I'm new to algebraic geometry and struggle with the notion of the dimension of an affine variety. In Robin Hartshorne's book Algebraic Geometry the dimension is introduced using the Zariski topology:
Definition: Let $X$ be a topological space, we define its dimension as the supremum over all integers $n$ s.t. there exists an ascending chain of closed irreducible components (w.r.t. the topology) $Z_0 \subset Z_1 \subset \dots \subset Z_n\subseteq X$. The dimension of an affine variety is then defined to be the dimension as its topological space.

Then he shows that for an affine variety $V \subseteq K^n$ the Krull dimension of the coordinate ring $K[x]/\mathcal{I}(V)$ (defined as the supremum of heights of prime ideals) equals the dimension of $V$ as defined above.

Question 1: My understanding is that Hartshorne assumes the field $K$ to be algebraically closed throughout the book and I'm wondering whether this particular result changes when using e.g. $K=\mathbb{R}$. My understanding from quickly going through section 3.3 in https://perso.univ-rennes1.fr/michel.coste/polyens/SAG.pdf is that this should indeed hold. Can anyone help me out or ideally refer to some literature about this?

Question 2: I stumbled upon the supposed counter-example $\mathcal{V}(\sum_{i=1}^n x_i^2)\subseteq K^n$, which I don't quite understand. My reasoning is as follows:
Let $f=\sum_{i=1}^n x_i^2\subseteq \mathbb{R}[x_1,\dots,x_n]$, then $V:=\mathcal{V}(f)\subseteq \mathbb{R}^n=\{(0,\dots,0)\}$ should have dimension 0 in the Zariski topology.
The coordinate ring is given by $$ \mathbb{R}[x_1,\dots,x_n]/\mathcal{I}(V) = \{[f]:f\in \mathbb{R}[x_1,\dots,x_n]\} $$ with $g\in[f] \iff f-g\in \mathcal{I}(V)$. Here we have that $h\in\mathcal{I}(V) \iff h(0,\dots,0)=0$, which happens if the constant part of $h$ denoted by $a_0^h$ is zero (because $h(0,\dots,0)=a_0^h$). Thus, two polynomials are in the same equivalence class if they have the same constant part, i.e. $g\in[f] \iff a_0^g = a_0^f$. Since $\forall f\in \mathbb{R}[x_1,\dots,x_n] : a^f_0\in\mathbb{R}$ we have $\mathbb{R}[x_1,\dots,x_n]/\mathcal{I}(V) =\mathbb{R}$. If I'm not mistaken, the Krull dimension of $\mathbb{R}$ as a ring should also be 0.
Is there anything that I have missed?

  • $\begingroup$ You shouldn't think to $V(\sum_{i=1}^n x_i^2)$ as a subset of $\Bbb{R}^n$. You can think to it as a subset of $\Bbb{C}^n$ quotiented by the Galois action of $Gal(\Bbb{C/R})^n$ (quotienting by the Galois action is to keep a correspondence between subvarieties, defined over $\Bbb{R}$, and prime ideals of $\Bbb{R}[x_1,\ldots,x_n]/(\sum_{i=1}^n x_i^2)$) $\endgroup$
    – reuns
    Commented Dec 6, 2022 at 17:34

1 Answer 1


Question 1: Hartshorne does not actually assume $K = \overline{K}$ throughout his book, only in chapter 1. For an affine variety $X$ with coordinate ring $R$, whether or not $\dim X = \dim R$ holds depends on your definition of $X$.

Question 2: This counter example actually does work. To simplify things, consider the vanishing set $X = V(x^2 + y^2)$ over $\mathbb R$ and its coordinate ring $R = \mathbb R[x,y]/(x^2 + y^2)$. The Zariski topology on $\mathbb R^2$ still makes sense in the absence of algebraic closure, and as you noted, $X$ is simply the origin and thus has dimension $0$.

However, the coordinate ring $R$ does not have Krull dimension $0$, it has Krull dimension $1$. Recall that Krull dimension is defined as the maximum length of chains of prime ideals. By the correspondence theorem, the prime ideals of $R$ are precisely the prime ideals of $\mathbb R[x,y]$ which contain the ideal $(x^2 + y^2)$: $$\operatorname{Spec}(R) = \left\{\mathfrak p \subseteq \mathbb{R}[x,y] ~:~ \mathfrak p \text{ is prime and } x^2 + y^2 \in \mathfrak p\right\}.$$ Since $\mathbb R[x,y]$ is a UFD, ideals generated by irreducible polynomials are prime, and hence the ideal $(x^2 + y^2)$ is prime in $\mathbb R[x,y]$. It corresponds to the prime ideal $(0)$ in $R$. The element $x^2 + y^2$ is also contained in the maximal ideal $(x,y) \subseteq \mathbb R[x,y]$, so $(x,y)$ corresponds to a prime in $R$ which strictly contains $0$. This implies that $$ (0) \subsetneq (x,y)$$ is a chain of prime ideals in $R$, and since it has length 1, we conclude that $\dim R \geq 1$ (there might be a longer chain of primes than the one we found), which for our purposes is enough. Your proposed counter example does actually work.

Is there a fix to this? As I said, whether or not Krull dimension and Zariski dimension agree for affine varieties more generally depends on your definition of "variety", and indeed, the failure of these two notions to agree for non-algebraically closed fields might motivate you to look for a better definition. In chapter 2, Hartshorne introduces the notion of a scheme as a replacement for varieties which, among many other things, frees us from assuming all fields are algebraically closed. In fact, it allows us to consider affine variety-like objects over not just fields, but arbitrary rings as well. It turns out that topological dimension and Krull dimension always agree for affine schemes.

  • 1
    $\begingroup$ +1, good answer, good perspective - and there's even more stuff lurking under "what is $X$ really". For instance, if you took the coordinate ring of the vanishing set by defining it as $\Bbb R[x_1,\cdots,x_n]/I(V(x^2+y^2))$, you'd get $\Bbb R$ as expected. This is called taking the real radical of $I$. This is occasionally useful in real algebraic geometry, which is at times nearly a separate subject from what's normally thought of as algebraic geometry. $\endgroup$
    – KReiser
    Commented Dec 6, 2022 at 18:53
  • $\begingroup$ Thanks for the good answer, that makes more sense to me now. Follow up question: Can we say anything about the relation of these two notions if we know for $V=\mathcal{V}(I)\subseteq \mathbb{C}^n$ that $V=V\cap\mathbb{R}^n$? $\endgroup$
    – joap
    Commented Dec 7, 2022 at 9:01
  • $\begingroup$ @KReiser Indeed, this is another good answer to "is there a fix for this" and is far more grounded than immediatley deferring to schemes. $\endgroup$
    – LéKitty
    Commented Dec 7, 2022 at 19:46
  • $\begingroup$ @joap I'm not sure I understand your question, but if you are proposing that we first take the vanishing set in $\mathbb C^n$ and then intersect with $\mathbb R^n$, ask yourself: will you end up with something different than if you just took the = zero locus in $\mathbb R^n$ in the first place? $\endgroup$
    – LéKitty
    Commented Dec 7, 2022 at 19:50
  • 1
    $\begingroup$ In this case, it's actually not that - the ideal $(x^2+y^3)\subset\Bbb R[x,y]$ is radical! One thing you'll come to appreciate as you continue in algebraic geometry is that keeping track of $I$ is the "right" thing to do, and that there is a real advantage to the extra technical machinery that's been invented for doing algebraic geometry over rings that aren't algebraically closed fields. $\endgroup$
    – KReiser
    Commented Dec 8, 2022 at 13:55

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