Relation between reduced finite algebra, prime ideal and field extension

Question

Is it true that if $L$ is a reduced finite dimensional commutative algebra over a field $K$ (which is finite or of characteristic $0$) and if $\mathfrak p$ is a prime ideal of $L$, then $L/\mathfrak p$ is a field extension of $K$? If yes, how to prove it?

Also, can someone explain the relation between an algebra over a field and a field extension? I got a bit confused about the former concept. Help is much appreciated.

Answer 1

We have a homomorphism $K \hookrightarrow L \twoheadrightarrow L/p$. Any homomorphism between fields is automatically a field extension. So $p$ must be maximal, then this is a field extension. In the finite-dimensional case, this is always the case.

A field extension is nothing else but an algebra over a field, which also happens to be a field itself.

Answer 2

Equivalently: If $L$ is an integral domain, finite-dimensional over $K$, is $L$ a field?

The answer is yes: If $a\in L$ is nonzero, then $x\mapsto ax$ is a linear transformation of finite-dimensional $K$-vector spaces. It is injective, hence surjective, and it follows that $a$ is a unit.