Vakil Rmk 5.2.2: Identify the following space with $\operatorname{Spec}(\mathbb C[x])$: $$X:=\operatorname{Spec}(\mathbb C[x,y_1, y_2,\ldots]/(y_1^2, y_2^2,\ldots, (x-1)y_1, (x-2)y_2,\ldots))$$ and then show that the nonreduced points of $X$ are precisely the closed points corresponding to positive integers and that the complement of this set is not Zariski open.


First let's consider $$X':=\operatorname{Spec}(\mathbb C[x,y_1, y_2,\ldots]/(y_1^2, y_2^2, \ldots))$$

Because we are modding out $\operatorname{Spec}(\mathbb C[x,y_1, y_2, \ldots])$ by the ideals $(y_i^2)$, we get $\mathbb A^1$ with a fuzz that is the intersection of the fuzz of each one of the $(y_i^2)$. This intersection is parallelepiped around the $x$-axis. So we see that we get a map $X' \to \operatorname{Spec}(\mathbb C[x])$, which implies that we also have $X \to \operatorname{Spec}(\mathbb C[x])$.

Now we look at what happens when we further mod out by $((x-i)y_i)$. These ideals are a cross of hyperplanes $x = i$ with $y_i = 0$. They surely add some fuzz at the integer points, but I can't really visualize it. Algebraically, they are adding 1-order differential information, since now we can consider a Taylor expansion around $(i, 0_1, 0_2,\ldots y_i, 0_{i+1}, \ldots)$ up to terms $a_0 + a_xx + a_yy$.


1 Answer 1


It is probably easiest for the first part to consider the reduced scheme, since it is homeomorphic to X. Then you will see instantly that $X_{red}$ is simply $$\text{Spec } \mathbb{C}[x].$$

For the second part:

If $p \in X$ is of the form $(x-a)$ for $a \in \mathbb{Z}$, then if we mod out with p in X we get: $$\text{Spec } \mathbb{C}[y_1,y_2, \ldots]/(y_1^2,y_2^2 \ldots,y_1, \ldots , \hat{y_a},\ldots) = \text{Spec } \mathbb{C}[y_a]/(y_a^2).$$ Here $\hat{y_a}$ means we are leaving out $y_a$. So it is a non-reduced point. If you, conversely have a non-reduced point p, you know that it can't eliminate all of the $y_i^2$. By going to $X_{red}$ you can check that each closed point must contain $(x-a)$ for $a \in \mathbb{C}$. The only points containing these that gives a non-reduced ring when quotiening out are the ones of the form $(x-a)$ where $a \in \mathbb{Z}$. So you are done with this part.


Let us call the set of all primes of the form $(x-a)$ where $a \in \mathbb{Z}$ for $K$. To show that the complement of K is not Zariski open is the same as saying that K is not Zariski closed. Once again, you have a homeomorphism $X_{red} \rightarrow X$, and in the former, any proper closed set contains a finite number of points, while your K has infinitely many points. So, K can't be Zariski closed.

I admire your careful pictorial analysis of the situation, but I would only suggest that as a first part to gain intuition on what is really going on! After that, you should try to be more formal. What I have done is essentially taking your intuition of the situation and made it formal.

  • $\begingroup$ Thank you Dedalus! You nailed it. I couldn't really see before what happened when taking the quotient by (x-a), but obviously it means substituting x with a everywhere and the quotient falls out as you put. $\endgroup$
    – Rodrigo
    Jun 30, 2013 at 12:09

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