Galois Theory HT25, Normality

There are a few equivalent definitions:

• 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)
• The minimal polynomial $m _ {F, \alpha}$ splits completely in $K$ for every $\alpha \in K$.
• $K/F$ is the splitting field of some polynomial (useful consequence).

@important~

It is the splitting field of some polynomial.

$K/F$ normal implies splitting field: Let $\alpha _ 1, \ldots, \alpha _ k$ be a $F$-basis for $K$, and consider the polynomial

\[f(x) = \prod^k_{i = 1}m_{\alpha_i}(x)\]

Then $f$ must split in $K$ since each of the minimal polynomials split, and by definition $K$ is generated by the roots of $f$.

Splitting field of some polynomial implies $K/F$ normal:

This is the same as the proof that $K/F$ Galois implies $K/F$ normal.

$K/F$ 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 $[K(\theta _ 1) : K] = [K(\theta _ 2) : K]$ for any two roots $\theta _ 1, \theta _ 2$, since then if $\theta _ 1 \in K$, all these field extensions are trivial and hence all roots lie in $K$.

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

\[\begin{aligned} \text{}[K(\theta _ i) : F] &= [K(\theta _ i) : K] [K:F], \\\\ [K(\theta _ i): F] &= [K(\theta _ i) : F(\theta _ i)][F(\theta _ i): F] \end{aligned}\]

corresponding to the two paths through the diagram starting from $K(\theta _ i)$. By equating these two, we conclude

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

where $(\star)$ is justified by:

• $[F(\theta _ 1) : F] = [F(\theta _ 2) : F]$: This follows from the fact $g = m _ {F, \theta _ 1} = m _ {F, \theta _ 2}$ and thus $[F(\theta _ 1):F] = \deg g = [F(\theta _ 2) : F]$.
• $[K(\theta _ 1) : F(\theta _ 1)] = [K(\theta _ 2) : F(\theta _ 2)]$: This follows from the above implying the existence of the isomorphisms $\varphi : F(\theta _ 1) \to F(\theta _ 2)$ and its extension to $\varphi^\ast : K(\theta _ 1) \to K(\theta _ 2)$.

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

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Isomorphism results:

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

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

First, note that the only root of $t^3 - 1$ in $\mathbb Q(x)$ is $1$. Suppose that $p(x) / q(x) \in \mathbb Q(x)$ where $(p(x) / q(x))^3 = 1$. Then $p(x)^3 = q(x)^3$ and $\deg p = \deg q$.

Suppose that $p(x)$ and $q(x)$ are non-constant, then without loss of generality we may assume $p, q$ are coprime. But $p(x)$ and $q(x)$ must share a common factor, a contradiction.

Hence either $p(x)$ or $q(x)$ is constant but since $\deg p = \deg q$, they must both be constant and $p(x) / q(x) \in \mathbb Q$. But then $p(x) / q(x) = 1$, since this is the only third root of $1$ in $\mathbb Q$.

Now we contradict separability by considering the minimal polynomial $m _ {\mathbb Q(x^3), x}$, which is $t^3 - x^3$. Clearly $x \in \mathbb Q(x)$ is a root, but any other root must be of the form $\mu \cdot x$ where $\mu$ is a third root of unity in $\mathbb Q(x)$. But then $t^3 - x^3$ does not split completely in $\mathbb Q(x)$.