Lemma 90.4.5. If $A \to B$ is a ring map, then $H^0(L_{B/A}) = \Omega _{B/A}$.

Proof. We will prove this by a direct calculation. We will use the identification of Lemma 90.4.3. There is clearly a map from $\Omega _{\mathcal{O}/A} \otimes \underline{B}$ to the constant sheaf with value $\Omega _{B/A}$. Thus this map induces a map

$H^0(L_{B/A}) = H^0(L\pi _!(\Omega _{\mathcal{O}/A} \otimes \underline{B})) = \pi _!(\Omega _{\mathcal{O}/A} \otimes \underline{B}) \to \Omega _{B/A}$

By choosing an object $P \to B$ of $\mathcal{C}_{B/A}$ with $P \to B$ surjective we see that this map is surjective (by Algebra, Lemma 10.131.6). To show that it is injective, suppose that $P \to B$ is an object of $\mathcal{C}_{B/A}$ and that $\xi \in \Omega _{P/A} \otimes _ P B$ is an element which maps to zero in $\Omega _{B/A}$. We first choose factorization $P \to P' \to B$ such that $P' \to B$ is surjective and $P'$ is a polynomial algebra over $A$. We may replace $P$ by $P'$. If $B = P/I$, then the kernel $\Omega _{P/A} \otimes _ P B \to \Omega _{B/A}$ is the image of $I/I^2$ (Algebra, Lemma 10.131.9). Say $\xi$ is the image of $f \in I$. Then we consider the two maps $a, b : P' = P[x] \to P$, the first of which maps $x$ to $0$ and the second of which maps $x$ to $f$ (in both cases $P[x] \to B$ maps $x$ to zero). We see that $\xi$ and $0$ are the image of $\text{d}x \otimes 1$ in $\Omega _{P'/A} \otimes _{P'} B$. Thus $\xi$ and $0$ have the same image in the colimit (see Cohomology on Sites, Example 21.38.1) $\pi _!(\Omega _{\mathcal{O}/A} \otimes \underline{B})$ as desired. $\square$

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