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?
Best Answer
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.