Equality of transcendence degree and local dimension for non-algebraically closed fields

Question

In Atiyah-Macdonald, the authors prove that if $V$ is an irreducible variety over an algebraically closed field $k$, then the local dimension of $V$ (i.e. the Krull dimension of the localization of the coordinate ring at any point) is equal to the transcendence degree of the function field of $V$ over $k$ (Theorem 11.25). Then, in Exercise 3 of Chapter 11 the authors ask the reader to extend this result to non-algebraically-closed fields. I don't see why algebraically-closed is necessary for their proof, so I want to ask why the following does not work:

Proof: Let $k[x_1, \dots, x_n]$ be a polynomial ring in $n$ variables over any field $k$. Let $\mathfrak{p}$ be a prime of $k[x_1, \dots, x_n]$, and $A(V) = k[x_1, \dots, x_n] / \mathfrak{p}$. Let $k(V)$ denote the field of fractions of $A(V)$. Let $P = (a_1, \dots, a_n)$ be a point of the variety $V$ defined by $\mathfrak{p}$. Then the maximal ideal $(x_1 - a_1, \dots, x_n - a_n)$ of $k[x_1, \dots, x_n]$ descends to a maximal ideal $\mathfrak{m}$ of $A(V)$. By Noether's normalization lemma, we can find $y_1, \dots, y_d$ algebraically independent elements in $A(V)$ such that $A(V)$ is integral over $B = k[y_1, \dots, y_d]$. Thus $\mathfrak{n} = \mathfrak{m} \cap B$ is a maximal ideal of $B$. Further, $B$ is a UFD since a polynomial ring over any field is a UFD, so $B$ is integrally closed. Thus Atiyah Macdonald Lemma 11.26 shows that $\dim A_\mathfrak{m} = \dim B_\mathfrak{n}$, but $\dim B_\mathfrak{n} = d$ by the example on page 124 of Atiyah Macdonald. QED.

This is essentially the exact proof of Theorem 11.25 in A&M - it seems to me they are only using that $k$ is algebraically closed for the weak Nullstellensatz, but even when $k$ is not algebraically closed, the points of $V$ still correspond to maximal ideals. The correspondence just might not be surjective. Is there some other reason we need $k$ to be algebraically closed that I am missing in the above proof?

Answer 1

I think the problem in your proof does, in the end, lie in the fact that not every maximal ideal of $k[x_1, \ldots, x_n]$ is necessarily of the form $(x_1 - a_1, \ldots, x_n - a_n)$ for a non-algebraically-closed field $k$. The last step of your proof, where you claim that $\dim{B_\mathfrak{n}} = d$, is not actually demonstrated anywhere in Atiyah-MacDonald for an arbitrary maximal ideal $\mathfrak{n}$. What is demonstrated (page 121) is that the ring $k[x_1, \ldots, x_n]$ localized at a maximal ideal of the form $(x_1 - a_1, \ldots, x_n - a_n)$ is of dimension $n$, which implies, of course, that $\dim{k[x_1, \ldots, x_n]_{\mathfrak{m}}} = n$ for all maximal ideals $\mathfrak{m}$ when $k$ is algebrically closed. (Furthermore, the example you cite on page 124 has nothing to do with the dimension of $B_\mathfrak{n}$; rather, it stipulates that $B_{(x_1, \ldots, x_n)}$ is regular.)

To modify your proof as it is, you would either have to show that every maximal ideal in $B$ is of the form $(x_1 - a_1, \ldots, x_n - a_n)$ (which is untrue), or somehow show that $\dim{B_\mathfrak{n}} = d$ for an arbitrary maximal ideal $\mathfrak{n} \subseteq k[x_1, \ldots, x_d]$ independently.

Edit: Perhaps the third sentence of your proof ("Then the maximal ideal $(x_1−a_1,\ldots,x_n−a_n)$ of $k[x_1,\ldots,x_n]$ descends to a maximal ideal $\mathfrak{m}$ of $A(V)$") was an attempt to remedy this error; if it was, I couldn't decipher exactly what you meant.

