1

If $f:A\to B$ is a homomorphism of rings such that $f':\text{spec} B \to \text{spec} A$ is a homeomorphism does it follows that the spectra are isomorphic as schemes?

I was able to reduce this problem to trying to show that given a prime $\mathfrak{q}$ in $B$ and setting $\mathfrak{p} = f^{-1} \mathfrak{q}$, that the map $A_{\mathfrak{p}} \to B_{\mathfrak{q}}$ is an isomorphism of rings. But I do not see why this has to be true, even in the special case when $B = A$ modulo its nilradical.

Nicolas Bourbaki
  • 385
  • In fact, a simple example to show that this is not true is to look at what happens when you mod out the nilradical! – Mariano Suárez-Álvarez Jul 12 '15 at 04:45
  • 1
    Your example at the end is a counterexample: if $X$ is a scheme, then $X_\mathrm{red}$ is a homeomorphic scheme that is not isomorphic (I guess Mariano beat me to this). See Hartshorne, Exercise II.2.3(b). Moreover, Exercise II.2.18(d) says that if we do have a homeomorphism, all we get is a surjection on local rings. – Takumi Murayama Jul 12 '15 at 04:47
  • @MarianoSuárez-Alvarez Can you explain more? (By the way, I am trying to prove that the reduced sheaf of a scheme is a scheme, and I thought this approach might work). – Nicolas Bourbaki Jul 12 '15 at 04:47
  • What exactly you want details on? That the map induced by reduction modulo the nilradicalis a homeo follows more or less by using the definition of the topologies induced and the fact that all prime ideals contain the nilradical. That the map is not an iso of schemes is simply because one is reduced and the other isn't. – Mariano Suárez-Álvarez Jul 12 '15 at 05:13
  • @MarianoSuárez-Alvarez Please delete your comment, I will delete my previous two afterwards. I was making a silly mistake, I kept on trying to prove that spec(A) is iso to spec(A_red) (which is false), while I had to prove that spec(A)_red is iso to spec(A_red). – Nicolas Bourbaki Jul 12 '15 at 05:24
  • I won't, as what I wrote answers quite precisely what you actually asked! – Mariano Suárez-Álvarez Jul 12 '15 at 05:29

1 Answers1

5

I might be mistaken, but can't you take the inclusion map $\mathbb{Q}\to\mathbb{R}$? Then the corresponding map $\text{Spec}(\mathbb{R})\to\text{Spec}(\mathbb{Q})$ of spectrums will be clearly homeomorphism (because both spectrums are single points!), but the spectra cannot be isomorphic as schemes, because the ring of global sections are $\mathbb{R}$ and $\mathbb{Q}$, respectively, which are non-isomorphic rings.

Prism
  • 11,162