If $A$ is reduced, Spec $A$ has no embedded points

I've partly solved the following exercise of Vakil's FOAG, but I am not sure I got the last part right. Could some take a look?

5.5.C. EXERCISE (ASSUMING (A)). Show that if $A$ is reduced, Spec $A$ has no embedded points. Hints: (i) first deal with the case where $A$ is integral, i.e., where Spec $A$ is irreducible. (ii) Then deal with the general case. If $f$ is a nonzero function on a reduced affine scheme, show that $\operatorname{Supp}f = \overline{D(f)}$: the support is the closure of the locus where $f$ doesn’t vanish. Show that $\overline{D(f)}$ is the union of the irreducible components meeting $D(f)$, using (i).

My solution goes as follows:

(i) Since $A$ is integral, an embedded point is given by a non-unit $f$ such that $\operatorname{Supp}f \not \supset V(f)$. This is because Supp $f\supset D(f)$, so that if it also contained $V(f)$, it would contain everything. On the other hand, if $\operatorname{Supp} f\not \supset V(f)$, then Supp $f$ can't be an irreducible component of Spec $A$ and must be the closure of a union of embedded points.

But Supp $f\not \supset V(f)$ means that $\exists \mathfrak p \ni f$ with $f_\mathfrak p =0 \iff f\cdot a =0$ for some $a\not \in \mathfrak p$. But this is not possible because $A$ was assumed to be a domain.

(ii) In this post is a proof of $\operatorname{Supp}f = \overline{D(f)}$ with the above conditions. I claim that, since there are only finitely many minimal primes (noetherianness is a standing assumption in this chapter), we can cover Spec $A$ by finitely many clopen $D(g_i)$, each one of which corresponding to the closure of a minimal prime. It follows that $(g_i) = A$, and we can write $$\overline{D(f)}=\overline{D(fA)}=\overline{D(f(g_i))}=\overline{D((fg_i))}=\bigcup\overline{D(fg_i)}$$

Now $\overline{D(fg_i)}$ is the $\operatorname{Supp} f \cap \operatorname{Spec} A_{g_i}$, which is the spectrum of an integral domain. By (i) this cannot correspond to a proper irreducible subset of $\operatorname{Spec} A_{g_i}$, so (ii) follows.

The claim can be proven by induction, and I show the case where $A$ has three minimal primes $\mathfrak p, \mathfrak q, \xi$. By symmetry, it suffices to show that we can pick an element in $\mathfrak p$ that is not in $\mathfrak q, \xi$. So pick $p_1, p_2 \in \mathfrak p$ such that $p_1 \not \in \mathfrak q$ and $p_2 \not \in \xi$. If either one of the $p_i$ is not in the other two primes, we are done. Otherwise $p_1+p_2$ does the job.

Solution 1:

Let me give an algebraic answer to your question: to me, an embedded point of $\text{Spec}(A)$ is a nonminimal associated prime (your definition does not quite match up with this: points of $\text{Spec}(A)$ do not correspond to elements of $A$).

Here are two reasons why a reduced ring $A$ cannot have embedded primes:

1) Any embedded prime is the annihilator of a nilpotent element. Indeed, every associated prime $p$ is of the form $\text{ann}(x)$, for some $x \in A$. If $p = \text{ann}(x)$ is embedded, then $px = 0$ is contained in every minimal prime, but $p$ is not, so $x$ is in every minimal prime, hence $x$ is nilpotent (this even shows $x^2 = 0$).

2) The associated primes of any ideal $I$ are the radicals of the primary ideals appearing in a minimal primary decomposition of $I$. Since $A$ is reduced, $0 = \sqrt{0} = p_1 \cap \ldots \cap p_n$, where $p_i$ are the minimal primes of $A$. But then every associated prime of $A$ is one of the $p_i$, hence minimal.

Solution 2:

Here's an answer that follows Vakil's treatment (an algebraic answer is nice but Vakil intends for this problem to be solved with the geometric definition, and so it's nice to have a solution of this form). Note that the answer given in the question is not a correct approach, it assumes we can take open and closed sets $D(g_i)$ corresponding to the irreducible components of $\operatorname{Spec}(A)$, which is not true in general (for example, if $\operatorname{Spec}(A)$ is connected but reducible).

I think we can in fact ignore part i) of the hint and just use part ii).

Firstly, since $\operatorname{Spec}(A)$ is Noetherian, it has finitely many irreducible components $X_i$. Certainly if $D(f)$ intersects $X_i$, then this intersection is an open subset of $X_i$, whence contains the generic point $\{x_i\}$. Thus $\operatorname{Supp}(f)$ is a closed subset containing the $x_i$ and so contains all of $X_i$, so one containment is proved*. For the other containment, suppose $\mathfrak{p}$ lies in no irreducible component meeting $D(f)$. Since there are only finitely many such components, their union is closed and $\mathfrak{p}$ lies in the complement, which is open. But then $\mathfrak{p}$ lies in an open set disjoint from $D(f)$, so $\mathfrak{p}$ lies in the complement of $\overline{D(f)} = \operatorname{Supp}(f)$.

This exercise puzzled me for a while because I can't work out where $i)$ came into it. I think you can use it in an alternate argument for the second containment, but I'm not sure exactly how. Perhaps I'm missing something obvious, or perhaps an entirely different proof. If anyone has thoughts on this I'd appreciate a comment about it.

*This part of the proof made no use of the fact that $X$ was reduced (or even Noetherian, since we didn't use the finiteness of the $X_i$ in any way) and the idea is good to have since it will be repeated in the next exercise.