Lemma 91.4.5. 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 91.4.3. 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

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 21.39.1) $\pi _!(\Omega _{\mathcal{O}/A} \otimes \underline{B})$ as desired. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)

There are also: