I have some problems solving the following exercise from Liu's book Algebraic Geometry and Arithmetic Curves, exercise 3.15 from chapter 2.

Let $X$ be a quasi-compact scheme, $A=O_X(X)$. Let us consider the morphism $f:X\rightarrow Spec(A)$ induced by the identity on $A$. Show that $f(X)$ is dense in $Spec(A)$.

So I want to prove that for every distinguished open $D(g)$ of $Spec(A)$, the intersection $f(X)\cap D(g)\neq\emptyset$. Following my intuition, I would like to prove that the image of $X_g=\{x\in X\,|\,g_x\in O_{X,x}^*\}$ is in $D(g)$. (I have this idea, because $O_X(X_g)\simeq O_X(X)_g=A_g$, which is equal to $O_{Spec A}(D(g))$.)

I don't know how to really prove this, and I don't see where to use the quasi-compactness condition.

Thank you in advance!


The map $\mathscr{O}_X(X)_g\rightarrow\mathscr{O}_X(X_g)$ is injective for quasi-compact $X$, but it might not be surjective if $X$ is not also quasi-separated.

Anyway, let $X=\bigcup_{i=1}^nU_i$, $U_i=\mathrm{Spec}(A_i)$, be a finite affine open covering of $X$. The map $f:X\rightarrow\mathrm{Spec}(A)$ takes a point $x$ and maps it to the ideal obtained as the inverse image of $\mathfrak{m}_x\subseteq\mathscr{O}_{X,x}$ along $A=\mathscr{O}_X(X)\rightarrow\mathscr{O}_{X,x}$. So you want to show that for each non-nilpotent $g\in A$, there exists $x\in X$ such that $g\notin f(x)$, i.e., such that $g_x\notin\mathfrak{m}_x$. Alternatively, if $g_x\in\mathfrak{m}_x$ for all $x$, you want to prove that $g$ is nilpotent. This is where quasi-compactness comes in.

The assumption $g_x\in\mathfrak{m}_x$ for all $x$ implies that, if $g_i$ is the restriction of $g$ to $U_i$, then $D(g_i)=\mathrm{Spec}(A_i)$. So $g_i$ lies in the nilradical of $A_i$, i.e., $g_i$ is nilpotent. Therefore $g_i^{k_i}=0$ for some $k_i\geq 1$.

Let $k=\max_{1\leq i\leq n}k_i$. Then $g^k\vert_{U_i}=g_i^k=0$ for all $i$, so $g^k=0$. This means that $g$ is nilpotent.

In fact, this is very similar to the argument used to prove that $\mathscr{O}_X(X)_g\rightarrow\mathscr{O}_X(X_g)$ is injective when $X$ is quasi-compact. It occurs to me now that what you want also follows from this injectivity, as you surmise. If $g\in\mathfrak{m}_x$ for all $x$, then $X_g=\emptyset$, so $\mathscr{O}_X(X_g)=0$, and injectivity of the aforementioned map forces $\mathscr{O}_X(X)_g=0$, i.e., $g$ is nilpotent.

  • $\begingroup$ Maybe it is obvious, but I don't see why $f$ is the map you describe. ($f$ takes a point $x$ and maps it to the ideal obtained as the inverse image of $m_x⊆O_{X,x}$ along $A=O_X(X)→O_{X,x}$) $\endgroup$ – Alies Jan 5 '13 at 8:35
  • $\begingroup$ Dear @Alies, I wouldn't say it is obvious, but it follows from the compatibility of the stalk $f_x^\sharp:A_{f(x)}\rightarrow \mathscr{O}_{X,x}$ with the restriction map $A\rightarrow\mathscr{O}_X(X)$ (in this case the latter is the identity map) together with the fact that $f_x^\sharp$ is a local homomorphism. In fact for any morphism $f:X\rightarrow\mathrm{Spec}(B)$, $B$ any ring, $f(x)$ is equal to the inverse image of $\mathfrak{m}_x\subseteq\mathscr{O}_{X,x}$ under the induced ring map $B\rightarrow\mathscr{O}_X(X)\rightarrow\mathscr{O}_{X,x}$. $\endgroup$ – Keenan Kidwell Jan 5 '13 at 9:35
  • $\begingroup$ It's easier to understand this with a commutative diagram, but I don't know how to make those on MSE. But the relevant diagram can be found in the proof of Lemma 3.23 of Liu's book (replacing $B$ and its local ring there by $\mathscr{O}_X(X)$ and $\mathscr{O}_{X,x}$). This result is stated clearly and proved in the Stacks Project as Lemma 21.6.1 (permanent tag 01HY). Also see my answer to this question: math.stackexchange.com/questions/56854/… $\endgroup$ – Keenan Kidwell Jan 5 '13 at 9:39
  • $\begingroup$ Here's a link to the Stacks Project result, which is probably the best place to look for this particular fact: stacks.math.columbia.edu/tag/01HY $\endgroup$ – Keenan Kidwell Jan 5 '13 at 9:41
  • $\begingroup$ thank you very much, this helped me a lot! $\endgroup$ – Alies Jan 5 '13 at 11:29

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.