I am sorry, in the whole text below $k$ is just meant to be $\mathbb{C}$.

This question is closely related to my previous one here.

I am considering the two rings $k[X]=k[x,y,z]/\langle xy,xz,yz\rangle$ and $k[Y]=k[x,y]/\langle xy(x-y)\rangle$. The question is if $X\subseteq\mathbb{C}^3$ and $Y\subseteq\mathbb{C}^2$ are isomorphic as schemes, and a hint is to define and compute the tangent space of such a scheme at the origin.

If I regard $X$ as an affine scheme $\operatorname{Spec}k[X]$, and if $\mathcal{O}_X$ is its structure sheaf, then the origin 'is' the maximal ideal $\mathfrak{p}=\langle x,y,z\rangle\subseteq k[X]$. Then $\mathcal{O}_{X,\mathfrak{p}}\cong k[X]_\mathfrak{p}$, which is a local ring with maximal ideal $\mathfrak{m}:=\mathfrak{m}_{X,\mathfrak{p}}=\mathfrak{p}k[X]_\mathfrak{p}$. I can then define the tangent space of $X$ at the origin as the dual space of $\mathfrak{m}_{X,\mathfrak{p}}/\mathfrak{m}_{X,\mathfrak{p}}^2$ I guess.

But what is the standard procedure to compute such tangent spaces? Can one do this without 'computing' the ring $k[X]_\mathfrak{p}$ first (maybe by finding a basis of the tangent space by looking at $\mathfrak{m}$ directly)? In any case I would like to know what $k[X]_\mathfrak{p}$ and $k[Y]_{\langle x,y\rangle}$ are, but as usual I can't think of a way to 'simplify' these. Does it ever help that localization commutes with taking quotients in actual computations like these?

The maximal ideal $\mathfrak{m}$ consists of fractions $\frac{f}{g}$, where $f\in\mathfrak{p},g\notin\mathfrak{p}$, i.e. where $f$ doesn't have a constant term while $g$ has a non-zero one. $\mathfrak{m}^2$ has as generators products of two elements of the above form, in particular the numerator has only degree-2 or higher terms. So elements in the quotient have a numerator with only degree-1 terms? This would be consistent with the naive $\mathfrak{m}/\mathfrak{m}^2=\mathfrak{p}_\mathfrak{p}/(\mathfrak{p}_\mathfrak{p})^2=(\mathfrak{p}/\mathfrak{p}^2)_\mathfrak{p}$, from which I don't really know if I am allowed to calculate like this.

As you can see, I'm confused ;) Answers to the various questions above would be very much appreciated, as wells as hints on how to go on with computing the tangent space, and on how to compute the local rings above. My (very vague) guess would be that we get a basis of the dual of the tangent space of $X$ at the origin by taking $x/1,y/1,z/1$, hence a 3-dimensional vector space, which for $Y$ can never happen, as this should be the only way to show $X$ and $Y$ are not isomorphic via tangent spaces at a certain point!? But I could also be totally wrong here.

Edit: Again, I forgot the geometry: Can I guess from the geometric picture ($X$ is the union of the three coordinate axes in $\mathbb{C}^3$, while $Y$ consists of the two coordinate axes plus the line $y=x$ in $\mathbb{C}^2$) what the tangent spaces at these two (clearly singular) points should be?

Thank you in advance!


1 Answer 1


Your intuition is 100% correct and I'll just help you write down the argument formally.

First let us simplify our lives. The Zariski cotangent space is the right and natural invariant : dualizing in order to get the tangent space is useless and can only lead to loss of information.

a) Let us begin with the plane curve $Y=Spec(A/I)$ with $A=k[x,y],\;I=(xy(x-y)) $ .
Call $M=(x,y)\subset k[x,y], \;\mathfrak m=(\bar x, \bar y)\subset A$ and $\mathfrak n=\mathfrak m A_{\frak m}\subset A_{\frak m}$.
The cotangent space at $O$ (corresponding to $\mathfrak m$) of $Y$ is $\mathfrak n/\mathfrak n^2=\mathfrak m/\mathfrak m^2=(\bar x, \bar y)/(\bar x^2, \bar x \bar y,\bar x^2)\cong M/M^2=(x, y)/( x^2, xy, x^2)\cong k^2$ .

