K$ is a splitting field of $f$, then $[K : F] < \infty$.

Since $f$ is irreducible in $F[t]$, $\langle f \rangle$ must be a maximal ideal in $F[t]$ and hence $K := \frac{F[t]}{\langle f\rangle}$ is a field. Then $\alpha := t + \langle f \rangle$ is a root of $f$ in $K$.

Since $\alpha$ generates $K$ as a ring together with $F$, $K = F[\alpha]$.

Since $\alpha$ is algebraic over $F$, it follows

\[F(\alpha) = F[\alpha] = K\]

and $K / F$ is the required extension.

Induct on $d := \deg f$.

Base case. $d = 1$. Then $F$ is already a splitting field.

Inductive step. $d > 1$. Suppose that $g$ is an irreducible factor of $f$. By the result that says:

If $g \in F[t]$ is irreducible, there exists a simple extension $F(\alpha)$ where $\alpha$ is a root of $g$.

we can construct $L := F(\alpha)$. Since $g \mid f$, $f(\alpha) = 0$ and hence $(t - \alpha) \mid f$.

Define $h := \frac{g}{(t - \alpha)} \in L[t]$. Since $\deg h < d$, by induction there exists a splitting field $K$ of $h$.

Since $a \in L \subset K$, $f = (t - \alpha) \cdot h$ splits completely in $K[t]$.

Since the roots of $f$ in $K$ generate $K$ together with $F$, it follows that $K$ is a splitting field of $f$.

There is a unique extension of $\varphi$ to

\[\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha)\]

such that $\tilde \varphi (\alpha) = \tilde \alpha$.

This could be summarised by the diagram:

Existence. By assumption, $\varphi(m _ {F, \alpha})(\tilde \alpha) = 0$. Then $\varphi(m _ {F, \alpha}) \mid m _ {\tilde F, \tilde \alpha}$, but as $m _ {\tilde F, \tilde \alpha}$ is irreducible, it follows that $\phi(m _ {F, \alpha}) = m _ {\tilde F, \tilde \alpha}$.

We have the isomorphisms

\[\begin{aligned} &\theta : \frac{F[t]}{\langle m_{F, \alpha} \rangle} \stackrel\cong\longrightarrow F(\alpha) \\\\ &\tilde \theta : \frac{\tilde F[t]}{\langle m_{\tilde F, \tilde \alpha}\rangle} \stackrel\cong\longrightarrow\tilde F(\tilde \alpha) \end{aligned}\]

And $\phi : F \to \tilde F$ extends to an isomorphism

\[\phi' : F[t] \stackrel\cong\longrightarrow\tilde F[t]\]

by fixing $t$. As $\phi’$ sends $\langle m _ {F, \alpha}\rangle$ to $\langle m _ {\tilde F, \tilde \alpha} \rangle$, by the universal property of quotients this map descends to give

\[\overline \phi : \frac{F[t]}{\langle m_{F, \alpha}\rangle} \longrightarrow \frac{\tilde F[t]}{\langle m_{\tilde F, \tilde \alpha}\rangle}\]

This can be summarised with the following commutative diagram:

Chaining isomorphism together, we have

\[\begin{aligned} &\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha) \\\\ &\varphi^\ast := \tilde \theta \circ \overline \varphi \circ \theta^{-1} \end{aligned}\]

gives the existence of the required isomorphism.

Uniqueness. Suppose that $\psi : F(\alpha) \to \tilde F(\tilde \alpha)$ is another isomorphism with the property that $\psi(\alpha) = \tilde \alpha$ (this is what we are assuming in the construction).

Then $\psi$ agrees with $\phi^\ast$ on $F$ and $\alpha$.

Since $F$ and $\alpha$ generate $F[\alpha]$, and $F[\alpha] = F(\alpha)$ ($\alpha$ is algebraic) it follows that $\psi = \phi^\ast$, so we have uniqueness as required.

We proceed by induction on $[K : F]$.

Base case: $[K:F] = 1$. Then $K \cong F$ already, and so we have at least one isomorphism.

Inductive step: Let $g$ be a monic irreducible factor of $f$ in $F[t]$ with $\deg g \ge 2$. Then $g$ is separable in $F[t]$. (@todo, does this assume that $F$ is a field of characteristic zero?).

$\varphi$ is an isomorphism and it extends naturally to an isomorphism $\varphi’ : F[t] \to \tilde F[t]$. Then $\tilde g := \varphi’(g)$ is also separable (since if it weren’t this would also imply $g$ wasn’t).

Since $\tilde K$ is a splitting field of $\tilde f$ and $\tilde g \mid \tilde f$, $\tilde g$ has exactly $n := \deg \tilde g$ roots, say $\beta _ 1, \ldots, \beta _ n \in K$ by the result that all roots of separable irreducible polynomials are simple.

Choose a root $\alpha \in K$ of $g$. Since $g(\alpha) = 0$, by the “division property” of minimal polynomials, $m _ {F, \alpha} \mid g$. But then by assumption $g$ is monic and irreducible over $F$, so it must actually be the case $m _ {F, \alpha} = g$.

