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?
Best Answer
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 $K$.