Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Does anyone know any example that invalidates the following affirmation:

If a morphism $f:A\to A$ induces the identity $\hat f:\operatorname{Spec} \left( A \right) \to \operatorname{Spec} \left( A \right)$ then $f = \operatorname{id} _A $.

share|cite|improve this question

closed as off-topic by Andrés E. Caicedo, Amzoti, TMM, amWhy, Danny Cheuk Jul 15 '13 at 1:35

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." – Andrés E. Caicedo, Amzoti, TMM, Community
If this question can be reworded to fit the rules in the help center, please edit the question.

Take $A$ to be a field. This is not research-level. – Qiaochu Yuan Jul 14 '13 at 17:21
No, nobody knows such an example that would invalidate a true assertion! – Georges Elencwajg Jul 14 '13 at 17:25
The question is ambiguous. Do you mean to assume that $\hat{f}$ is the identity on the topological space $Spec(A)$ or on the scheme $Spec(A)$? If the former, your "affirmation" is false; if the latter, it's true. – WillO Jul 14 '13 at 18:26
To sum up the answers: If $f,g : A \to B$ are homomorphisms of commutative rings, such that $\tilde{f} =\tilde{g}$ as morphisms of schemes, then $f=g$. But it does not suffice to assume that $\tilde{f}=\tilde{g}$ as continuous maps. In other words, it depends on which category you are working with. – Martin Brandenburg Jul 14 '13 at 18:59

Let us recall that the category of affine schemes is equivalent to the category of commutative rings (with unit). In particular we have a bijection of sets

$$\mbox{Hom}_{\mathfrak{Rings}}(A,B) \leftrightarrow \mbox{Hom}_{\mathfrak{ScH}}(\mbox{Spec}(B),\mbox{Spec}(A)).$$

defined as follows. For any $\varphi : \mbox{Spec}(B) \to \mbox{Spec}(A)$ we have an associated map on sheaves $\varphi^{\sharp} : \mathcal{O}_{\mbox{Spec}(A)} \to \varphi_\ast \mathcal{O}_{\mbox{Spec}(B)}$. Taking global sections we obtain a ring homomorphism $A \to B$. Now consider the case $A = B$. Because the the identity map $f : A \to A$ induces the identity map on $\mbox{Spec}(A)$, then whatever map that induces the identity on the spectra has to be the identity (by injectivity of the correspondence above).

Edit: I should clarify that in my answer above I am treating $\operatorname{Spec} A$ as an affine scheme with its structure sheaf and not just as a topological space. In the latter case my answer will not be true anymore; take $A = k(x)$ that has automorphism group $\mbox{PGL}(2)$.

share|cite|improve this answer

The question is ambiguous. Do you mean to assume that $\hat{f}$ is the identity on the topological space $\operatorname{Spec}(A)$ or on the scheme (or ringed space) $\operatorname{Spec}(A)$? If the former, then any non-identity automorphism of a field provides a counterexample. If the latter, BenjaLim's answer applies.

share|cite|improve this answer
YACP: Fair enough. Editing to add the adjective "non-identity". – WillO Jul 14 '13 at 19:30

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