Let k be a field and let X be scheme over k. Let K be a field extension of k and denote by $X_K$ the basechange of X to Spec K. Under what conditions is the canonical map of Picard groups $Pic(X)\to Pic(X_K)$, induced by the projection, injective? I know that this is true if X is geometrically integral and proper over k, but what if X is only of finite type, separated and geometrically integral over k?
3 Answers
This is not true. Let $E$ be an elliptic curve over $\mathbb{R}$, such that $E(\mathbb{R})$ has one connected component. Let $u$ be a point of $E(\mathbb{C}) \setminus E(\mathbb{R})$. Writing $\sigma$ for complex conjugation; $u + \sigma(u)$ is in $E(\mathbb{R})$. Two of the solutions to $2v=u + \sigma(u)$ are real and the other two are complex conjugate to each other; let $v$ and $\sigma(v)$ be the complex conjugate pair. So
$$2v=2\sigma(v) = u + \sigma(u)$$
in the group law of $E$.
Now, let $X = E \setminus \{ v, \sigma(v) \}$ and consider the line bundle $L:=\mathcal{O}(u + \sigma(u))$. Since this is a $\sigma$ invariant divisor, the line bundle $L$ is defined over $\mathbb{R}$. I claim that $L$ is nontrivial, but becomes trivial after base extending to $\mathbb{C}$.
Proof that $L$ is nontrivial: If not, there would be a real function $f$ on $X$, with zero divisor precisely $u + \sigma(u)$. Extending to a meromorphic function on $E$, we would have $$u + \sigma(u) = k v + (2k) \sigma(v).$$ But, since $f$ is $\sigma$ invariant, it has poles of the same order at $v$ and $\sigma(v)$, so $u + \sigma(u) = v + \sigma(v)$. Using the linear relation $u + \sigma(u) = 2v$, we deduce that $v = \sigma(v)$, contradicting that $v$ was chosen not to be real.
Proof that $L \times_{\mathbb{R}} \mathbb{C}$ is trivial: The relation $u + \sigma(u) = 2v$ means there is a meromorphic function on $E$ with zeroes at $u$ and $\sigma(u)$, and a double pole at $v$. Restricting this function $X$, we get a trivialization of $L$.

$\begingroup$ This is a nice answer. (I previously had answered the question incorrectly, and with no justification.) In the case of curves (as above), it is related to an exact sequence of Mike Rosen: see equation (2) on p. 6 of math.uga.edu/~pete/ellipticded.pdf $\endgroup$ Dec 9, 2009 at 19:03

7$\begingroup$ In the case of Galois extension K/k one can explicitly describe the kernel of $Pic(X) \to Pic(X_K)$. I believe it's equal to $H^1(G, K[X]^*)$. It is a part of 5term exact sequence coming from a HochshildSerre spectral sequence, associated to Galois covering $X_K \to X$. $\endgroup$ Dec 9, 2009 at 22:08

1$\begingroup$ In particualar, when $K[X]^* = K^*$, the kernel is zero by Hilbert 90. $\endgroup$ Dec 9, 2009 at 22:09

1$\begingroup$ David, this is a really great answer, thanks a lot for this beautiful example (I'm the one who asked this question before I had a real account here). $\endgroup$ Jan 17, 2010 at 21:59
Perhaps it is worth recording a simple example: let $X_k= \text{Spec}(k[x,y]/(x^2+y^21))$. Then $\text{Pic}(X_{\mathbb R}) = \mathbb Z/(2)$ while $\text{Pic}(X_{\mathbb C}) = \mathbb 0$, see Fossum's book "Divisor class groups of Krull domains", Prop 11.8.

$\begingroup$ This is one of my favorite examples in algebra: see Theorem 12 of math.uga.edu/~pete/ellipticded.pdf. $\endgroup$ Oct 15, 2010 at 17:46

$\begingroup$ @Pete: thanks! Your note looks beautiful, I will try to read more. $\endgroup$ Oct 15, 2010 at 17:50
On an integral, proper scheme X over k a line bundle L is trivial if and only if both L and its dual have a nonzero global section. This can be checked after an arbitrary field extension. The proof of this criterion uses that the global section of O_X are a field. See for example the proof of Corollary II.5.6 in Mumford's book on abelian varieties.