b) Let us now study $X=Spec(B/J)$ with $B=k[x,y,z],\;J=(xy, xz,yz) $
Call $P=(x,y,z)\subset k[x,y,z], \;\mathfrak p=(\bar x, \bar y, \bar z)\subset B$ and $\mathfrak q=\mathfrak p B_{\frak q}\subset B_{\frak p}$.
The cotangent space at $O'$ (corresponding to $\mathfrak p$) of $X$ is
$\mathfrak q/\mathfrak q^2=\mathfrak p/\mathfrak p^2=(\bar x, \bar y,\bar z)/(\bar x^2, \bar x \bar y,...,\bar z^2)\cong M/M^2=(x, y,z)/( x^2, xy, ,..., z^2)\cong k^3$

Since no point of $Y$ has a cotangent space of dimension 3, and since $X$ has such a point $O'$, the schemes $X$ and $Y$ are not isomorphic.
[Note that the calculation in a) was actually not necessary: a subscheme of $\mathbb A^2_k$ has cotangent space of dimension $\leq 2$ at any point. However I wanted to make that computation as a warm-up for b)]
Edit: as an answer to Frederik's comment below, let me mention that if $Y\subset Z$ is a subvariety and $y\in Y$ , we have $dim(Y)\leq dim_k (T^*_y(Y))\leq dim_k(T^*_y(Z))$ and if $ Z=\mathbb A^n_k$, then $dim_k(T^*_y(\mathbb A^n_k))=n$

Commutative algebra reminder
I have used that if $\mathfrak m\subset A$ is a maximal ideal of a ring and if $\mathfrak n=\mathfrak m A_{\frak m}\subset A_{\frak m}$, then $\mathfrak n/\mathfrak n^2=\mathfrak m/\mathfrak m^2$. You can find this on page 24 of Milne's freely available algebraic geometry notes (with however the superfluous condition that $A$ be an integral noetherian domain).

  • $\begingroup$ Hello @Georges, and thank you once again for a very encouraging and helpful answer! I'm going to check Milne's proof of that CA fact now. As a side-question: So the actual computation of those tangent spaces becomes more difficult once you drop e.g. 'noetherian'. Does that definition of (co)tangent space still make sense in case of a general affine scheme? And couldn't one define this cotangent space even for non-closed points of such an affine scheme just the same way? $\endgroup$ Commented Mar 14, 2012 at 12:46
  • $\begingroup$ I hope I'm not being too demanding here, but another - hopefully short - question came up. Regarding how one proves that those two schemes (and thus rings) are not isomorphic, I guess 'computing' (simplifying) the actual localized rings themselves and checking if they are isomorphic is rather impracticable in general (working with the maximal ideal seems simpler)? $\endgroup$ Commented Mar 14, 2012 at 12:56
  • $\begingroup$ @Georges: I have a (potentially dumb) question. Hartshorne states in prop 5.2A that $\dim_k m/m^2 \geq \dim A$ for a noetherian local ring $A$. You seem to state the reversed inequality in the paragraph with header "conclusion". How is this? $\endgroup$ Commented Mar 14, 2012 at 12:59
  • $\begingroup$ @Fredrik: That confused me a bit, too, but I think it's not a contradiction here. In the case of $Y$, the only non-regular point is the origin, where the cotangent space has dimension 2, which is greater than $\dim(Y)=\dim k[Y]=1$. The dimension of the cotangent space can still have an upper bound, while being bounded by $\dim k[Y]$ from below, or not? At least that's how I understood it :) $\endgroup$ Commented Mar 14, 2012 at 13:13
  • $\begingroup$ Dear @Rand, it is unfortunate that Milne mentions noetherianity: it is totally irrelevant for his lemma (and I think it isn't even used in his own proof !) $\endgroup$ Commented Mar 14, 2012 at 16:20

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