For reference I'm trying to understand section 12.2.16 of Vakil's FOAG. Let $X$ be a finite type $k$-scheme and $p$ a closed point with residue field $k'/k$, a finite separable extension of degree $d$. Let $\pi: X_{\bar{k}} \to X$ be the projection morphism. It's easy to see that $k' \otimes_k \bar{k} \cong \bar{k}^d$ as rings, so the fibre over $p$ is $\pi^{-1}(p) = \sqcup_d \mathrm{Spec}(\bar{k})$. Call these $d$ pre-images $p_1, \dots, p_d$. For the sake of this question, since regularity can be checked pointwise, we may assume $X = \mathrm{Spec} A$ and $p$ corresponds to a maximal ideal $\mathfrak{m}$ of $A$. One can show that the fibre is the closed subscheme defined by $\mathrm{Spec}((A \otimes_k \bar{k})/(\mathfrak{m} \otimes_k \bar{k}))$. Vakil claims we have the equality
$$
\sum_{i=1}^{d} \dim_{\bar{k}}(T_{X_{\bar{k}}, p_i}) = d \cdot \dim_{k'}(T_{X, p}).
$$
I can't figure out how to prove it. Write $r := \dim_{k'}(T_{X, p})$. He suggests using the fact that $$(\mathfrak{m} \otimes_k \bar{k})/(\mathfrak{m}^2 \otimes_k \bar{k}) \cong \mathfrak{m}/\mathfrak{m}^2 \otimes_k \bar{k} = \bar{k}^{dr},$$ which means I only have to show that the dimension of $(\mathfrak{m} \otimes_k \bar{k})/(\mathfrak{m}^2 \otimes_k \bar{k})$ as a $\bar{k}$-module is the same as the sum of the dimensions of the tangent spaces of the preimages of $p$. Does anyone have any hints? I would like to understand intuitively why this equality of dimensions should be true. Once I figure this out, I can conclude that $X$ is regular at $p \iff X_{\bar{k}}$ is regular at $p_1, \dots, p_d$!
Regularity of finite type $k$-schemes by base changing to $\bar{k}$ and an equality of dimensions of tangent spaces
algebraic-geometry
Best Answer
The key step in proving it the way Vakil is after is understanding why $(\mathfrak{m}\otimes_{k'}\overline{k})/(\mathfrak{m}\otimes_{k'}\overline{k})^2$ decomposes as the direct sum of the tangent spaces at each of the $p_i$. First, we note the fact that the $p_i$ are exactly the preimages of $p$ (combined with the fact that $k'$ is separable) implies that $(\mathfrak{m}\otimes_{k'}\overline{k})=\mathfrak{m}_1\mathfrak{m}_2\cdots\mathfrak{m}_d$ where the ideals on the right hand side are the ideals of the points $p_1,p_2,\cdots,p_d$.
Next, since for commutative rings and comaximal ideals $I_1,\cdots,I_n$ we have $I_1I_2\cdots I_n=I_1\cap I_2\cap\cdots\cap I_n$, we may apply the Chinese remainder theorem to get that $A'/(\mathfrak{m}_1\mathfrak{m}_2\cdots\mathfrak{m}_d) \cong \prod A'/\mathfrak{m}_i$. So $$(\mathfrak{m}\otimes_{k'}\overline{k})/(\mathfrak{m}\otimes_{k'}\overline{k})^2= (\mathfrak{m}\otimes_{k'}\overline{k}) \otimes_{A'} A'/(\mathfrak{m}\otimes_{k'}\overline{k})$$ and we get that the RHS is isomorphic to $(\mathfrak{m}_1\cdots\mathfrak{m}_d)\otimes_{A'} \prod A/\mathfrak{m}_i $.
Since tensor product and product commute for finitely presented modules, it remains to see what happens to each $(\mathfrak{m}_1\cdots\mathfrak{m}_d)\otimes_{A'} A/\mathfrak{m}_i $. But this is isomorphic to $\mathfrak{m}_i/\mathfrak{m}_i^2$ via the map sending $x_1x_2\cdots x_d\mapsto x_1x_2\cdots x_d\mod \mathfrak{m}_i^2$, and since finite products agree with direct sums in module categories, we get the desired isomorphism between $(\mathfrak{m}\otimes_{k'}\overline{k})/(\mathfrak{m}\otimes_{k'}\overline{k})^2$ and $\bigoplus \mathfrak{m}_i/\mathfrak{m}_i^2$.