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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?

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    $\begingroup$ 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. $\endgroup$ Aug 18 '19 at 15:35
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    $\begingroup$ 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. $\endgroup$ Aug 19 '19 at 10:21
  • $\begingroup$ @JasonStarr and Rijul Saini Thanks for the answer. $\endgroup$
    – Ron
    Aug 19 '19 at 18:01
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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).

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    $\begingroup$ Isn't the special fiber of that map given by $\operatorname{Spec} k[x]/(x^2) \to \operatorname{Spec} k$? $\endgroup$
    – Marc Paul
    Aug 18 '19 at 12:04

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