# Isomorphism in fibers and flatness

Let $$X$$, $$Y$$ be (reduced) affine varieties and $$f:X \to Y$$ is a finite morphism which is an isomorphism over an open dense subset (for example a normalization map). Let $$A$$ be a local noetherian ring and $$f_A:X_A \to Y_A$$ be a finite morphism which coincides with $$f$$ over the special fiber and $$X_A, Y_A$$ are both $$A$$-flat with special fiber isomorphic to $$X$$ and $$Y$$, respectively (in other words, $$f_A$$ is a deformation of the morphism $$f$$). Does there exist a non-empty (or dense) open subset $$U_1$$ of $$X_A$$ and $$U_2$$ of $$Y_A$$, such that $$f_A$$ maps $$U_1$$ isomorphically to $$U_2$$? If not, is it true if $$A$$ is artinian?

• Look at Hartshorne's Deformation Theory Page 39. Roughly, it is not true for all A, and true for artin local if true over dual numbers. Aug 18 '19 at 15:35
• Define $U_2$ to be the open complement of the closed support of the cokernel of the map $f_A^\#:\mathcal{O}_{Y_A} \to \mathcal{O}_{X_A}$. If $A$ is an Artin local ring, this open is dense in $Y_A$. If $A$ is not an Artin local ring, it can happen that $U_2$ is not dense in $Y_A$, just as stated by @RijulSaini. Aug 19 '19 at 10:21
• @JasonStarr and Rijul Saini Thanks for the answer.
– Ron
Aug 19 '19 at 18:01

This should be false without additional assumptions. Take a field $$k$$. Let $$A=k[[x^2]]$$, $$Y_A=\mathrm{Spec}\:A$$, $$X_A=\mathrm{Spec}\:k[[x]]$$. It is not hard to see that $$X_A$$ is flat over $$A$$. The map over the closed point is the identity $$\mathrm{Spec}\:k\rightarrow \mathrm{Spec}\:k$$. The only proper non-empty open subset of $$Y_A$$ is the set consisting of the generic point, and the map is not an isomorphism over it (inducing a field extension of degree 2).
• Isn't the special fiber of that map given by $\operatorname{Spec} k[x]/(x^2) \to \operatorname{Spec} k$? Aug 18 '19 at 12:04