3
$\begingroup$

Let $A$ be a commutative Noetherian ring and $B$ a finitely generated $A$-algebra. Then the set $$U\colon=\{P\in\operatorname{Spec}B\mid B_P\ \mathrm{is\ flat\ over}\ A\}$$ is open in $\operatorname{Spec}B$. (See, e.g., page 187 of Matsumara's Commutative Ring Theory.)

Is there a known example of a non finitely generated $A$-algebra $B$, where the set $U$ as defined above is not open?

$\endgroup$
  • 3
    $\begingroup$ Let $A = \mathbf{Z}$, $B = \prod_p \mathbf{F}_p$. Assume the non-flat locus $Y$ in Spec($B$) is closed. It contains the evident clopen points Spec($\mathbf{F}_p$), so if $J$ is an ideal in $B$ cutting out $Y$ then $J$ has vanishing image in each direct factor ring $\mathbf{F}_p$, so $J=0$. Thus, $B$ would be nowhere flat over $A$, so $B \otimes_A \mathbf{Q}$ would vanish and hence $B[1/N]=0$ for some $N > 0$. But this is false. $\endgroup$ – user76758 Feb 2 '14 at 4:42
  • $\begingroup$ @user76758: How do you go from the assertion that $B\otimes_A \mathbb{Q}$ vanishes to the assertion that there exists an integer $N$ that annihilates $B$? Of course you can directly show that $B\otimes_A \mathbb{Q}$ is nonzero using ultrafilters, etc. But I do not see how to directly conclude that $B$ is annihilated by a single integer $N$. $\endgroup$ – Jason Starr Feb 2 '14 at 5:28
  • $\begingroup$ @user76758: Here is a direct way of getting a contradiction from vanishing of $B\otimes_A \mathbb{Q}$, rather than using ultrafilters or the vanishing of $B[1/N]$ (which I still do not immediately see). The natural ring homomorphism $\mathbb{Z} \to B$ is clearly injective. Thus, by flatness of $\mathbb{Q}$ over $\mathbb{Z}$, also the induced homomorphism $\mathbb{Q} \to B\otimes_A \mathbb{Q}$ is injective. Therefore $B\otimes_A \mathbb{Q}$ is nonvanishing. $\endgroup$ – Jason Starr Feb 2 '14 at 5:35
  • 2
    $\begingroup$ @JasonStarr: The constant sequence $(1)_p$ is obviously non-torsion. $\endgroup$ – ACL Feb 2 '14 at 6:06
  • 1
    $\begingroup$ @JasonStarr: I was looking at the distinguished element 1 in $B$ (which is what makes $B$ more tractable than a random $A$-module). $\endgroup$ – user76758 Feb 2 '14 at 7:21
4
$\begingroup$

Here is one example. Let $S$ denote the set of positive prime integers. Let $A$ be $\mathbb{Z}$. Let $R$ be the countably generated polynomial ring over $\mathbb{Z}$, $$R = \mathbb{Z}[\{x_p:p\in S\}].$$ Let $I\subset R$ be the ideal generated by $\{px_p : p\in S\}$. Let $B$ be $R/I$. Then the ideal $\mathfrak{p}$ of $B$ generated by $\{x_p : p\in S\}$ is prime, since $B/\mathfrak{p}$ is just the integral domain $\mathbb{Z}$. For every integer $p\in S$, $p$ is not in $\mathfrak{p}$. Hence, the localization of $B$ at $\mathfrak{p}$ factors through $B\otimes_{\mathbb{Z}}\mathbb{Q}$. But, of course, this localization is already $(B/\mathfrak{p})\otimes_{\mathbb{Z}}\mathbb{Q}$, which is $\mathbb{Q}$. Since $\mathbb{Q}$ is flat over $\mathbb{Z}$, thus $\mathfrak{p}$ is in $U$.

Every Zariski open subset of $\text{Spec}(B)$ containing $\mathfrak{p}$ contains a basic open subset of the form, $$D(b)=\{\mathfrak{q} \in \text{Spec}(B) : b\not\in \mathfrak{q}\},$$ for some $b\in B\setminus \mathfrak{p}$. Since $b$ is not in $\mathfrak{p}$, $b$ equals $n+c$ for some $c\in \mathfrak{p}$ and for some nonzero $n\in \mathbb{Z}$. Of course $c$ is a polynomial in only finitely many of the variables $x_p$. Also $n$ is divisible by only finitely many primes. Hence there exists a prime $q$ such that $c$ does not involve $x_q$, and, also, $q$ does not divide $n$.

Consider the unique surjective $\mathbb{Z}$-algebra homomorphism, $$ u_q : R \to (\mathbb{Z}/q\mathbb{Z})[x_q],\ \ u_q(x_q) = x_q, \ \ u_q(x_p) = 0,\ p\neq q. $$ Clearly $\text{Ker}(u_q)$ contains $px_p$ for every $p\neq q$, since $u_q(x_p)$ equals $0$. But since $q$ equals $0$ in $\mathbb{Z}/q\mathbb{Z}$, also $\text{Ker}(u_q)$ contains $qx_q$. Hence $u_q$ factors through a unique surjective ring homomorphism, $$ \overline{u}_q: B \to (\mathbb{Z}/q\mathbb{Z})[x_q].$$ Since $(\mathbb{Z}/q\mathbb{Z})[x_q]$ is an integral domain, the ideal $\mathfrak{q}:=\text{Ker}(\overline{u}_q)$ is a prime ideal of $B$. Also, since $\overline{u}_q(c)$ equals $0$, $\overline{u}_q(b)$ equals $\overline{u}_q(n)$. Since $q$ does not divide $n$, $\overline{u}_q(b)$ is nonzero. Thus $\mathfrak{q}$ is in $D(b)$.

