# Identity on topological space but not on scheme

I have this question just out of curiosity.

If X is a scheme, then a morphism $f: X \rightarrow X$ can be the identity on the underlying topological space of X, but not the identity on the structure sheaf. For example, f can be the Frobenius morphism.

Does someone know an example of such a morphism which is not a Frobenius? One can simply think in terms of affine schemes and hence ring: does someone know an explicit example of a ring endomorphism (sending 1 to 1) $\phi: X \rightarrow X$ such that $\phi^{-1}(I) = I$ for all prime ideals $I$ ? Is there a nice class of rings for which such an endomorphism is forced to be Frobenius?

• There is the composition $k[\epsilon]/\epsilon^2\to k\to k[\epsilon]/\epsilon^2$, where the first arrow is the morphism of $k$-algebras sending $\epsilon$ to $0$ et the second one is the morphism making $k[\epsilon]/\epsilon^2$ into a $k$-algebra. – Damian Rössler May 16 '13 at 9:17
• ($k$ is a field in my last comment). – Damian Rössler May 16 '13 at 9:18
• Notice that an automorphism of a field will also do the trick. – Damian Rössler May 16 '13 at 11:19
• Of course! thanks for the interesting examplers. – Nadim Rustom May 16 '13 at 15:23
• Just so this question doesn´t hang around forever, maybe someone can post as an answer. Nadim you can even do this, just copying and pasting the comments (with attribution) and accepting your own answer. The tradition is to do this self-answer as community wiki so you don´t get reputation for someone else´s work. – David White May 16 '13 at 18:52

There is the composition $k[ϵ]/ϵ^2→k→k[ϵ]/ϵ^2$, where the first arrow is the morphism of k-algebras sending $ϵ$ to 0 et the second one is the morphism making $k[ϵ]/ϵ^2$ into a $k$-algebra. – Damian Rössler
This answers a slightly different question, but if $X$ is the curve $x^2 = y^3$ in $\mathbb C^2$, then its normalization is $\mathbb C$. The normalization map $\mathbb C \to X$ is a homeomorphism of topological spaces, but is not an isomorphism of schemes. (If $A$ is the subalgebra of $\mathbb C[t]$ generated by $t^2$ and $t^3$, then $A = \mathcal O(X)$, and the normalization map $\mathbb C \to X$ is dual to the inclusion $A \to \mathbb C[t]$.)