Lemma 93.8.3. In Example 93.8.1 let $P$ be a $k$-algebra. Then
\[ T\mathcal{D}\! \mathit{ef}_ P = \text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P) \quad \text{and}\quad \text{Inf}(\mathcal{D}\! \mathit{ef}_ P) = \text{Der}_ k(P, P) \]
Proof. Recall that $\text{Inf}(\mathcal{D}\! \mathit{ef}_ P)$ is the set of automorphisms of the trivial deformation $P[\epsilon ] = P \otimes _ k k[\epsilon ]$ of $P$ to $k[\epsilon ]$ equal to the identity modulo $\epsilon $. By Deformation Theory, Lemma 91.2.1 this is equal to $\mathop{\mathrm{Hom}}\nolimits _ P(\Omega _{P/k}, P)$ which in turn is equal to $\text{Der}_ k(P, P)$ by Algebra, Lemma 10.131.3.
Recall that $T\mathcal{D}\! \mathit{ef}_ P$ is the set of isomorphism classes of flat deformations $Q$ of $P$ to $k[\epsilon ]$, more precisely, the set of isomorphism classes of $\mathcal{D}\! \mathit{ef}_ P(k[\epsilon ])$. Recall that a $k[\epsilon ]$-algebra $Q$ with $Q/\epsilon Q = P$ is flat over $k[\epsilon ]$ if and only if
\[ 0 \to P \xrightarrow {\epsilon } Q \to P \to 0 \]
is exact. This is proven in More on Morphisms, Lemma 37.10.1 and more generally in Deformation Theory, Lemma 91.5.2. Thus we may apply Deformation Theory, Lemma 91.2.2 to see that the set of isomorphism classes of such deformations is equal to $\text{Ext}^1_ P(\mathop{N\! L}\nolimits _{P/k}, P)$. $\square$
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