MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $R \rightrightarrows U \to X$ be a presentation of an algebraic space by schemes. Does this induce an exact sequence $|R| \rightrightarrows |U| \to |X|$ on underlying points?

The reason I ask is that this is stated as a lemma in the Stack Project (Lemma 33.4.4). However, the proof uses:

"Since $j = (s,t):R \to U \times U$ is a monomorphism we see that $|R| \to |U|\times|U|$ is injective."

But this can't be true in general. For instance $|U \times U| \to |U|\times|U|$ is not injective when $U$ is the affine line since all generic points in $|U\times U|$ corresponding to dimension 1 curves map to the same pair of generic points. Of course, this equivalence relation is not ├ętale. The question is: Does ├ętaleness prevent such collapses to happen? In that case, how?

Edit: It is pretty clear that $|R| \rightrightarrows |U|$ becomes a pre-equivalence relation with $|U| \to |X|$ as quotient. In fact, I think this is the only property used later in the text, which makes the error unproblematic.

Then the only question is: Is $|R| \to |U| \times_{|X|}|U|$ injective?

share|cite|improve this question
It's all ok. Idea: for schemes, $|X' \times_X X"| \rightarrow |X'| \times_ {|X|} |X"|$ is surjective (because $k' \otimes_k k" \ne 0$ for fields), which is why it "works" for etale cover $U$ of a scheme $X$. For general alg. space $X$ define $|X|$ via equiv. rel. on field-valued pts $x$, $x'$ that the scheme $x \times_X x'$ is non-empty (recovers D. Knutson's defn in qcqs case via his "atoms"); of course, have to check this is an equiv. rel. You ask if $|R| = |U| \times_ {|X|} |U|$. Crux is $R = U \times_ X U$, so $R$ contains non-empty $u \times_ X u'$. QED – BCnrd Apr 18 '10 at 16:15
Daniel, just to be clear, I'm saying the assertion of the lemma is ok, not that the proof given is ok. (As you point out, the argument is wrong. You should email deJong to point out the need to revise the argument.) – BCnrd Apr 18 '10 at 16:21
Sorry, but I don't think I follow your reasoning at all. I agree on your definition of a point, but I don't see how $|U\times_X U| \ to $|U|\times_{|X|}|U|$ becomes injective. I edit my question to clarify. – Daniel Bergh Apr 20 '10 at 13:12
No, $|R| \to |U| \times_{|X|}|U|$ is not injective. Take $X = \mathop{\rm Spec} \mathbb R$, $U = \mathop{\rm Spec} \mathbb C$. – Angelo Apr 20 '10 at 14:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.