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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!

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1 Answer 1

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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.

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    $\begingroup$ 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). $\endgroup$
    – Sardines
    Apr 14 at 23:32
  • $\begingroup$ @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. $\endgroup$
    – CJ Dowd
    Apr 14 at 23:50

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