Let $k$ be a field, and $A$ be a finitely generated $k$-algebra with $\text{dim}(A)\leq 1$. Then for any maximal ideal $\mathfrak{m}$ of $A$, does this inequality $[A/\mathfrak{m}:k]<\infty$ hold?

Actually, what I really want to know is following; for a scheme $X$ of finite type over $k$ with $\text{dim}(Z)=1$ for any irreducible component $Z$ of $X$. Let $x\in X$ be a closed point, and $k(x)$ be the residue field at $x$. Then $[k(x):k]<\infty$.


Yes, $[A/\mathfrak{m}:k]<\infty$.

This follows from Zariski's version of the Nullstellensatz: a finitely generated algebra $B$ over a field $k$ which is itself a field satisfies $[B:k] \lt \infty$ . Apply to $B=A/\mathfrak m$.
Notice that the hypothesis $\text{dim}(A)\leq 1$ is irrelevant.

The generalization to schemes about which you "really want to know" immediately follows by taking an affine open neighbourhood of the closed point $x$ you are interested in.
(Once again considerations on the dimension of the scheme are irrelevant)

Zariski's result is Corollary 5.24, page 67 of Atiyah-Macdonald's Introduction to Commutative Algebra.

In case you want your very own copy of Atiyah-Macdonald's book, the good people of INTERNATIONAL__BOOKS/DVD'S will send you one, used but in very good condition, at the bargain price of \$6,785.78
Ah yes, you have to add $3.99 for delivery.


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.