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Let $X$ be an algebraic variety over the field $k$. Hartshorne then defines its differential forms as $$\Omega^1_X=\delta^{\star}\left(\dfrac{\mathcal{J}} {\mathcal{J}^2} \right)$$ where $\delta: X \to X \times X$ is the standard diagonal morphism and $\mathcal{J}$ is the ideal sheaf of the closed subvariety $\delta(X) \subseteq X \times X$.

He also says that if $X$ is affine and $B$ is its coordinate ring, one can identify $\mathcal{J}$ with $I={\rm ker }(\alpha : B \otimes_k B \to B)$, where $$\alpha (x \otimes y)=xy .$$This is clear to me. So, $\frac{\mathcal{J}}{\mathcal{J}^2}$ is the sheaf $\widetilde{\dfrac{I}{I^2}}$ on $X \times X$.

Then , he concludes that $\Omega^1_X=\widetilde{\dfrac{I}{I^2}}$, viewing $\dfrac{I}{I^2}$ as a $B$ module as $b \cdot (b_1 \otimes b_2)=bb_1 \otimes b_2$ . This last part is really not clear to me. I can't see why the two definitions agrees.

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By Hartshorne II 5.2, if $\alpha : A \to B$ is a ring homomorphism and $f : {\rm Spec}(B) \to {\rm Spec}(A)$ is the corresponding morphism of affine schemes, and if $M$ is an $A$-module, then $f^\star(\widetilde M) = (M\otimes_A B)^\sim$, where $B$ gets its $A$-module structure from the ring homomorphism $\alpha$. We can apply this to our setup. Here, $A = B \otimes_k B$ and $M = I/I^2$, and the morphism $f$ is the diagonal morphism $\delta$ corresponding to the ring homomorphism $\alpha : A \to B$. So $\delta^\star((I/I^2)^\sim)$ is equal to $\left( (I / I^2) \otimes_A B\right)^\sim$.

But since the ring morphism $\alpha : A \to B$ is surjective, and since $I$ is the kernel of $\alpha$, it is clear that $B$ is really the same thing as $A / I$. So $(I / I^2) \otimes_A B \cong (I/I^2) \otimes_A (A/I)$, and this is isomorphic to $I/I^2$ as an $A$-module via the isomorphism $I/I^2 \to (I/I^2)\otimes_A (A/I)$ given by $[x] \mapsto [x]\otimes_A 1$ for every $[x] \in I/I^2$. (Here, $x$ is an element of $I$, and $[x]$ denotes its equivalence class modulo $I^2$. This old post may help, if you need further convincing.)

However, to characterise $\delta^\star((I/I^2)^\sim)$ as a quasicoherent sheaf on ${\rm Spec}(B)$, we really need to think of $((I/I^2) \otimes_A B)$ as a $B$-module rather than merely as an $A$-module. So we need to understand how to multiply an element $[x] \in I/I^2$ by an element $b \in B$. Identifying $[x]$ as the element $[x]\otimes_A 1 \in (I/I^2)\otimes_A(A/I)$, the multiplication is given by $b.([x]\otimes_A 1)=[x]\otimes_A b$. Now notice that $b=\alpha(b \otimes_A 1)$, where $\alpha : A \to B$ is the diagonal homomorphism. So if $x = \sum_i b_{1,i} \otimes_A b_{2,i}$, then $[x]\otimes_A b= [(b\otimes_A 1).(\sum_{i}b_{1,i} \otimes_A b_{2,i})]\otimes_A 1=[\sum_i bb_{1,i} \otimes_A b_{2,i}]\otimes_A 1$, which is identified with the element $[\sum_i bb_{1,i} \otimes_A b_{2,i}] \in I/I^2$ via the isomorphism $I/I^2 \cong (I/I^2)\otimes_A(A/I)$. This is precisely the multiplication rule that you described.

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