The tensor product of the canonical line bundle and k(x) for a closed point x

Question

I am reading the book “Fourier-Mukai transforms in algebraic geometry” by Daniel Huybrechts. At the beginning of the page 91, it is written that if $X$ is a smooth projective variety with a canonical bundle $\omega_X$, then for a closed point $x\in X$, we have $k(x)\cong k(x)\otimes \omega_X$. My question is that why is this true?

Answer 1

For any affine scheme $\operatorname{Spec} A$ and quasi-coherent sheaves $\mathcal{F}\cong \widetilde{F}$, $\mathcal{G}\cong \widetilde{G}$ for $A$-modules $F,G$, the tensor product $\mathcal{F}\otimes\mathcal{G}$ is exactly the sheaf $\widetilde{F\otimes G}$. This implies that for any scheme $X$, quasi-coherent sheaves $\mathcal{F},\mathcal{G}$ and affine open $U\subset X$, we have $\mathcal{F}(U)\otimes_{\mathcal{O}_X(U)}\mathcal{G}(U)\cong (\mathcal{F}\otimes\mathcal{G})(U)$. We'll use this result to compute in your case.

Now suppose $U\subset X$ is an affine open subscheme where $\omega_X$ is free. If $x\notin U$, then $k(x)(U)=0$ and so $k(x)\otimes\omega_X$ is a skyscraper sheaf supported at $x$. If $x\in U$, then we're looking at the tensor product $k(x)\otimes_{\mathcal{O}_X(U)} \mathcal{O}_X(U)$, which is exactly $k(x)$, so the stalk at $x$ is exactly $k(x)$ and in fact the tensor product $k(x)\otimes\omega_X\cong k(x)$.