# 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. May 16, 2013 at 9:17
• ($k$ is a field in my last comment). May 16, 2013 at 9:18
• Notice that an automorphism of a field will also do the trick. May 16, 2013 at 11:19
• Of course! thanks for the interesting examplers. May 16, 2013 at 15:23
• In fact any nonzero morphismen between two fields will do. Jun 7, 2013 at 11:09

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 $A \to \mathbb C[t]$.)