# If $\mathbb K:\mathbb F$ is Galois, then it is normal and separable

I'm trying to proof this proposition (below), I already find a proof in a pdf file but it was difficult to understand so I though would be better to do it myself but I got stuck trying to do it. Please help me, please!

Proposition: If $[\mathbb K:\mathbb F]<\infty$ and it is Galois, then it is normal and separable.

Proof: We need to show that every irreducible $p(x)\in\mathbb F$ with one root $\alpha\in\mathbb K,$ splits over $\mathbb K$. (This is for normality) and that for every $\alpha\in\mathbb K$ is separable.

So let $p(x)\in\mathbb F$ be irreducible with one root $\alpha\in\mathbb K$ and let $\alpha\in\mathbb K.$

By hypothesis, $G(\mathbb K:\mathbb F)^+=\sigma(\mathbb F)≈\mathbb F$ and also we have that $\mathbb K$ is a splitting field of a separable $p(x)\in\mathbb F.$

Now let $\sigma\in G(\mathbb K:\mathbb F)^+$ and consider the polynomial $x-\alpha\in\mathbb K.$ Thus $\sigma(x-\alpha)=x-\sigma(\alpha)$.

So we can get something like this $f(x)=\prod_{\sigma\in G(\mathbb K:\mathbb F)^+}(x-\alpha)=\prod_{\sigma\in G(\mathbb K:\mathbb F)^+}(x-\sigma(\alpha))$.

Now $f(x)\in\mathbb K$ and splits on it too, can I considerer $f$ to be the same irreducible polynomial as $p(x)?.$ If I can consider $f(x)=p(x),$ then $\mathbb K:\mathbb F$ would be normal.

The only thing to show now would be to proof that each $\sigma_i(\alpha)$ is different to then be able to conclude that each root of $f(x)$ it's different on $\mathbb F.$ How can I do this?

Any kind of help would be very appreciated :)

• What definition of $K/F$ Galois are you using?
– cat
Commented Jul 26, 2017 at 23:56
• @cat :) $K/F$ is Galois if $G(\mathbb K:\mathbb F)^+=\sigma(\mathbb F)$, where $\sigma$ is the monomorphism Commented Jul 26, 2017 at 23:58
• This is a very standard theorem in any textbook on basic algebra. The proof really depends on how the author has constructed the preliminaries it's hard to prove from scratch without knowing exactly what theorems you have at hand and exactly what definitions you're operating with. For example I am very familiar with Galois theory but I don't recall seeing that notation before $G(\Bbb K:\Bbb F)^{+}$. What do you mean $\sigma$ is "the" monomorphism? There is no canonical monomorphism associated with an extension. Commented Jul 27, 2017 at 0:08
• @GregoryGrant no, it's not a very standard theorem in any textbook on basic algebra (I've searched in many), if it is, please mention the name of the book. Haha yes my professor it's a little bit unusual with his notations, $G(\Bbb K:\Bbb F)^{+}$ it's the fixed Galois group of the extension. And $\sigma$ it's the monomorphism between $\mathbb K$ and $\mathbb F$, i.e. $\sigma:\mathbb F\to\mathbb K$ Commented Jul 27, 2017 at 0:19
• Hungerford is just called "Algebra". I thought it was one of the best written math textbooks I ever read. But it is not a very light treatment, he does a lot of infinite extension stuff. But very nicely. Commented Jul 27, 2017 at 1:11

This is a not so uncommon mistake. If $G = \{\sigma_1, \dots, \sigma_n\}$ then $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$ are not all distinct. If they were then the minimal polynomial of $\alpha$ would have $n$ roots: $\sigma_1(\alpha), \dots, \sigma_n(\alpha)$. Consequentially, $[\mathbf{F}(\alpha) : \mathbf{F}]$ would be $n$, which is true for most $\alpha$ but not all $\alpha$.

