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Oct 21, 2010 at 9:20 comment added David Holmes Thank you both for your speedy comments, and apologies for my poor editing. My edit was intended to be to assume $Y$ proper over $S$, but I am still interested in maps $X(Y) \rightarrow X(Y_v)$. Laurent, your example seems to occur because (in your notation) $X$ is not proper over the base; if we replace your $X$ by $\mathbb{P}^1_S$ the example seems no longer to work? Brian, your example certainly shows the answer to my question to be `no' in general, no matter what I assume to be proper and smooth. Thanks again to both of you!
Oct 20, 2010 at 15:57 comment added Laurent Moret-Bailly Assuming the intended question was about $Y(X)\to Y(X_v)$, here is an elementary affine example: take $X=$ the affine line over $S$ with coordinate $x$. If $t$ is a uniformizer, then $1-tx$ is not invertible on $X$ hence defines a nonempty closed subset $Z$, with empty closed fiber. Take $Y=X-Z$. Then $Y_v=X_v$ but $Y(X)$ is empty.
Oct 20, 2010 at 15:35 comment added BCnrd Dear Laurent: the counterexample was chosen using $Y = X$ so as to be robust with respect to change in the intended meaning.
Oct 20, 2010 at 14:36 comment added Laurent Moret-Bailly David, I don't understand your edit. But don't you mean $Y(X)\to Y(X_v)$?
Oct 20, 2010 at 13:59 comment added BCnrd Set $Y = X$ to be an elliptic curve $E$ over $S$. The surjectivity from $S$-points to $v$-points then reduces the question (via translations) to asking if endomorphisms of $E_v$ (as an elliptic curve, not merely as a curve) lift to endomorphisms of $E$. That often fails, e.g., elliptic curves over finite fields always have complex multiplication over the finite field.
Oct 20, 2010 at 13:53 history edited David Holmes CC BY-SA 2.5
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Oct 20, 2010 at 13:46 history asked David Holmes CC BY-SA 2.5