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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 $[k(x):k]<\infty$.

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  • $\begingroup$ For any $k$-scheme $X$ locally of finite type and a point $x\in X$, the following are equivalent: (i) $x$ is a closed point. (ii) The field extension $k\hookrightarrow \kappa(x)$ is finite. (iii) The field extension $k\hookrightarrow \kappa(x)$ is algebraic. This is Prop. 3.33 of Görtz, Wedhorn, Algebraic Geometry I: Schemes, 2nd ed. $\endgroup$
    – Elías Guisado Villalgordo
    Mar 8 at 12:43

1 Answer 1

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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)

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

Edit
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.

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