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Suppose that $G$ is a linear group of positive dimension, defined over some field $k$. Is that true, that $G$ admits a (closed) one-dimensional subgroup?

I'm pretty much sure this is true in characteristic 0, or at least for $k=\mathbb{C}$. It seems that the main obstacle in positive characteristic is that there may not exist elements of infinite order.

EDIT1: As @Daniel Loughran pointed out in the answer below, one need at least to assume that $k$ is algebraically closed (or it may not even be necessary, see the comments of @Marguax abaut $k$ being separably closed, or real closed in characteristic 0).

EDIT2: Since my first statement of the problem seems to be genereting a lot of confusion (after getting a comment from @Jim Humphreys I'm even a little ashamed), here is the final version of the question:

Suppose that $G$ is a linear algebraic group, of positive dimension, defined over an algebraically closed field $k$ (but of arbitrary characteristic). Is it true, that $G$ posses a one dimensional (closed) subgroup? (so in fact, either $\mathbb{G}_a$ or $\mathbb{G}_m$)

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    $\begingroup$ As indicated in the answers, your language is too loose to be clear: the meaning of "linear group defined over some field" needs to be made more precise, at which point the question is easy to answer. $\endgroup$
    – Jim Humphreys
    Commented Sep 26, 2013 at 0:21
  • $\begingroup$ It is not necessary to assume $k = \overline{k}$ if allowing any characteristic; separably closed is enough. In char. 0 "real closed" is enough; e.g., the conclusion holds for $k = \mathbf{R}$. By the way, why do you highlight points of infinite order? That doesn't seem germane, since every nontrivial torus over a field $k$ not algebraic over a finite field (even such a $k$-torus with no nontrivial proper $k$-subtori) contains a $k$-point generating a Zariski-dense subgroup, so one cannot produce a 1-dimensional smooth closed $k$-subgroup by using any old point of infinite order. $\endgroup$
    – Marguax
    Commented Sep 26, 2013 at 2:29

3 Answers 3

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Every positive dimensional linear algebraic group $G$ over an algebraically closed field has a one dimensional subgroup.

Case 1 $G$ is reductive. In that case, $G$ contains a torus $T$. Since we are over an algebraically closed field, $T \cong \mathbb{G}_m^r$ for some $r>0$. In particular, $\mathbb{G}_m \subseteq T \subseteq G$.

Case 2 $G$ has a nontrivial unipotent radical $U$. In turn, $U$ has a nontrivial center $Z$. We have $Z \cong \mathbb{G}_a^s$ for some $s>0$ (not quite right, see Peter McNamara's correction below). In particular, $\mathbb{G}_a \subseteq Z \subseteq U \subseteq G$.

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    $\begingroup$ In Case 2, if the field is of positive characteristic, then Z is not necessarily a power of the additive group. (counterexample: length two Witt vectors). However it will still have a filtration by additive groups, so you still have a G_a as a subgroup $\endgroup$
    – Peter McNamara
    Commented Sep 26, 2013 at 0:56
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    $\begingroup$ The same conclusion holds over any separably closed field. Indeed, if $G\ne 1$ is a smooth connected affine group over a field $k$, $G$ contains a torus $T \ne 1$ or $G$ is unipotent. The first case is clear for $k=k_s$ ($T$ splits). For any $k$ and unipotent $G\ne 1$, WLOG $G$ is commutative, and also $p$-torsion if char($k)=p>0$. Thus, $G_K=\underline{V}$ for a finite $K/k$ and finite-dimensional $K$-vector space $V \ne 0$. For $j:G\hookrightarrow {\rm{R}}_{K/k}(G_K)$ and $W\subsetneq V$ complementary to Lie($G$) inside Lie(R$_{K/k}(G_K))=V$, $G\rightarrow \underline{V/W}$ is etale. QED $\endgroup$
    – Marguax
    Commented Sep 26, 2013 at 2:21
  • $\begingroup$ I still have a problem to understand why a reductive group must contain a nontrivial torus. I suppose this is trivial to specialists, but for me this subject is totally new. $\endgroup$
    – Tomasz Lenarcik
    Commented Sep 26, 2013 at 11:05
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    $\begingroup$ Here's a proof that every positive-dimensional L.A.G. $G$ has a positive dimensional abelian subgroup, over an algebraically closed field. You can suppose $G$ closed connected of $GL_n$, and the field uncountable. If the function $\chi$ "characteristic polynomial" on $G$ is constant, every element in $G$ is unipotent, so $G$ is unipotent. [...->] $\endgroup$
    – YCor
    Commented Sep 26, 2013 at 14:04
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    $\begingroup$ [--->] Otherwise $\chi$ has uncountable image, so contains polynomials with at least one root not root of unity. Hence $G$ has an element of infinite order and hence the Zariski closure of the subgroup it generates is a positive-dimensional (and not unipotent!) abelian subgroup of $G$. $\endgroup$
    – YCor
    Commented Sep 26, 2013 at 14:04
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Here is a counter-example for $k$ not algebraically closed (for simplicitly I assume that $k$ is perfect). Recall that there is an anti-equivalence of categories between the category of algebraic tori over $k$ and the category of free $\mathbb{Z}$-modules with a continuous action of $Gal(\overline{k}/k)$ (given by the character group $\widehat{T}$).

