# Base-point free linear sheaf and extending the base field

This is a question from Ravi Vakil's notes I've been stuck on, namely 18.2.I.

Let $X$ be a scheme over a field $k$, and let $K/k$ be any field extension. Let $\mathcal{L}$ be an invertible sheaf on $X$. Then I would like to show $\mathcal{L}$ is base-point free if and only if the pullback of $\mathcal{L}$ to $X_K$, say $\mathcal{L}\otimes K$, is base-point free.

I know that $H^0(X, \mathcal{L})\otimes K\cong H^0(X_K, \mathcal{L}\otimes K)$. In particular, a basis of $H^0(X,\mathcal{L})$ over $k$ yields a basis of $H^0(X_K, \mathcal{L}\otimes K)$ after tensoring with $K$.

If $\mathcal{L}$ is base-point free, then so too must the pullback be base-point free. Otherwise there exists a base point $p$ in $X_K$, whose image in $X$ is $q$. As $q$ is not a base-point of $\mathcal{L}$, there exists some global section $s$ such that $s$ doesn't vanish at $q$, and thus the pullback of $s$ doesn't vanish at $p$.

I'm not sure how to proceed showing the converse, namely that $\mathcal{L}\otimes K$ being base-point free implies $\mathcal{L}$ is. If I try to proceed by contradiction I run into the issue where a point of #X# may not have a point in $X_K$ lying over it (as $K/k$ is not necessarily algebraic).

Could someone point me in the right direction?

• Try to compute the vanishing of the pullback of one of these global sections in the basis of $H^0(X,L)$. There is a formula in the big list of properties of the pullback in chapter 16 in Ravi. Then study how the intersections of the vanishing behave after pullback also. I think that works. Also, I think there will always be a point in $X \times_k K$ lying above any given point $p$ of $X$ - this just because $k(p) \otimes_k K$ is never the zero ring, so when you compute the scheme theoretic preimage it is not empty. – Lorenzo Najt Apr 7 '16 at 21:24

To show that any point of $X$ has a point in $X_K$ lying over it, just recall that surjectivity is stable under base change (and $\operatorname{Spec}(K)\rightarrow \operatorname{Spec}(k)$ is surjective).
To see why this finishes the argument (for future askers of this question), notice that we have a map $\Gamma(X,\mathcal{L})\rightarrow \Gamma(X_K,\mathcal{L}\otimes K)$ and for any $p \in X_K$ mapping to $q \in X$, we have a diagram:
$$\require{AMScd} \begin{CD} \Gamma(X,\mathcal{L}) @>>> \Gamma(X_K,\mathcal{L}\otimes K)\\ @VVV @VVV \\ \mathcal{L}_q @>>> (\mathcal{L}\otimes K)_p \end{CD}$$
Then $q$ is a basepoint of $\mathcal{L}$ iff the left hand map is the $0$ map (because $\mathcal{L}$ is a line bundle and so any non-zero element of the stalk will generate it) and since neither of the horizontal maps are $0$, this would imply that $p$ is a basepoint of $\mathcal{L} \otimes K$, a contradiction of our hypothesis, so no points in the image of $X_K$ are basepoints and we're done by surjectivity.