Pick some $i = 1, \ldots, n$. Then $\phi’(m _ {F, \alpha})(\beta _ i) = \phi’(g)(\beta _ i) = \tilde g(\beta _ i) = 0$, so the conditions of the lemma that

If

• $\varphi : F \to \tilde F$ is an isomorphism between two fields
• $K / F$ and $\tilde F / \tilde K$ are finite extensions
• $\alpha \in K$, $\tilde \alpha \in \tilde K$
• $\varphi(m _ {F, \alpha})(\tilde \alpha) = 0$ Then there is a unique extension of $\varphi$ to

\[\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha)\]

are satisfied. Hence we obtain a unique isomorphism $\phi _ i : F(\alpha) \to \tilde F(\beta _ i)$ which extends $\phi : F \to \tilde F$ and sends $\alpha$ to $\beta _ i$.

Let $m := [K : F(\alpha)]$. As $m < [K : F]$, we can apply the inducitve hypothesis to $K/F(\alpha)$ and $\tilde K / \tilde F(\beta _ i)$ together with the isomorphism to find at least $m$ different extensions

\[\phi_i^{(j)} : K \to \tilde K \quad\text{ for }j = 1, \ldots, m\]

It remains to show that these are all distinct. If $\phi _ {i}^{(j)} = \phi _ {i’}^{j’}$ for some $1 \le i, i’ \le n$ and $1 \le j, j’ \le m$, we have

\[\begin{aligned} \beta_i &= \phi_i^{(j)}(\alpha) \\\\ &= \phi_i^{(j')}(\alpha) \\\\ &= \beta_{i'} \end{aligned}\]

So $i = i’$ and by induction $j = j’$ (@todo, explain this?). Thus there are at least $mn$ different extensions. Since $[F(\alpha):F] = \deg m _ {F, \alpha} = \deg g = \deg \tilde g = n$, it follows that $[K : F] = [K : F(\alpha)] [F(\alpha) : F] = mn$ and we are done.

Suppose:

• $K/F$ is a Galois extension

Then:

\[|\text{Gal}(K/F)| \ge [K : F]\]

Proof: Since $K/F$ is a Galois extension, by definition $K$ is the splitting field for some polynomial $f \in F[t]$. In the above result, take $\tilde F := F$ and $\varphi : F \to \tilde F$ as the identity map. Then there are at least $[K : F]$ distinct automorphisms of $K$ extending $1 : F \to F$.

Automorphisms of this type are exactly the $F$-linear automorphisms of $K$, and hence

\[|\text{Gal}(K/F)| \ge [K : F]\]

as required.

Yes.

We show this in three steps:

1. Show that if $L$ is another extension of $F$ such that $f$ splits completely, there exists at least one injective homomorphism $K \to L$.
2. Show that if $L$ is a splitting field of $f$, then this injective homomorphism is actually an isomorphism.
3. Conclude that any two splitting fields of $f$ are isomorphic.

(1): Let $\tilde K$ be the subfield of $L$ generated by $F$ together with the roots of $f$. Then by the result above, there exists at least one isomorphism $\varphi : K \to \tilde K$ extending the identity map on $F$. If $i : \tilde K \to L$ is the inclusion map, then $i \circ \varphi : K \to L$ is the required monomorphism.

(2): If $L$ is a splitting field of $F$, it follows that $\tilde K$ as defined above equals $L$, and hence this injective homomorphism must actually be an isomorphism.

(3): Since there are isomorphism between these two splitting fields, and these splitting fields are isomorphic, it folows that any two splitting fields of $f$ are isomorphic.

\[|\text{Gal}(K/F)| = [K : F]\]

Suppose:

• $K/F$ is a Galois extension

Then:

\[|\text{Gal}(K/F)| \ge [K : F]\]

---

Suppose:

• $K/F$ is a finite field extension
• $G = \text{Gal}(K/F)$

Then:

\[|\text{Gal}(K/F)| \le [K : F]\]

$K^G$ is a subfield containing $F$, so Every $K^G$-linear automorphism of $K$ is also $F$-linear.

If $\sigma : K \to K$ is $F$-linear, then $\sigma \in G$ and so $\sigma$ fixes $K^G$ pointwise. Therefore $\sigma$ is also $K^G$-linear.

\[F = K^G\]

We have a result which says

\[G = \text{Gal}(K/K^G)\]

Then, by the result on upper bounds on the size of the Galois group,

\[|G| \le [K : K^G]\]

But then

\[|G| \le [K : K^G] \le [K : F] = |G|\]

and so $[K : K^G] = [K : F]$ and hence $[K^G : F] = 1$, and so $K^G = F$.

\[f_X := \prod_{y \in X} (t - y) \in L[t]\]

$K/F$ is Galois.

Let $\{z _ 1, \ldots, z _ n\}$ be an $F$-basis for $K$. Since $G$ is finite (by the result that says finite field extensions lead to finite Galois groups), it follows that the set

\[X := \bigcup^n_{i = 1} G\cdot z_i\]

is finite as well as $G$-stable.

Then we can construct the polynomial $f _ X$ where

\[f_X := \prod_{y \in X} (t - y)\]

which is a separable polynomial with coefficients in $K^G$ (this is nontrivial and comes from a previous result). Since $K^G = F$, it follows $f _ X \in F[t]$. And since $f _ X$ splits completely over $K$ and since $K$ is generated by the roots of $f _ X$ in $K$, it is the splitting field of $f _ X$ over $F$.