Define the subgroup $G_\alpha = \{\sigma \in G : \sigma(\alpha) = \alpha\}$, called the stabilizer of $\alpha$. Also define $G\cdot\alpha = \{\sigma(\alpha) : \sigma \in G\}$, called the orbit of $\alpha$. The set $G\cdot\alpha$ consists of the distinct values of $\sigma(\alpha)$ as $\sigma$ runs over all the elements of $G$.

The minimal polynomial of $\alpha$ is

$$f_\alpha(x) := \prod_{\beta \in G \cdot \alpha} (x - \beta).$$

The polynomial given by

$$\prod_{\sigma \in G} (x - \sigma(\alpha))$$

is called the characteristic polynomial of $\alpha$ and is equal to $f_\alpha(x)^{[\mathbf{K} : \mathbf{F}(\alpha)]}$.

### Why is $f_\alpha$ the minimal polynomial of $\alpha$?

• First, note that $f_\alpha(\alpha) = 0$, which follows since $\alpha = \operatorname{id}(\alpha) \in G \cdot \alpha$.

• Second, note that $f_\alpha \in \mathbf{F}[x]$, which follows since for all $\tau \in G$ we have $$\tau \cdot (G \cdot \alpha) = \{\tau(\beta) : \beta \in G \cdot \alpha\} = \{\tau\sigma(\alpha) : \sigma \in G\} = \{\sigma(\alpha) : \sigma \in G\} = G\cdot \alpha.$$ Hence, $$\tau \cdot f_\alpha(x) = \tau \cdot \prod_{\beta \in G \cdot \alpha} (x - \beta) = \prod_{\beta \in G \cdot \alpha} (x - \tau(\beta)) = \prod_{\gamma \in \tau\cdot(G \cdot \alpha)} (x - \gamma) = \prod_{\gamma \in G \cdot \alpha} (x - \gamma) = f_\alpha(x).$$ Therefore $f_\alpha(x) \in \mathbf{K}^G[x] = \mathbf{F}[x]$. Here $\mathbf{K}^G$ is the fixed field of $G$ and equals $\mathbf{F}$ since $\mathbf{K}/\mathbf{F}$ is Galois.

• Third note that $f_\alpha$ is minimal. Indeed if $f(\alpha) = 0$ then $f(\sigma(\alpha)) = \sigma(f(\alpha)) = \sigma(0) = 0$ for all $\sigma \in G$. Thus $\sigma(\alpha)$ is a root for all $\sigma \in G$. Thus $f_\alpha \mid f$.

Finally, we note that $f_\alpha$ splits over $\mathbf{K}$ and is separable, by construction.

## The proof that $$\prod_{\sigma \in G} (x - \sigma(\alpha)) = f_\alpha(x)^{[\mathbf{K} : \mathbf{F}(\alpha)]}.$$ Only read if interested.

The Galois Correspondence Theorem says:

There is a one-to-one correspondence between subgroups $H$ of $G$ and subextensions $\mathbf{K}/\mathbf{K'}/\mathbf{F}$ given by $H \mapsto \mathbf{K}^{H}$ (the fixed field of $H$) and $\mathbf{K}' \mapsto \operatorname{Gal}(\mathbf{K}/\mathbf{K}')$. Moreover, $|H| = [\mathbf{K} : \mathbf{K}^H]$ and $[G : \operatorname{Gal}(\mathbf{K}/\mathbf{K}')] = [\mathbf{K}' : \mathbf{F}]$.

The correspondence theorem tells us that $|G_\alpha| = [\mathbf{K} : \mathbf{K}^{G_\alpha}]$. But we know that $G_\alpha$ is exactly the group that fixed $\alpha$. Thus $G_\alpha$ fixes $\mathbf{F}(\alpha)$, i.e. $$|G_\alpha| = [\mathbf{K} : \mathbf{F}(\alpha)]. \tag{1}$$

Decompose $G$ as a disjoint union of cosets of $G_\alpha$: $$G = \sigma_1G_\alpha \cup \cdots \cup \sigma_rG_\alpha \tag{2}$$ Then $\sigma_1(\alpha), \dots, \sigma_r(\alpha)$ are all distinct because otherwise $\sigma_i(\alpha) = \sigma_j(\alpha)$ implies $\sigma_j^{-1}\sigma_i(\alpha) = \alpha$ implies $\sigma_j^{-1}\sigma_i \in G_\alpha$ implies $\sigma_iG_\alpha = \sigma_jG_\alpha$. Since the cosets are disjoint, this forces $i = j$.

