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If $S$ is a Henselian local scheme with closed point $v$ and $X$ a smooth $S$-scheme, then it is well known that the canonical map $X(S) \rightarrow X(v)$ is surjective.

Suppose now that $Y$ is smooth over $X$. Is the canonical map $X(Y) \rightarrow X(Y_v)$ surjective (all maps are $S$-maps)? I am happy to make additional assumptions on $X$ and $S$, though preferably not too many on $Y$... I am mainly interested in the case $S = Spec(\mathbb{Z}_p)$ or finite extensions.

Thanks,

David

edit: oops, for the map $X(Y) \rightarrow X(Y_v)$ to make sense I need to assume a bit more about $S$, I think I also need $Y/S$ proper.

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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. –  BCnrd Oct 20 '10 at 13:59
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David, I don't understand your edit. But don't you mean $Y(X)\to Y(X_v)$? –  Laurent Moret-Bailly Oct 20 '10 at 14:36
    
Dear Laurent: the counterexample was chosen using $Y = X$ so as to be robust with respect to change in the intended meaning. –  BCnrd Oct 20 '10 at 15:35
    
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. –  Laurent Moret-Bailly Oct 20 '10 at 15:57
    
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! –  David Holmes Oct 21 '10 at 9:20

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