$E$ is a splitting field of a polynomial over $K$ then $E/K$ is finite normal extension

Question

Let $E$ is a splitting field of a polynomial over $K$ then $E/K$ is finite normal extension.

Now the polynomial is of finite degree and will have finitely many roots, so the extension $E/K$ is finite and also since $E$ is the splitting field of the polynomial over $K$ so it contains all the roots of the polynomial and hence a normal extension.

Is the reasoning correct?

Answer 1

Theorem: Let be $K$ field extension of $k$, the the following are equivation:

$(1)$ $K$ normal extension of $k$ and finite

$(2)$ $K$ splitting fields of $k$ for $f\in k[x]$

proof $2) \to 1)$ Let be $\alpha_1,..,\alpha_n $ roots of polynomial $f(x)$ then $K=k(\alpha_1,..,\alpha_n)$ and let $g(x)$ irreducible polynomial in $k[x]$ and have root $\beta \in K$ and $L$ splitting field $g(x)$ and $\beta'\in L$ onther root of $g(x)$ , then exitst isomorphism $\sigma: k(\beta) \to k(\beta'); \sigma(\beta)=\beta' $, and we note that $K$ is spilliting field of $f(x)$ over $k(\beta) $ and $K(\beta')=k(\alpha_1,..,\alpha_n,\beta')$ is spilliting field of $f(x)$ over $k(\beta')$, then exists isomorphism $\tau:K \to K(\beta') ; \tau(a)=a ; a\in k(\beta) , \tau(\beta)=\beta'$, then $\tau(\alpha_1),..,\tau(\alpha_n)$ are roots of $f(x)$ in $K(\beta')$. and $\beta \in K; \beta=h(\alpha_1,..,\alpha_n)$, then $\beta'=\tau(\beta)=\tau(h(\alpha_1,..,\alpha_n))=h(\tau(\alpha_1),..,\tau(\alpha_n))\in K$, so all roots of $g(x)$ in $K$ then splitting in $K[x]$ and $K$ normal extension of $k$. we note that $K$ finite extension of $k$

Answer 2

Your reasoning that $E$ is finite over $K$ is correct.

For normality, it depends on what definition of a normal extension you are using. There are a few equivalent ones, and from your proof it is not clear to me which definition you are using.

Here is one definition we can take for $E$ to be normal over $K$ (from wikipedia). Let $\overline{K}$ be an algebraic closure of $K$ containing $E$. Then $E$ is normal over $K$ if every embedding, $\sigma$, of $E$ into $\overline{K}$ that fixes $K$ (means $\sigma\rvert_K = id\rvert_K$), sends $E$ to $E$, that is, $\sigma(E) = E$.

Let's say that $E$ is the splitting field of the polynomial $f\in K[x]$. Then $E = K(\alpha_1,\alpha_2,\ldots,\alpha_n)$, where the $\alpha_i$ are the roots of $f$. One way to write the proof would be to argue as follows. Let $\sigma$ be an embedding of $E$ into $\overline{K}$ that fixes $K$. Then $\sigma$ permutes the roots of $f$, so for all $i$, $\sigma(\alpha_i) = \alpha_j$ for some $j$. The $\alpha_j$'s are in $E$, so $\sigma$ sends $E$ to $E$. Thus $E$ is normal over $K$.