# Splitting field of an irreducible and separable polynomial is Galois

I've solved the following problem but I have some doubts about my proof.

Problem statement

$K$ is a field. Given $p(x)\in K[x]$ an irreducible and separable polynomial over $K$ and $E$ its splitting field over $K$, prove that $E/K$ is Galois.

Proof

We denote $n=deg(p)$. As $p(x)=\lambda\,(x-\alpha_1)\cdots(x-\alpha_n)$ with $\lambda\in K$ and $\alpha_i$ the roots of $p(x)$ in $E$, by the minimal property of the splitting field we have that $E=K\left(\alpha_1,\ldots,\alpha_n\right)$.

We write $K_0=K$ and $K_i = K_{i-1}(\alpha_i)$ for $i=1,\ldots,n$ and we consider the following chain of fields:

$$K=K_0 \subset K_1 \subset \ldots \subset K_{n-1} \subset K_n = E.$$

So we will prove the statement by induction over $i$ by showing that $[K_i:K]$ is the number of morphisms $\varphi_i:\,K_i\longrightarrow E$ that $\varphi_i\mid_K=id$ for $i=1,\ldots,n$.

If $i=1$, then $p(x)=Irr(\alpha_1,K)$ and the number of morphisms $\varphi_1$ is defined by the number of roots of $\underline{\varphi_1}\left(Irr(\alpha_1,K)\right)=\underline{id}(p(x))=p(x)$ (where $\underline{\varphi_i}:K_i[x]\longrightarrow E[x]$ sends a polynomial $q(x) = a_0+a_1\,x+\ldots+a_m\,x^m$ to $\underline{\varphi_i}\left(q(x)\right):=\varphi_i(a_0)+\varphi_i(a_1)\,x+\ldots+\varphi_i(a_m)\,x^m$).

Since $p(x)$ is irreducible and separable all roots are distinct and the number of roots is

$$n=deg(p)=deg(Irr(\alpha_1,\,K)) = [K(\alpha_1):K] = [K_1:K].$$

Now we assume that it is satisfied for $i=n-1$ and we prove it for $i=n$.

As $Irr(\alpha_n,K_{n-1})\mid Irr(a_n,K)$, we have that

$$\underline{\varphi_{n-1}}\left(Irr(\alpha_n,K_{n-1})\right)\mid\underline{\varphi_{n-1}}\left(Irr(a_n,K)\right)= \underline{id}\left(p(x)\right) = p(x).$$ Since $p(x)$ splits over $E$, $\underline{\varphi_{n-1}}\left(Irr(\alpha_n,K_{n-1})\right)$ also splits over $E$ and the number of roots in $E$ is

$$deg\left(\underline{\varphi_{n-1}}\left(Irr(\alpha_n, K_{n-1})\right)\right) = deg(Irr(\alpha_n, K_{n-1})) = [K_n:K_{n-1}].$$

Thus, the number of ways of defining $\varphi_n:\,K_n\longrightarrow E$ that satisfies $\varphi_n\mid_{K_{n-1}} = \varphi_{n-1}$ is $[K_n:K_{n-1}]$.

Finally, applying the induction hypothesis and the formula for degrees, we have

$$\left|\right\{\varphi_n:K_n\rightarrow E\mid \varphi_n\mid_K =id\left\}\right| = [K_n:K_{n-1}]\,[K_{n-1}:K] = [K_n:K] = [E:K].$$

Doubts

1. First of all, it is correct to assume $E=K\left(\alpha_1,\ldots,\alpha_n\right)$? I've found some texts that refer to various splitting fields for the same polynomial. However, I thought that there was a unique splitting field, the one that satisfies the minimal property. Where is my mistake?

2. Also, there is a step in the proof that I borrowed from another proof. It's when it says that since $\underline{\varphi_{n-1}}\left(Irr(\alpha_n,K_{n-1})\right)$ divides $p(x)$ it also splits in $E$. I don't really understand this step.

Finally, do you see any other mistake?

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What is your definition of "Galois extension"? For me it is exactly what you've given: the splitting field of a separable polynomial, so thus a normal, algebraic and separable extension = Galois extension! You must have another definition...? – DonAntonio Dec 31 '12 at 13:06
Well, you're right. I'm using the definition: an extension $L/K$ is Galois when $\mbox{Gal}(L/K)=\mbox{Aut}_K(L)=[L:K]$. – synack Dec 31 '12 at 13:10
This somewhat depends on the order in which this material is presented. If you already have proved the existence of algebraic closures of fields, then you can fix an algebraic closure $\overline{K}$ at the beginning which gives you a unique splitting field $E$ in between $\overline{K}/K$ exactly as you wrote (by picking the roots from $\overline{K}$). Otherwise, the splitting field is only unique up to isomorphism. – Matt Dec 31 '12 at 15:22
+1 for answering the first doubt. In both cases that you expose, the splitting field of a polynomial over a field $K$ is the minimal field that contains all the roots of the polynomial. And, of course, is unique up to isomorphism, but there cannot be another splitting field between the splitting field and $K$. So, when they are speaking about splitting fields they are always referring to the same one, right? – synack Dec 31 '12 at 18:26