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This is tagged algebraic geometry becomes it comes from Hartshorne exercise V.1.8(b), where we have a group $\operatorname{Num} X :=G$ and an inclusion $G \hookrightarrow \mathbb{Q}^n$. Furthermore, we have a nondegenerate bilinear pairing $\langle \cdot,\cdot \rangle$ on $\mathbb{Q}^n$, so that $\langle G, G\rangle \subseteq \mathbb{Z}$. As suggested in the comments, let's also assume that $G$ generates $\mathbb{Q}^n$ over $\mathbb{Q}$.

I'd like to show that $G$ is finitely generated under these assumptions. I had a couple ideas for how to do this. One was to expand $G$ to be the group of all elements with integral inner product, but this is hard to describe, and I'm not sure one can. Another is to do induction, where the base case uses something like this, and then viewing $G$ as an extension of two finitely generated abelian groups.

However, it's still not completely clear how to conclude. Does anyone know how to finish this kind of problem?

Thanks!

Daniel
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  • This is certainly not true in full generality: for instance, $G$ could the $\mathbb{Q}$-submodule generated by an element $x$ such that $\langle x,x\rangle=0$. Maybe you want to assume that $G$ generates $\mathbb{Q}^n$ over $\mathbb{Q}$? – Eric Wofsey Mar 26 '23 at 22:09
  • Yes, let's include that assumption. That seems okay, since we could just assume the vector space it lives in is $G \otimes \mathbb{Q}$. – Daniel Mar 26 '23 at 22:11
  • How did you get around the issues with the isomorphism $H^1(D,\Omega_D)\cong k$ from exercise III.7.4 only being defined up to scaling? Or did that only mess with part (a)? – Hank Scorpio Mar 26 '23 at 22:23
  • @HankScorpio I didn't see that as an issue (perhaps mistakenly) since in III.7.4(a), (d) says that $t_D(\eta(P)) = t_D(c(\mathcal{O}(P)))= 1$, where I interpreted this $t_D$ to be the isomorphism in question. I did not do this exercise in III.7, so I might be missing a subtlety there. – Daniel Mar 26 '23 at 22:56
  • See Shubhodip Mondal's answer and the comments here. The essential problem is that Hartshorne defines a dualizing sheaf as a coherent sheaf $\omega$ along with a trace morphism $t:H^n(X,\omega)\to k$ so that $$Hom(\mathscr{F},\omega)\times H^n(X,\mathscr{F}) \to H^n(X,\omega) \to k$$ gives an isomorphism $Hom(\mathscr{F},\omega)\cong H^n(X,\mathcal{F})'$, so you can just scale $t$ by any nonzero constant and preserve all the properties in the definition but change the result of $t(-)$. (I think.) – Hank Scorpio Mar 26 '23 at 23:08

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$G$ generates $\mathbb{Q}^n$, so there is a basis $(e_1,\ldots,e_n)$ of $\mathbb{Q}^n$ made with elements of $G$. Then let $(f_1,\ldots,f_n)$ be the dual basis with respect to $\langle \cdot,\,\cdot\rangle$ (say, $\langle f_i,\,e_j\rangle =\delta_{ij}$).

Then, for all $i$, $\langle G,\, e_i \rangle \subset \mathbb{Z}$, so $G$ is contained in the $\mathbb{Z}$-submodule generated by the $f_i$. In particular, $G$ is finitely generated.

Aphelli
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