Let $X$ be a proper scheme over a field $k$. Let $T$ be a scheme over $k$. Is it true that morphisms $T \times X \to \mathbb{A}^1$ are in bijection with morphisms $T \to \Gamma (X, \mathcal{O}_X)$ (where the finite-dimensional vector space $\Gamma (X, \mathcal{O}_X)$ is interpreted as a scheme as usual)?

Or, perhaps, is something weaker (put condition on $T$, say) or stronger (more general scheme instead of $\mathbb{A}^1$, say) true?

Thank you

  • $\begingroup$ what do you mean by interpreting $\Gamma(X,O)$ as a scheme as usual? Do you mean taking Spec of it, or Spec of Sym of its dual? If the latter, then it's false (eg take X to be projective space and T = Spec k). $\endgroup$
    – pro
    Jan 19 '16 at 19:53
  • $\begingroup$ I mean Spec of Sym of dual. Why would that be wrong for T a point? It would say that morphisms from $X$ to $\mathbb{A}^1$ are the same as elements of the set $\Gamma(X,\mathcal{O}_X)$, which is correct, no? $\endgroup$
    – Sasha
    Jan 19 '16 at 22:58
  • $\begingroup$ isn't Sym k = k[x]? $\endgroup$
    – pro
    Jan 19 '16 at 23:20
  • $\begingroup$ sorry, I was confused. I take back what I wrote! $\endgroup$
    – pro
    Jan 20 '16 at 0:37

You can just use all the universal properties one at a time. We only need that $X \to \operatorname{Spec} k$ is qcqs and that $\Gamma(X,\mathcal O_X)$ is finite-dimensional; both these assumptions are satisfied if $X$ is proper over $k$. We also need that $T \to \operatorname{Spec} k$ is flat, which is always true if $k$ is a field.

Remark. Note that $\Gamma(T \times X, \mathcal O_{T \times X}) = \Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X)$. Indeed, this follows from flat base change (Tag 02KH), since $T \to \operatorname{Spec} k$ is flat and $X \to \operatorname{Spec} k$ is qcqs.

Remark. If $V$ is a vector space and $W$ is a finite-dimensional vector space, then the natural map \begin{align*} V \otimes W &\to \operatorname{Hom}_k(W^\vee, V)\\ v \otimes w &\mapsto (\phi \mapsto \phi(w) v) \end{align*} is a natural isomorphism. For example, one can prove the case for $W$ of dimension $1$, and use that every $W$ decomposes as a finite direct sum of $1$-dimensional vector spaces. (Note, however, that this is false for $W$ infinite-dimensional, already if $V = k$.)

Lemma. Let $X$ be a qcqs $k$-scheme such that $\Gamma(X, \mathcal O_X)$ is finite-dimensional. Let $T$ be a $k$-scheme. Then $$\operatorname{Hom}_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)).$$

Proof. By the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ given by $\Gamma$ and $\operatorname{Spec}$, we get $$\operatorname{Hom}_{\textrm{Sch}/k}(T \times X, \mathbb A^1) = \operatorname{Hom}_k^{\operatorname{alg}}(k[x],\Gamma(T \times X, \mathcal O_{T \times X})).$$ By the universal property of $k[x]$, the latter is just $\Gamma(T \times X, \mathcal O_{T \times X})$. By the two remarks above, this equals $$\Gamma(T, \mathcal O_T) \otimes \Gamma(X, \mathcal O_X) = \operatorname{Hom}_k(\Gamma(X,\mathcal O_X)^\vee, \Gamma(T,\mathcal O_T)).$$ The universal property of the symmetric algebra turns this into $$\operatorname{Hom}_k^{\operatorname{alg}}(\operatorname{S}(\Gamma(X, \mathcal O_X)^\vee), \Gamma(T,\mathcal O_T)).$$ Finally, using the adjunction $\operatorname{Ring}^{\operatorname{op}} \rightleftarrows \operatorname{Sch}$ again, this finally becomes $$\operatorname{Hom}_{\textrm{Sch}/k}(T, \operatorname{Spec} \operatorname{S}(\Gamma(X,\mathcal O_X)^\vee)),$$ which proves the claim. $\square$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.