Of course for every prime $p\neq q$, since $\overline{u}_q(p)$ is nonzero, also $p$ is not in $\mathfrak{q}$. Thus the localization $B\to B_{\mathfrak{q}}$ factors through $B\otimes_{\mathbb{Z}} \mathbb{Z}_{(q)}$, which is clearly just $\mathbb{Z}_{(q)}[x_q]/\langle qx_q \rangle$. Of course the image of $\mathfrak{q}$ in this localization is the principal ideal $\langle q \rangle$. In particular, $x_q$ is not in this prime ideal. Hence, the localization inverts $x_q$, and thus annihilates $q$. So $B_{\mathfrak{q}}$ is simply the field $(\mathbb{Z}/q\mathbb{Z})(x_q)$ of rational functions in the variable $x_q$ over the field $\mathbb{Z}/q\mathbb{Z}$.

The field $(\mathbb{Z}/q\mathbb{Z})(x_q)$ is not flat over $\mathbb{Z}$, since $q$ is a zerodivisor. Thus $\mathfrak{q}$ is not in $U$. Therefore $D(b)$ is not contained in $U$. Since this holds for every $b\in B\setminus \mathfrak{p}$, $U$ contains no Zariski open neighborhood of $\mathfrak{p}$, even though $U$ contains $\mathfrak{p}$. Therefore $U$ is not a Zariski open subset of $\text{Spec}(B)$.

Edit. In fact, it is not hard to see that $U$ is precisely $\{\mathfrak{p}\}$ for this ring.

Second Edit. I realize now that the ring $B$ above is "almost" a subring of the ring proposed by user76758 in the comments (I did not see that proposal until after I posted). Let $J$ be the ideal in $R$ generated by $px_p$ and $x_p^2-x_p$ for every $p$ in $S$. Then $C=R/J$ is still a counterexample, for essentially the same reason as above. Also $C$ is isomorphic to the $\mathbb{Z}$-subalgebra of $\prod_p \mathbb{F}_p$ generated by every element $\overline{x_p}$ that has coordinate $1$ in the $p$-factor and that has $0$ in every other factor.

$\endgroup$
  • $\begingroup$ Dear Jason: Thank you for this detailed and interesting example. The ring $B$ has infinitely many minimal prime ideals (the $\mathfrak{p}$ and all the ideals $\mathfrak{q}$). I tried to see if what is happening in your example has anything to do with having infinitely many minimal prime ideals, but I couldn't find a connection. $\endgroup$ – Mahdi Majidi-Zolbanin Feb 2 '14 at 14:57
  • $\begingroup$ What I mean is, I tried to re-state your argument using the fact that $B$ has infinitely many minimal prime ideals, but I didn't succeed. $\endgroup$ – Mahdi Majidi-Zolbanin Feb 2 '14 at 15:42
4
$\begingroup$

Here's a slightly more "geometric" example.

Take a smooth affine $3$-fold $X = \mathrm{Spec}(R)$ over the complex numbers, and let $X^+ \to X$ be an absolute integral closure of $X$ in the sense of Artin, i.e., $X^+$ is the normalisation of $X$ in an algebraic closure of its function field. We may view $X^+$ as a directed inverse limit of finite (normal) covers $Y \to X$ along finite surjective transition maps. Let $U$ be the flat locus of $\pi:X^+ \to X$.

Claim: $U = \pi^{-1}(X - X^0)$, where $X^0 \subset X$ is the set of closed points.

Proof: To show $\supset$, note that a finite normal extension of a regular local ring of dimension $\leq 2$ is automatically flat. To show $\subset$, one must check that $\pi$ is not flat at any closed point of $X$. This can be shown using local cohomology (using crucially the characteristic $0$ assumption).

Finally, it remains to observe that $U$ is not open: if it were open, it would arise as the inverse image of an open $V \subset Y$ for some finite cover $Y \to X$ occurring in the inverse limit defining $X^+$. However, any such $V$ must contain a closed point of $Y$. As $X^+ \to Y$ is surjective, $U$ contains a point mapping to a closed point of $X$ (through $Y$), which cannot happen by the claim above.

$\endgroup$
  • $\begingroup$ This is very nice, though I wonder: how does local cohomology tell us about flatness properties for the absolute integral closure? $\endgroup$ – user76758 Feb 2 '14 at 20:49
  • $\begingroup$ If $R \to S$ is a flat map, and $I \subset R$ is an ideal, then flat base change gives $H^i_I(R) \otimes S \simeq H^i_{IS}(S)$. In the above example, this means that $X^+$ should have trivial local cohomology in non-top degrees if $\pi$ was flat. On the other hand, it's easy to show that some large enough finite cover $Y \to X$ has non-trivial local cohomology in degree $2$, and this persists on passage to $X^+$ (due to characteristic $0$ + trace maps). $\endgroup$ – anonymous Feb 2 '14 at 20:57
  • $\begingroup$ @anonymous: Nice!! $\endgroup$ – user76758 Feb 4 '14 at 3:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.