Non existence of cyclic infinite linear algebraic groups

Let $G$ be a linear algebraic group defined over some algebraically closed field $\mathbb{K}$ and also over some subfield $k\subset \mathbb{K}$. There is thus a natural group structure on the set of $k$-points of $G$. It seems impossible to me that this group is cyclic and infinite. Is this true?

EDIT: user48841 explained that this happens with elliptic curves. I am in fact interested in linear algebraic groups.

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Interesting question. In characteristic 0, a linear group is unirational (this is due to Chevalley), so if $k$ is uncountable $G(k)$ must be very large. Even if $k$ is countable I would expect "too many" points, but I don't know how to get a proof from that. – abx Mar 28 '14 at 8:59
Yes, the difficult case is when the field is infinite but countable (and I am also interested in positive characteristic by the way, but I think that this should not change the result). – Jérémy Blanc Mar 28 '14 at 9:49
"infinite cyclic" sounds a bit focused and unstable... the same question with "infinite finitely generated" would be more natural. – YCor Mar 29 '14 at 9:28
For my purpose, I only need cyclic, but you are right that finitely generated seems natural too. – Jérémy Blanc Mar 29 '14 at 9:30

There are elliptic curves $E/\mathbb{Q}$ with $E(\mathbb{Q}) \simeq \mathbb{Z}$.

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Ah yes, thanks for the answer, which is perfect with respect to my question, but not with what I had in mind. In fact, in my case I was thinking about a linear algebraic group. What happens in this case? – Jérémy Blanc Mar 28 '14 at 7:23
Defintely $\mathbf{G}_a$ and $\mathbf{G}_m$ do not have finitely generated subgroup as its $\mathbf{Q}$-points, for get cyclicity. Is there any clue in exponential mapping ? – P Vanchinathan Mar 28 '14 at 8:57
This was indeed the beginning of my intuition: the case of additive and multiplicative groups. But what happens if the group is more complicated? What do have in mind with "exponential mapping" ? – Jérémy Blanc Mar 28 '14 at 9:48

We may assume, by replacing $G$ by the Zariski closure of $G(k)$, that $G(k)$ is Zariski dense. By assumption, $G(k)$ is $\mathbb Z$; hence $G$ is abelian and after replacing $G$ by the connected component of identity of $G$, we may assume that $G$ is connected. \vskip 5mm

Now a conneced abelian algebraic group is a product of ${\mathbb G}_a$ and ${\mathbb G}_m$. Hence the assumptions mean that $k$ or $k^*$ must be $\mathbb Z$; this is impossible for a subfield of $\mathbb C$.

In detail, $G$ is a product of copies of ${\mathbb G}_a$ and copies of ${\mathbb G}_m$. Hence $G(K)$ is the product of (copies of) $K$ and $K^*$. Assume that a factor like a product of ${\mathbb G}_a$ occurs. Then $G(k)$ contains the subgroup $k\supset {\mathbb Q}$. However, $G(k)$ being $\mathbb Z$, any infinite subgroup must be $\mathbb Z$ but $\mathbb Q$ is not $\mathbb Z$.

A similar but more involved reasoning may be used to tackle the case when $G$ is a product of the multiplicative group ${\mathbb G}_m$ over $K$.

Since the questioner has asked for details, if we assume that $G$ over $K$ does not have the additive group components, then over the smaller field $k$, $G$ is isogenous to a product of ${\mathbb G}_m$ and the groups $R^1_{l/k}({\mathbb G}_m$ where $R_1$ refers to the group of norm one elements. This is a theorem due to Ono (for reference, one may see Borel-Tits). The ${\mathbb G}_m$ factor may easily be taken care of. As to the other group $R^1$, one has to work a little to say that the group of norm one elements in $l$ cannot be isomorphic to $\mathbb Z$, for any char zero $l/k$.

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Thanks for the answer, but I do not get all details. Why is $G$ abelian if $G(k)$ is? Why "the assumptions mean that $K$ or $K^*$ must be $\mathbb{Z}$" ? A priori, I expected that we can have some forms of tori for instance. How do you deal with these? – Jérémy Blanc Mar 28 '14 at 11:42
For the first part, how about replacing $G$ with the centralizer of a generator $g$ of $G(k)$? – Aurel Mar 28 '14 at 12:53
@Aurel: "how about replacing $G$ with the centralizer of ... $g$" That seems like a good idea, but the centralizer need not be commutative, e.g., for $g$ in the center of $G$, the centralizer of $g$ is all of $G$. – Jason Starr Mar 28 '14 at 13:03
@JérémyBlanc: Just to restate: first replace $G$ by the centralizer $Z_G(g)$, and then replace $Z_G(g)$ by its center $Z(Z_G(g))$, which is an Abelian algebraic group that contains $g$. – Jason Starr Mar 28 '14 at 13:35
@Aakumadula: It sounds to me like you do indeed have a proof, yet the rest of us do not understand it the way that you explained it. Could you explain the proof carefully in the special case that $G_K$ is isomorphic to $\mathbb{G}_{m,K}$? – Jason Starr Mar 29 '14 at 11:51

[An extended comment, in community-wiki mode.] There is a closely related fact: Take $G$ to be an algebraic torus such as the multiplicative group, defined over a field $k$ which is not an algebraic extension of a finite field. (The characteristic doesn't matter, but this condition does.) Then there exists an element of infinite order in $G_k$ which generates a dense (cyclic) subgroup of $G$.

For a direct proof when $G$ is $k$-split, see Proposition 8.8 in Borel's book Linear Algebraic Groups (Springer, GTM 126). He remarks also that the split assumption can be dropped, using a more delicate argument from Tits' lectures at Yale.

What I've just quoted does not exactly fit your question. since it doesn't state that $G_k$ itself is cyclic, but it does illustrate a sort of algebraic parallel to the existence of a topological generator for a compact torus.

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Thanks for the nice comment. – Jérémy Blanc Mar 29 '14 at 7:12