Choose an algebraic torus $T$ of dimension larger than one, whose character group $\widehat{T}$ is simple as a Galois module (these exist if e.g. $k=\mathbb{Q}$). Suppose that $T$ contains a one dimensional irreducible algebraic subgroup $C$. Then it is well known that such a group is isomorphic to either $\mathbb{G}_a,\mathbb{G}_m$ or an elliptic curve over $\overline{k}$. As $T$ is affine and every element is semi-simple, we see that $C$ is itself an algebraic torus. The inclusion $C \subset T$ therefore induces a non-trivial homorphism $\widehat{T} \to \widehat{C}$. The kernel of this homorphism is a Galois sub-module of $\widehat{T}$. However by assumption $\widehat{T}$ was simple, thus we deduce that $\widehat{C} = 0$ or $\widehat{T}$, which is a contradiction.

Conclusion: Such a torus $T$ cannot contain a one dimensional algebraic subgroup.

Edit: As Marguax points out I do in fact need $\widehat{T} \otimes_{\mathbb{Z}} \mathbb{Q}$ to be simple as a Galois module. I believe that such examples may also be given be given by the norm one torus $$R^1_{K/k} \mathbb{G}_m : N_{K/k}(x_1,\ldots,x_n) = 1.$$ Here $k \subset K$ is a cyclic Galois field extension of prime degree $n > 2$ and $N_{K/k}$ denotes the norm form for $K/k$. This may be defined by choosing an isomorphism $K \cong k^n$ and considering the usual field norm as a homogeneous polynomial of degree $n$ on $k^n$.

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    $\begingroup$ Can you provide an more explicit example of such a torus? $\endgroup$
    – Tomasz Lenarcik
    Commented Sep 25, 2013 at 21:32
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    $\begingroup$ You meant that $\widehat{T}_{\mathbf{Q}}$ is simple as a Galois module. Your argument works for $k$ (perfect or not) not separably closed or real closed, as such $k$ has a finite Galois extension $k'$ not a compositum of quadratic fields (so some $k$-simple isogeny factor of R$_{k'/k}(\mathbf{G}_m)$ has dimension $> 1$). Deeper is that a nontrivial smooth connected unipotent group over a field $k$ has a 1-dimensional smooth connected (central) $k$-subgroup. A connected semisimple example is the norm-1 unit group of a central division algebra of dimension $\ell^2$ for a prime $\ell > 2$. $\endgroup$
    – Marguax
    Commented Sep 25, 2013 at 21:35
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    $\begingroup$ @Tomasz: Let $k'/k$ be a finite Galois extension of prime degree $p$. The torus $T = {\rm{R}}_{k'/k}(\mathbf{G}_m)/\mathbf{G}_m$ of dimension $p-1$ has rationalized character group given by the permutation action of cyclic group of order $p$ on the hyperplane of rational $p$-tuples whose coordinates sum to 0, and that's irreducible over $\mathbf{Q}$ (!), so $T$ is $k$-simple. So use $p > 2$. $\endgroup$
    – Marguax
    Commented Sep 25, 2013 at 21:38
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Over an algebraically closed field, the answer is yes. I assume that by "linear group" you mean a smooth affine algebraic group scheme. Such a group has a filtration whose quotients (modulo finite group schemes) are successively a unipotent group, a torus, and a semisimple group, so it is only a question of checking each of these cases.

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