On the infinitesimal lifting property of non-singular affine schemes

Question

Let $k$ be an algebraically closed field (not necessarily of characteristic $0$), $X$ a non-singular affine closed subscheme in $\mathbb{A}^n_k$ for some $n \ge 2$. Denote by $I_X$ the ideal of $X$ in $\mathbb{A}^n_k$. Let $X'$ be a first order infinitesimal deformation of $X$ i.e., $X'$ is flat over $\mathrm{Spec}(k[t]/(t^2))$ having $X$ as the closed fiber. Denote by $I_{X'}$ the ideal of $X'$ in $\mathbb{A}^n_k \times \mbox{Spec}(k[t]/(t^2))$. By the infinitesimal lifting property, we know that $X'$ is isomorphic to $X \times_k \mbox{Spec}(k[t]/(t^2))$. But this isomorphism is not canonical. The question is whether there is an isomorphism between $X \times_k \mbox{Spec}(k[t]/(t^2))$ and $X'$ such that the induced morphism of ring $$\frac{k[X_1,...,X_n]}{I_X} \otimes_k k[t]/(t^2) \to \frac{(k[t]/(t^2))[X_1,...,X_n]}{I_{X'}}$$ sending $X_i$ to $X_i$? Furthermore, can we get such an isomorphism for which $t$ maps to $t$ as well?

Answer 1

The answer is no. Easiest nontrivial case: $n=1$, $I_X=(X_1)$, so $X=Spec(k[X_1]/(X_1))=Spec(k)$, take $I_{X'} = (X_1 - t)$, so that $X' = Spec(k[X_1, t]/(X_1 - t, t^2))$. Now you are asking for an isomorphism $k[X_1, t]/(X_1, t^2)\to k[X_1, t]/(X_1 - t, t^2)$ sending $X_1$ to $X_1$, but this cannot exist as $X_1=0$ on the left while $X_1\neq 0$ on the right.