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?

  • $\begingroup$ It depends on your definitions and setup. If you're taking a "classical" viewpoint, then $V(x^2+y^2)\subset\Bbb A^2_{\Bbb R}$ is a 1-dimensional variety which has only one point. If you're thinking appropriately scheme-theoretically, the conclusion is fine (see for instance Hartshorne exercise II.3.20) but your proof needs refinement. $\endgroup$
    – KReiser
    Commented Jul 16, 2022 at 2:23
  • $\begingroup$ Hmmm, so in the classical viewpoint, does $V(x^2 + y^2) \subset \mathbb{A}^2_\mathbb{R}$ actually contradict the equality of transcendence degree of the function field and Krull dimension of the local ring at any point? If so then I guess that's a mistake in Atiyah Macdonald, and where does the argument in my post use the fact that $k$ is algebraically closed? $\endgroup$ Commented Jul 16, 2022 at 20:11
  • $\begingroup$ No, $\Bbb R[x,y]_{(x,y)}/(x^2+y^2)$ is of dimension one. (But if you were talking about local dimension as the dimension of the variety near the point, the classical setup runs in to some issues.) $\endgroup$
    – KReiser
    Commented Jul 16, 2022 at 20:18
  • $\begingroup$ I see, so defining the local dimension as the dimension of $\mathcal{O}_P$, where $\mathcal{O}$ is the coordinate ring of the variety, we should still have the equality for a non-algebraically-closed field, right? And am I missing anything in the proof above? It just seems odd to me that the authors would include an exercise where we can just immediately apply the proof given in the book, rather than just commenting that the assumption that $k$ is algebraically-closed wasn't really necessary to begin with $\endgroup$ Commented Jul 16, 2022 at 20:32
  • $\begingroup$ 1: Yes, that's what I was saying with the reference to Hartshorne. 2: As I said, your proof needs some refinement. For instance, over an algebraically closed field, the maximal ideals are not all of the form $(x_1-a_1,\cdots,x_n-a_n)$. $\endgroup$
    – KReiser
    Commented Jul 16, 2022 at 21:36

2 Answers 2


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.

  • $\begingroup$ Thanks for the thoughtful response! That line is saying that the ideal of functions vanishing at $(a_1, \dots, a_n) \in V$ is the ideal $\pi((x_1 - a_1, \dots, x_n - a_n))$, where $\pi$ is the quotient from $k[x_1, \dots, x_n]$ to $A(V)$. Apologies if this is unclear in the post, but my issue is - why do we care about all the maximal ideals that are not $\pi((x_1 - a_1, \dots, x_n - a_n))$ for some $(a_1, \dots, a_n)$ on the variety? We aren't trying to prove that $\dim V = \dim B_\mathfrak{m}$ for all maximal ideals $\mathfrak{m}$, only for the ideals given by some point on the variety $\endgroup$ Commented Jul 20, 2022 at 15:50
  • $\begingroup$ Maybe the issue is in using example 124 to see that the dimension of $B_\mathfrak{n} = d$? I was thinking that the example shows that $\dim B_{(y_1, \dots, y_d)} = d$ since the ring is regular so the dimension of that ring equals the minimum number of generators for $(y_1, \dots, y_d)$, which is clearly $d$. But if $\mathfrak{n}$ is not $(y_1 - b_1, \dots, y_d - b_d)$ for some $(b_1, \dots, b_d)$, then that example wouldn't apply, and even if $\mathfrak{m} = \pi(x_1 - a_1, \dots, x_n - a_n)$, $\mathfrak{n}$ might not be of the same form? $\endgroup$ Commented Jul 20, 2022 at 15:54
  • $\begingroup$ No, you have misunderstood the exercise. Atiyah-MacDonald asks you to show that $\dim{V} = \dim{A(V)_{\mathfrak{m}}}$ for an arbitrary maximal ideal $\mathfrak{m} \subseteq A(V)$. For a non-algebraically-closed field, these maximal ideals may not correspond to points on the variety. In particular, since $A$ is integral over $B$, every maximal ideal of $B$ is the contraction of some maximal in $A$ and vice versa, so you would have to show that $\dim{B_\mathfrak{n}} = d$ for every maximal ideal $\mathfrak{n} \subseteq B$. $\endgroup$
    – Emory Sun
    Commented Jul 21, 2022 at 2:28
  • $\begingroup$ Ok if that's the exercise then it's clear why you need to modify their proof. They define local dimension in the context of algebraically closed fields, so I suppose I assumed the generalization would be to only ideals that are vanishing sets of points on the variety, rather than to all maximal ideals $\endgroup$ Commented Aug 2, 2022 at 7:59

$\def\trdeg{\operatorname{trdeg}} \def\p{\mathfrak{p}} \def\m{\mathfrak{m}} \def\n{\mathfrak{n}}$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 $k[y_1,\dots,y_d]\to A$.


[AM] Atiyah, Macdonald, Introduction to Commutative Algebra


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