# Zariski tangent spaces over residue fields

I came up with the following question while looking at the definition of Zariski tangent spaces.

Let $$R$$ be a noetherian ring, and let $$\mathfrak{m}\subset R$$ be a maximal ideal. Write $$k:=R/\mathfrak{m}$$ as the residue field. Denote by $$\tilde{\mathfrak{m}}$$ the unique maximal ideal in the local ring $$R_\mathfrak{m}$$. Then it is well known that $$R_\mathfrak{m}/\tilde{\mathfrak{m}}$$ is canonically isomorphic to $$k$$. Hence, both $$\mathfrak{m}/\mathfrak{m}^2$$ and $$\tilde{\mathfrak{m}}/\tilde{\mathfrak{m}}^2$$ have canonical structures as $$k$$-vector spaces.

Question: it is true that $$\dim_k(\mathfrak{m}/\mathfrak{m}^2)=\dim_k(\tilde{\mathfrak{m}}/\tilde{\mathfrak{m}}^2)$$ ? If so, are they canonically isomorphic as $$k$$-vector spaces? If not, is there a way to add some extra hypotheses to make this true? Say, assume $$k$$ is algebraically closed, or assume $$R$$ is a finitely generated $$k$$-algebra, etc.

The reason why I ask this is the following: in my algebraic geometry class we proved that if $$X$$ is an affine variety over $$k$$, then the $$k$$-vector space of derivations at a point $$p$$ is isomorphic to $$\mathfrak{m}_p/\mathfrak{m}_p^2$$, where $$\mathfrak{m}_p\subset \mathscr{O}(X)=k[X]$$ is the maximal ideal of global regular functions vanishing at $$p$$. It is not the maximal ideal of local germs vanishing at $$p$$. This is different from the usual definition of Zariski tangent spaces, and I would like to reconcile these two definitions.

Any help would be greatly appreciated. Thanks!

Yes, these two cotangent spaces are naturally isomorphic as $$k$$-vector spaces. Think of it this way: $$\mathfrak{m}$$ is naturally an $$R$$-module; to avoid confusion, let's just call it $$M$$. We can perform two operations to $$M$$: quotienting by $$\mathfrak{m}$$ (inducing an $$R/\mathfrak{m} = k$$-module structure) or localizing at $$\mathfrak{m}$$ to get an $$R_{\mathfrak{m}}$$-module. The claim is that these operations commute in the sense that $$(M/\mathfrak{m}M)_{\mathfrak{m}} \simeq (M_{\mathfrak{m}})/\tilde{m}M_{\mathfrak{m}}$$ naturally. This is exactness of localization, which is a standard commutative algebra result; for example, see Atiyah-MacDonald Proposition 3.3. (Actually, you only need right-exactness for this, which follows from the weaker statement of right-exactness of the tensor product.) Then finally $$(M/\mathfrak{m}M)_{\mathfrak{m}} \simeq M/\mathfrak{m}M$$ as $$k$$-modules because localizing by $$\mathfrak{m}$$ does nothing, since $$k$$ is already a field.

All of this holds very generally. You do not even need noetherian hypotheses on $$R$$. In the case above, it's a coincidence that the ideal $$\mathfrak{m}$$ we are quotienting by is the same ideal that we are localizing by; this isn't required in general.

• Is it correct to summarize your argument as follows: if $M$ is an $R$-module with submodule $N$, and if $S$ is a multiplicative system, then $S^{-1}(M/N)\cong (S^{-1}M)/(S^{-1}N)$ as $S^{-1}R$-modules. Now take $N=\mathfrak{a}M$ for some ideal $\mathfrak{a}$, then $S^{-1}(M/\mathfrak{a}M)\cong (S^{-1}M)/(S^{-1}\mathfrak{a}M)$ as $S^{-1}R$-modules, hence also as $S^{-1}(R/\mathfrak{a})=(S^{-1}R)/(S^{-1}\mathfrak{a})$-modules (this last step has nothing to do with localization; it's just changing the underlying rings). Apr 14 at 23:32
• @Sardines Yes, that's right. Unfortunately, you might find that these identifications are swept under the rug in literature beyond a first course in commutative algebra, so it's good to keep all of this in mind. Apr 14 at 23:50