Let $K \subset \Bbb C$ be a field, let $P \in K[X]$ be a polynomial of degree $n$ and let $N$ be the splitting field of $P$ over $K$. Then $N$ is a normal extension of finite degree over $K$.
The proof of this is as follows;
Let $x_1, \dots, x_n$ be the (not necessarily distinct) roots of $P$ in $\Bbb C$. We know that $N = K[x_1, \dots, x_n]$ is an algebraic extension of $K$. Let $\sigma : N \to \Bbb C$ be a $K$-homomorphism. For $i = 1, \dots, n$ we have $0 = \sigma(P(x_i)) = P(\sigma(x_i))$, so $\sigma(x_i) \in \lbrace x_1, \dots, x_n\rbrace \subset N$. As $\lbrace x_1, \dots, x_n\rbrace$ generates $N$ we have $\sigma(N) \subset N$, and so since $\sigma$ was arbitrary, the image of every $K$-homomorphism is contained in $N$ and so $N$ is normal of finite degree.
I think I'm having difficulty understanding the conclusion that $\sigma(N) \subset N$. It seems that, since $\sigma(x_i) \in \lbrace x_1, \dots, x_n\rbrace \subset N$ and $\sigma$ fixes $K$, the image of $\sigma$ is simply all of $N$. Or is this only the case when all of the roots are distinct? Since if $x_i \neq x_j$ for all $i, j = 1, \dots, n$ then $\sigma(x_i) \neq \sigma(x_j)$. I'd be grateful if somebody could clarify this for me.