We also know that $\{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\} = G \cdot \alpha$. Certainly $$\{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\} \subseteq \{\sigma(\alpha) : \sigma \in G\} = G \cdot \alpha \tag{3}$$ and if $\sigma \in G$ then we can write $\sigma = \sigma_i\tau$ for some $i \in \{1,\dots,r\}$ and $\tau \in G_\alpha$ by $(2)$ and then $\sigma(\alpha) = \sigma_i(\tau(\alpha)) = \sigma_i(\alpha)$ (note: $\tau(\alpha) = \alpha$ by definition of $G_\alpha$). Thus $$G \cdot \alpha \subseteq \{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\}. \tag{4}$$ Combining $(3)$ and $(4)$ gives us $$\{\sigma_1(\alpha), \dots, \sigma_r(\alpha)\} = G \cdot \alpha \tag{5}.$$

Therefore \begin{align} \prod_{\sigma \in G} (x - \sigma(\alpha)) &= \prod_{i = 1}^r \prod_{\tau \in G_\alpha} (x - \sigma_i\tau(\alpha)) \tag{by (2)} \\ &= \prod_{i = 1}^r \prod_{\tau \in G_\alpha} (x - \sigma_i(\alpha)) \tag{def. of $G_\alpha$} \\ &= \left( \prod_{i = 1}^r (x - \sigma_i(\alpha)) \right)^{|G_\alpha|} \\ &= \left( \prod_{\beta \in G \cdot \alpha} (x - \beta) \right)^{|G_\alpha|} \tag{by (5)} \\ &= f_\alpha(x)^{|G_\alpha|} \tag{def. of $f_\alpha(x)$} \\ &= f_\alpha(x)^{[\mathbf{K} : \mathbf{F}(\alpha)]}. \tag{by (1)} \end{align}

• Why $\alpha = \operatorname{id}(\alpha) \in G \cdot \alpha$ ? Commented Jul 27, 2017 at 21:59
• @Anne $\operatorname{id}$ is the identity function defined by $\operatorname{id}(x) = x$ so $\operatorname{id}(\alpha) = \alpha$. The point of writing $\alpha$ this way is because $G \cdot \alpha$ is the set of anything that looks like "(function in $G$) applied to $\alpha$" so we are taking $\operatorname{id}$ for the function in $G$. Commented Jul 27, 2017 at 22:01
• I see, also I haven't notice but the identity is the only function in G so my question was kinda obvious Commented Jul 27, 2017 at 22:10
• At the beginning, 'if $\sigma_i(\alpha)$ were all different, then $[\mathbf{F}(\alpha) : \mathbf{F}]=n$ which it's true for most $\alpha$ but not all $\alpha$' why it's not true for all $\alpha$? I have a Theorem where it says is true for all $\alpha$ Commented Jul 27, 2017 at 22:30
• For example, let $\mathbf{K} = \mathbf{Q}(i)$ and $\mathbf{F} = \mathbf{Q}$ then $n = [\mathbf{Q}(i) : \mathbf{Q}] = 2$ and for $\alpha = 2 + 3i$ we have $[\mathbf{Q}(2 + 3i) : \mathbf{Q}] = 2$. But for $\alpha = 2$ we have $[\mathbf{Q}(2) : \mathbf{Q}] = 1$ since $\mathbf{Q}(2) = \mathbf{Q}$. Commented Jul 27, 2017 at 22:35