Answer 2

$\def\trdeg{\operatorname{trdeg}} \def\p{\mathfrak{p}} \def\m{\mathfrak{m}} \def\n{\mathfrak{n}}$Even the OP didn't ask for it, for completeness I will explain how one deduces the result over a not necessarily algebraically closed field from the algebraically closed case.

Let $k$ be a field, $\p\subset k[x_1,\dots,x_n]$ be a prime, $A=k[x_1,\dots,x_n]/\p$ and let $\m\subset A$ be maximal. We want to show that $\dim A_\m=\trdeg_k (\operatorname{Frac}A)$. Since $k[x_1,\dots,x_n]\subset \overline{k}[x_1,\dots,x_n]$ is an integral extension, it satisfies lying over [AM, Theorem 5.10], so let $\overline{\p}\subset\overline{k}[x_1,\dots,x_n]$ be a prime contracting to $\p$ in $k[x_1,\dots,x_n]$. Denote $\overline{A}=\overline{k}[x_1,\dots,x_n]/\overline{\p}$. The induced map $A\to\overline{A}$ is injective, and the image in $\overline{A}$ is polynomials that modulo $\overline{\p}$ have coefficients in $k$. Since $x_1,\dots,x_n$ generate $\overline{A}$ as a $\overline{k}$-algebra, Noether normalization lemma tells us that there are $f_1,\dots,f_d$ in the $\mathbb{Z}$-subalgebra of $k[x_1,\dots,x_n]$ generated by $x_1,\dots,x_n$ such that the map $\overline{k}[y_1,\dots,y_d]\to\overline{A}$, $y_i\mapsto f_i$, is injective and integral (since $\overline{k}$ is infinite, we may actually take these $f_i$'s in the $\mathbb{Z}$-submodule of $k[x_1,\dots,x_n]$ generated by $x_1,\dots,x_n$, but this is not important). In any case, the $f_i$'s lie in the image of $A\to\overline{A}$, so we get a commutative diagram $$ \require{AMScd} \begin{CD} k[y_1,\dots,y_d]@>>>\overline{k}[y_1,\dots,y_d]\\ @VVV@VVV\\ A@>>>\overline{A} \end{CD} $$ The map $k[y_1,\dots,y_d]\to A$ is injective, but also integral (for the composite $k[y_1,\dots,y_d]\to \overline{k}[x_1,\dots,x_n]\to \overline{A}$ of integral morphisms is integral). Since $A\to\overline{A}$ is an integral extension, there is a maximal ideal $\overline{\m}\subset\overline{A}$ contracting to $\m$ in $A$ [AM, Theorem 5.10, Corollary 5.8]. Denote $\n$ (resp., $\overline{\n}$) to the contraction of $\m$ in $k[x_1,\dots,x_n]$ (resp., in $\overline{k}[x_1,\dots,x_n]$). By [AM, Lemma 11.26], the ideals $\n$ and $\overline{\n}$ are maximal and $$ \dim(A_\m) =\dim(k[y_1,\dots,y_d]_\n) =\dim\left(\overline{k}[y_1,\dots,y_d]_\overline{\n}\right) =\dim\left(\overline{A}_{\overline{\m}}\right). $$ We already know that $d=\dim\left(\overline{k}[y_1,\dots,y_d]_\overline{\n}\right)=\trdeg_\overline{k}\left(\operatorname{Frac}\overline{A}\right)$. Thus, it suffices to see that $\trdeg_k(\operatorname{Frac}A)=\trdeg_\overline{k}\left(\operatorname{Frac}\overline{A}\right)$. But this follows from Lemma 2 of this answer applied to $k[y_1,\dots,y_d]\to A$.

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References

[AM] Atiyah, Macdonald, Introduction to Commutative Algebra