0
$\begingroup$

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 $F$?

$\endgroup$
0
$\begingroup$

We have a homomorphism $F \to R \to R/P$. A homomorphism between fields is automatically a field extension, hence $R/P$ contains $F$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user228113 Apr 26 '15 at 11:36
  • 1
    $\begingroup$ Both maps are homomorphisms in the cateory of rings with $1$. We do not even have to bother about that. $\endgroup$ – MooS Apr 26 '15 at 11:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.