# General identification of Zariski tangent space with space of derivations

I'm reading through Schlessinger's Functors of Artin Rings to get background on Schlessinger's Theorem. The notion of a tangent space is critical in the argument, but I do not have enough intuition for it yet.

Let $$\Lambda$$ be a complete noetherian local ring with maximal ideal $$\mu$$ and residue field $$k$$. We work over the category $$\mathbf{C}$$ of artinian local $$\Lambda$$-algebras with residue field $$k$$, and the category $$\hat{\mathbf{C}}$$ of complete noetherian local $$\Lambda$$-algebras $$A$$ such that the quotients by a power of the maximal ideal $$A/\mathfrak{m}^n$$ all lie in $$\mathbf{C}$$.

Schlessinger defines the Zariski cotangent space of $$A \in \hat{\mathbf{C}}$$ as the $$k$$-vector space $$t_A^{*} = \mathfrak{m}/(\mathfrak{m}^2 + \mu A)$$ with the Zariski tangent space $$t_A$$ being the dual $$k$$-vector space to $$t_A$$. He then claims that "by a simple calculation" we may identify $$t_A$$ with the space $$\operatorname{Der}_{\Lambda}(A, k)$$ of $$\Lambda$$-linear derivations from $$A$$ to $$k$$ (treating $$k$$ as an $$A$$-module).

I understand how this identification works in more familiar cases. For example, if $$\Lambda = \mathbb{C}$$ and $$A = \mathbb{C}[[X_1, \dots, X_n]]$$, then $$t_A^{*}$$ may be realized as the $$n$$-dimensional space spanned by $$X_1, \dots, X_n$$, and an element $$\varphi$$ of the tangent space is determined by the image $$\alpha_i = \varphi(X_i)$$ of each of these variables in $$\mathbb{C}$$. The derivation corresponding to this element is given by $$a = a_0 + \sum_{i = 1}^n a_{1,i} X_i + \dots \mapsto \sum_{i = 1}^n \alpha_i a_{1, i}.$$

I'm having trouble with how this construction would generalize to arbitrary complete noetherian local $$\Lambda$$-algebras; is there a canonical way of finding the "$$t_A^{*}$$-component" of $$A$$ using the hypotheses on $$A \in \hat{\mathbf{C}}$$, analogous to taking the linear terms in the example above?

Here is an elementary argument. We want to find a perfect pairing of $$k$$-vector spaces $$\mathfrak{m}/(\mathfrak{m}^2 + \mu A) \times \mathrm{Der}_\Lambda(A,k) \to k$$ or equivalently, a $$\Lambda$$-linear pairing $$\mathfrak{m} \times \mathrm{Der}_\Lambda(A,k) \to k$$ with trivial right kernel and left kernel equal to $$\mathfrak{m}^2 + \mu A$$.

The pairing is evaluation of a derivation on an element of $$\mathfrak{m}$$: $$(x, \partial) \mapsto \partial(x) \in k.$$ Note first that by assumption, the natural map $$\Lambda \to A$$ induces an isomorphism $$\Lambda/\mu \to A/\mathfrak{m}$$. In particular, every element of $$A$$ can be written as $$r + x$$ where $$r$$ is in the image of $$\Lambda \to A$$ and $$x \in \mathfrak{m}$$.

Suppose now that $$\partial$$ is a derivation such that $$\partial(x) = 0$$ for all $$x \in \mathfrak{m}$$. Then $$\partial(r + x) = \partial(r) + \partial(x) = 0$$ since $$\partial$$ is $$\Lambda$$-linear and so $$\partial$$ is the $$0$$ derivation. This proves that the right kernel of the pairing is zero.

Next note that both $$\mathfrak{m}$$ and $$\mu$$ act by $$0$$ on $$k$$. Therefore, by Leibniz rule and $$\Lambda$$-linearity, $$\partial(\mathfrak{m}^2 + \mu A) = 0$$ for all derivations $$\partial$$. Thus the pairing factors through a pairing $$\mathfrak{m}/(\mathfrak{m}^2 + \mu A) \times \mathrm{Der}_\Lambda(A,k) \to k.$$

To conclude that this pairing is perfect, we need to show that for any nonzero $$x$$ in $$\mathfrak{m}/(\mathfrak{m}^2 + \mu A)$$, there exists a derivation $$\partial$$ with $$\partial(x) \neq 0$$. Let $$\bar{A} = A/(\mathfrak{m}^2 + \mu A)$$ and denote by $$\bar{\mathfrak{m}} = \mathfrak{m}/(\mathfrak{m}^2 + \mu A)$$. Then composition with the surjection $$A \to \bar{A}$$ induces a bijection $$\mathrm{Der}_\Lambda(\bar{A},k) \to \mathrm{Der}_\Lambda(A,k)$$.

Now $$\bar{A}$$ is a local Artinian $$k$$-algebra with maximal ideal $$\bar{\mathfrak{m}}$$ satisfying that $$\bar{\mathfrak{m}}^2 = 0$$. Then $$\mathrm{Der}_\Lambda(\bar{A},k) = \mathrm{Der}_k(\bar{A},k)$$ and we can identify our pairing with the same pairing for $$\bar{A}$$ as a $$k$$-algebra, namely $$\bar{\mathfrak{m}} \times \mathrm{Der}_k(\bar{A},k) \to k$$ and now this is the familiar setting where we can check the pairing is perfect by hand, e.g. by picking a splitting $$\bar{A} = k \oplus \bar{\mathfrak{m}}$$ as vector spaces.

I should note that $$\mathfrak{m}/(\mathfrak{m}^2 + \mu A)$$ is the relative Zariski cotangent space for the map of schemes $$\mathrm{Spec} A \to \mathrm{Spec} \Lambda$$ and the above result is true more generally for any morphism of schemes $$f : X \to S$$. This is proved using more sophisticated ideas in the Stacks Project.