Proof Verification Regarding Image of $K$-homomorphisms of a Normal Extension Let $N$ be a normal extension of $K$, where $N,K\subseteq \mathbb{C}$, and suppose that $[N:K]=n$. Show that every $K$-homomorphism $\sigma:N\to \mathbb{C}$ satisfies $\sigma (N)=N$. 
My Proof: Since $N$ is normal, we can assume that it is the splitting field of some irreducible polynomial $P\in K[X]$, and given that $[N:K]=n$, we know that $P$ is of degree $n$. Let $\alpha_1,...,\alpha_n\in \mathbb{C}$ be the roots of $P$. By definition of a splitting field we write $N=K(\alpha_1,..., \alpha_n)$.
For any element $b\in N$ we can write $b$ as a linear combination of the generators of $N$, i.e. $b=a_0+a_1\alpha_1+...+a_n\alpha_n$ where $a_i\in K$. We consider the image of $b$ under $\sigma$. By the $K$-linearity of $\sigma$ we obtain:
$\sigma (b)= \sigma (a_0+a_1\alpha_1+...+a_n\alpha_n)=a_0+a_1\sigma (\alpha_1)+...+a_n\sigma (\alpha_n)$
Since $\sigma$ is a $K$-homomorphism, it sends the roots of $P$ to their conjugates i.e. for any $\alpha_i$ for $1\leq i \leq n$, $\sigma (\alpha_i)$ is a conjugate of $\alpha_i$. Since $N$ is normal, we have that for any $1\leq i \leq n$, $\sigma (\alpha_i)\in N$. Therefore $\sigma (N) \subseteq N$. Since $\sigma$ is a ring homomorphism, it is injective, and an injective map $\sigma:N\to N$ must necessarily be a bijection, therefore $\sigma (N)=N$.
I am somewhat unsure of the points I made in the last paragraph as well as the assumption regarding linear combinations. Any corrections or verification of this proof would be greatly appreciated.
Thank you  
 A: I'm afraid there are some weak steps in your argument. Listing those I spotted


*

*$N$ is a splitting field of some polynomial $P\in K[X]$ all right. However, you cannot conclude that $\deg P=[N:K]$. For example the splitting field of $x^3-2$ over $\Bbb{Q}$ is a degree six extension. You can select the polynomial $P$ to have degree $[N:K]$, if you select a primitive element $\alpha$ such that $N=K(\alpha)$, and then let $P$ be the minimal polynomial of $\alpha$ over $K$.

*The zeros $\alpha_1,\alpha_2,\ldots,\alpha_n$ generate $N$ as an extension field of $K$. You seem to think that they generate $N$ as a vector space over $K$. This is not necessarily so. Not even if you include $K$ itself. Consider the polynomial $P(x)=x^4-2$. Its zeros come in pairs of negatives of each other, so they span a vector space of dimension two only. When we say that the zeros $\alpha_i$ generate $N$, we mean that we get all the elements of $N$ as products and quotients of $K$-linear combinations of the zeros. So elements like $\alpha_1^2\alpha_3/(2+\alpha_2^2)$ are not necessarily included yet. Because all the extension in sight are algebraic, we do get all the elements of $N$ as polynomials $f(x_1,x_2,\ldots,x_n)\in K[x_1,x_2,\ldots,x_n]$ evaluated at $x_i=\alpha_i, i=1,2,\ldots,n$.

*A ring homomorphism is not necessarily injective. A homomorphism $f$ of fields OTOH is. This is because the kernel of a homomorphism of rings is an ideal, and the field has only the trivial ideals. As $f(1)=1$ we then necessarily have $\operatorname{ker}(f)=\{0\}$ proving injectivity of $f$.

A: Just to add: while $N$ being normal does imply that it is a splitting field of some $P\in K[x]$, $P$ need not be irreducible. More generally, a finite extension $N/K$ is normal iff $N$ is the splitting field of some $P\in K[x]$.

To complete your proof, let $N=K(\alpha_1,\cdots, \alpha_m)$ where the $\alpha_i$ are roots of $P$ in $N$ (and are all the roots in $\Bbb{C}$, since $P$ splits in $N$). Fix a $K$-homomorphism $\sigma:N\rightarrow \Bbb{C}$. $\sigma$ maps roots of $P$ to roots of $P$, and moreover is completely determined by the values $\sigma(\alpha_1),\cdots,\sigma(\alpha_m)$. But the roots of $P$ in $\Bbb{C}$ are precisely $\lbrace \alpha_1,\cdots,\alpha_n\rbrace$, so $\sigma(N)\subseteq N$.
$\sigma$ is injective since it's a $K$-homomorphism, and moreover is also a $K$-linear map. By the rank-nullity theorem, $N$ and $\sigma(N)$ are $K$-vector spaces of the same finite dimension, and so $\sigma(N)=N$.
Alternatively, avoiding the rank-nullity theorem, note that $\sigma$ permutes $\lbrace \alpha_1,\cdots,\alpha_n\rbrace$. Then, $\sigma(N)=K(\sigma(\alpha_1),\cdots, \sigma(\alpha_m))=K(\alpha_1,\cdots, \alpha_m)=N$.
