Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $k$ be an arbitrary field, and let $\varphi:A\to B$ be a morphism of abelian varieties over $k$.

If $k$ has characteristic zero, then $\varphi(A)$ has the structure of an abelian subvariety of $B$ which is defined over $k$.

Question: For an arbitrary field $k$, has $\varphi(A)$ the structure of an abelian subvariety of $B$ which is defined over $k$?

share|improve this question

2 Answers 2

up vote 8 down vote accepted

In general, if $f:G \rightarrow H$ is any homomorphism between smooth group schemes of finite type over a field $k$, the image $f(G)$ is always a smooth closed $k$-subgroup of $H$. (This is a special case of general results in SGA3, but it seems more instructive to give the direct argument in this case rather than wade through SGA3 for this purpose. The argument below has all of the same ideas as in the affine case treated in every book on linear algebraic groups, except that the argument is given in geometric terms without reference to coordinate rings so that it works without an affineness hypothesis. In the argument below we work with schemes throughout.)

Give the closure $Z$ of $f(G)$ the "schematic image" structure; this corresponds to the kernel of $O_H \rightarrow f_{\ast}(O_G)$. We want to show that $Z$ is a smooth closed $k$-subgroup of $H$ and $f:G \rightarrow Z$ is surjective. That is, we want (i) $Z$ is equi-dimensional with the coherent $\Omega^1_{Z/k}$ flat over $O_Z$ of constant rank equal to $\dim(Z)$, (ii) $m:H \times H \rightarrow H$ carries $Z \times Z$ into $Z$, (iii) $f:G \rightarrow Z$ is surjective. (It is obvious that $Z$ contains the identity point and is carried into itself under inversion since $f$ is a continuous map that is a $k$-homomorphism, so for the group aspect it is multiplication that we may focus upon.)

Since $f(G)$ is a constructible set whose formation commutes with extension of the ground field (in the sense that for an extension field $K/k$, the preimage of $f(G)$ under $H_K \rightarrow H$ is $f_K(G_K)$) and the formation of "schematic image" of $f$ commutes with extension of the ground field too (as the formation of $f_{\ast}(O_G)$ commutes with flat base change, due to quasi-compactness of $f$), we may apply extension of the ground field up to an algebraic closure to reduce to the case when $k$ is algebraically closed. In this case since $Z$ inherits reducedness from $G$, it is generically smooth (as $k$ is algebraically closed), and thus $Z$ is smooth once it is a subgroup of $H$ (by using $k$-point translations).

Since $k$ is algebraically closed and $Z$ is reduced (so $Z \times Z$ is reduced), $Z$ is a $k$-subgroup of $H$ provided that $Z(k)$ is a subgroup of $H(k)$ in the ordinary sense. For any $z \in Z(k)$, it suffices to show that left multiplication $\ell_z$ on $H$ carries $Z$ into itself. But $Z$ is the closure of $f(G)$, so it suffices to show that $\ell_z(f(G)) \subset Z$. In other words, we want $\ell_z \circ f:G \rightarrow H$ to land inside $Z$. Since $k = \overline{k}$ and $Z$ is closed in $H$, it is equivalent to show that $\ell_z \circ f$ carries $G(k)$ into $Z$. That is, for $g \in G(k)$ we want $z f(g) \in Z$. Thus, we are reduced to checking that for any $g \in G(k)$, right multiplication $\rho_{f(g)}$ carries $Z$ into itself. It suffices to show that it carries $f(G)$ into $Z$, and obviously it even carries $f(G)$ into $f(G)$ (since $f$ is a $k$-homomorphism).

That completes the proof that $Z$ is a smooth closed $k$-subgroup of $H$, and it remains to check (still with $k = \overline{k}$) that $f:G \rightarrow Z$ is surjective. The dense image $Y$ is constructible by Chevalley's theorem, and equality of constructible subsets of a finite type scheme over an algebraically closed field can be checked on rational points. It is easy to check that $Y(k) = f(G(k))$, and this is a subgroup of $Z(k)$. By constructibility, there is a dense open subset $U \subset Z$ contained in $Y$, so $U(k) \cdot U(k) \subset Y(k)$.

Thus, to conclude that $Y = Z$ it suffices to check that if $Z$ is any smooth finite type group over an algebraically closed field $k$ and if $U \subset Z$ is a dense open subset then $m(U \times U) = Z$. By constructibility of $m(U \times U)$, it is equivalent to check that its set of $k$-points $U(k) \cdot U(k)$ exhausts $Z(k)$. For any $z \in Z(k)$, the translate $z U^{-1}$ is a dense open subset of $Z$, so it meets $U$. The non-empty overlap must contain a $k$-point since $k = \overline{k}$, so $zu^{-1} = u'$ for some $u, u' \in U(k)$. Hence, $z = u' u \in U(k) \cdot U(k)$ as desired.

Remark: If you apply generic flatness and translation arguments over $\overline{k}$, it also follows that the natural map $G \rightarrow f(G)$ is flat and so $f(G)$ represents the fppf quotient group sheaf $G/(\ker f)$, thereby making contact with the quotient formulation in Cesnavicius' answer.

share|improve this answer
    
Very nice answer!!! Thank you very much for explaining everything in detail. –  Umit Demir Sep 8 '13 at 8:14

Yes. A relevant reference is [SGA3, VI_A, 6.7 (i)] (in case the numbering is different, I am referring to the new edition); using the notation of loc. cit., in your situation $G/N$ will inherit connectedness, smoothness, and properness from $G$, and hence will be an abelian variety over $k$.

share|improve this answer
    
Thanks a lot for your nice answer, and the precise reference. –  Umit Demir Sep 8 '13 at 8:16

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.