Let $X$ be a connected projective finite type noetherian scheme of dimension 1 over a field $k$.

  1. When can $H^0(X,\mathcal{O}_X)$ fail to be a field?
  2. If it is a field $k'$, then is $X$ also a curve over $k'$? Must $X$ be geometrically connected over $k'$?

Note that I don't assume $X$ to be smooth, regular, irreducible, or reduced, or even geometrically connected. Nor do I assume that $k$ is algebraically closed.

Certainly $H^0(X,\mathcal{O}_X)$ is a $k$-algebra. If $k$ is algebraically closed, then $H^0(X,\mathcal{O}_X)$is just $k$, hence a field.

What are some useful keywords/references to look up?

  • 2
    $\begingroup$ $\mathbb{P}^1_k\times \mathrm {Spec} k[\epsilon]$, where $\epsilon^2=0$ satisfies your conditions, but $H^0=k[\epsilon]$. $\endgroup$ – Mohan Apr 22 '17 at 1:08

Here are some relevant facts. Lets start with $X$ connected projective noetherian over $k$, with structure morphism $f : X\rightarrow\text{Spec }k$.

Firstly, $\text{Spec }H^0(X,\mathcal{O}_X) = \underline{Spec}_k f_*\mathcal{O}_X$, and hence the Stein factorization of $f$ gives us: $$X\stackrel{f'}{\rightarrow}\text{Spec }H^0(X,\mathcal{O}_X)\stackrel{g}{\rightarrow}\text{Spec k}$$ with $f'$ having geometrically connected fibers and $g$ being the normalization map of $\text{Spec }k$ inside $X$, hence finite. In particular, $f'$ is surjective since $f$ was. This already tells us quite a lot.

  1. Assume $X$ integral. Then $H^0(X,\mathcal{O}_X)$ is an integral $k$-algebra of finite dimension (since $X$ is projective), and hence is a field, and by the Stein factorization, we see that $X$ is a geometrically connected curve over $H^0(X,\mathcal{O}_X)$.

  2. Assume $X$ not necessarily integral, but at least reduced. Then $H^0(X,\mathcal{O}_X)$ is a reduced finite $k$-algebra, hence a product of fields, but $f'$ is surjective and $X$ is connected, so again $H^0(X,\mathcal{O}_X)$ must be a single field.

  3. Lastly, if $X$ is not reduced, then at least $A := H^0(X,\mathcal{O}_X)$ is an Artinian $k$-algebra, which must be local since $f'$ is surjective and $X$ connected. Let $m$ be the maximal ideal of $A$, and let $k' := A/m$, so $k'$ is a finite field extension of $k$. By the fiberwise criteria of flatness, $f'$ is flat, and so $X/A$ becomes a deformation of the geometrically connected curve $X_{k'} := X\times_A A/m$. In general without additional assumptions on $X$ or $X_{k'}$, this leads into the realm of deformation theory. However, a simple case arises when $X_{k'}$ has no (nontrivial) infinitesimal deformations - for example, if $X_{k'} = \mathbb{P}^1_{k'}$, in which case Mohan's example is the only possibility: ie, $X = \mathbb{P}^1_{k'}\times_{k'} A$.


Your Answer

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