Galois Theory HT25, Normality

Whenever $g \in F[t]$ is an irreducible polynomial such that $g$ has a root in $K$, $g$ splits completely in $K[t]$.

“One out, all out!”, Ian Stewart

@important~

It is the splitting field of some polynomial.

$K/F$ is Galois $\implies$ $K/F$ is normal

Choose a separable $f \in F[t]$ such that $K$ is a splitting field of $f$. Pick $g \in F[t]$ irreducible where $g$ has a root in $K$.

Since any polynomial has a splitting field, find a splitting field of $fg \in K[t]$, say $L$, where $L/F$. Suppose that $\theta _ 1$ and $\theta _ 2$ are zeroes of $g$ in $L$, and consider the following diagram:

We aim to conclude that $[K(\theta _ i) : K] = 1$, so that we know $\theta _ i \in K$.

By two applications of the tower law, we have that for $i = 1$ or $i = 2$,

\[[K(\theta _ i) : F] = [K(\theta _ i) : K] [K:F] =\]

and

\[[K(\theta _ i): F] = [K(\theta _ i) : F(\theta _ i)][F(\theta _ i): F]\]

corresponding to the two paths through the diagram:

Since $g$ is irreducible,

\[g = m _ {F, \theta _ 1} = m _ {F, \theta _ 2}\]

and hence:

\[[F(\theta _ 1):F] = \deg g = [F(\theta _ 2) : F]\]

by the lemma that says the degree of an algebraic extension like this agrees with the degree of the minimal polynomial.

By the result which says

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 $\varphi^\ast : F(\alpha) \to \tilde F(\tilde \alpha)$

by taking $\varphi$ as the identity map, we can find an $F$-linear isomorphism

\[\varphi : F(\theta _ 1) \to F(\theta _ 2)\]

where $\theta _ 1 \mapsto \theta _ 2$.

Since $K(\theta _ i)$ is a splitting field for $f$ over $F(\theta _ i)$ and $f$ is separable over $F(\theta _ i)$ (by the result that says if a polynomial is separable, it separable over any field extension), by the result that says

If:

• $\varphi : F \to \tilde F$ is an isomorphism of fields
• $f \in F[t]$ is a separable polynomial
• $K$ is a splitting field for $f$
• $\tilde f = \varphi(f)$
• $\tilde K$ is a splitting field for $\tilde f$

Then there are at least $[K : F]$ distinct isomorphisms extending $\varphi$.

We can extend $\varphi$ to an isomorphism of fields

\[K(\theta _ 1) \to K(\theta _ 2)\]

Therefore

\[[K(\theta _ 1) : F(\theta _ 1)] = [K(\theta _ 2) : F(\theta _ 2)]\]

and hence

\[\begin{aligned} \text{}[K(\theta _ 1) : K] &= \frac{[K(\theta _ 1) : F(\theta _ 1)][F(\theta _ 1) : F]}{[K : F]} \\\\ &= \frac{[K(\theta _ 2) : F(\theta _ 2)][F(\theta _ 2) : F]}{[K : F]} \\\\ &= [K(\theta _ 2) : F] \end{aligned}\]

and so if $\theta _ 1$ is a root of $g$ which lies in $K$,

\[[K(\theta _ 2) : K] = [K(\theta _ 1) : K]\]

for any other root of $\theta _ 2$ of $g$ in $L$.

Hence all roots of $g$ lie in $K$, so $L = K$ and $g$ splits completely over $K$.