Let $F$ be a field and $R$ a finitely generated $F$-algebra. Let $P$ be a maximal ideal of $R$. Then $\dim(R/P)$ as a vector space over $F$ is finite.

Let $F$ be a field and $R$ a finitely generated $F$-algebra. Let $P$ be a maximal ideal of $R$. Then $\dim(R/P)$ as a vector space over $F$ is finite.

$P$ is a maximal ideal of $R/P$ is a field. I know that the result directly follows from the weak nullstellensatz if we can show that $R/P$ is a finitely generated $F$-algebra, but how can we be sure that $R/P$ contains $F$?

We have a homomorphism $F \to R \to R/P$. A homomorphism between fields is automatically a field extension, hence $R/P$ contains $F$.
• I humbly think that your answer might improve if you point out that these map don't send $1\mapsto 0$. I agree that, strictly speaking, it should follow from the definition of ring-homomorphism, but in this case it's literally half of the solution. – user228113 Apr 26 '15 at 11:36
• Both maps are homomorphisms in the cateory of rings with $1$. We do not even have to bother about that. – MooS Apr 26 '15 at 11:40