Proving every prime ideal is maximal in a finite dimensional ring over a field. 
Let $R$ be commutative ring with identity that contains a field $K$ as a subring. If$ $R is a finite dimensional vector space over the field $K$, prove that every prime ideal in $R$ will be maximal.

My idea was to prove the integral domain $R/p$ (if $p$ is a prime ideal) was a field as for any ideal $P$, $R/P$ will be a field iff $P$ is maximal in $R$. But how can I use the fact that $R$ is finite dimensional over $K$? I don't understand.
 A: $A:=R/p$ is also finite dimensional and has no zero divisors. 
Take any nonzero $a\in A$ and consider its powers $1,a, a^2,\dots$, these are linearly dependent. Take the least $n\in\Bbb N$ giving $\lambda_na^n+\dots+\lambda_0=0$. 
Since we can cancel out $a$, the constant term $\lambda_0$ is nonzero. 
Then dividing by $-\lambda_0$ and pulling out $a$, we arrive to $a\cdot f(a)=1$ for a specific polynomial $f$, yielding an inverse for $a$. 
A: Hint:
Note that $R/\mathfrak p$ has finite dimension over $K$. You have to prove that any non-zero element $x$ in  $R/\mathfrak p$ is invertible. Consider the map
$$R/\mathfrak p\xrightarrow {\enspace{}\times x\enspace\:}R/\mathfrak p.$$
This is an endomorphism of the $K$-vector space  $R/\mathfrak p$, and it is injective since $R/\mathfrak p$ is an integral domain.
Note :
The more general following result is often used:

Let $A\subset B$ two integral domains, such that $B$ is a finitely generated $A$-module. Then $A$ is a field if and only if $B$ is.

