A line avoiding an Algebraic group

Question

Let $K$ be an algebraically closed field, and $G\subset (K,+)^3$ an algebraic subgroup (i.e. given as the zero sets of finitely many polynomial equations) of dimension 1.

Is it clear that there is a line $L\subset K^3$ such that $$L\cap G=\{0\}?$$

Answer 1

The story is rather complicated in characteristic $p$, but I think the answer of Drike takes care of it. On the other hand, Gregory Grant makes a claim that seems to me to be not right.

We’re working with the algebraic group $(K^3,+)$, which I think would usually be called $\mathbf G_{\mathrm a}^3$. Consider now the subgroup of $(K^3,+)$ given parametrically by $(t,t^p,t^{p^2})$, in other words as the common zeros of $X^p-Y$ and $X^{p^2}-Z$. Certainly a one-dimensional sub-algebraic-group. But I say that this one-dimensional subgroup is not contained in any plane through the origin.

Indeed, consider $\Bbb F_{P^3}$, and an element $\zeta\notin\Bbb F_p$ whose trace down to $\Bbb F_p$ is nonzero, call this trace $\xi$. Consider also the point $\mathfrak Z=(\zeta,\zeta^p,\zeta^{p^2})$ on our subgroup. Then $\mathfrak Z'=(\zeta^p,\zeta^{p^2},\zeta)$ and $\mathfrak Z''=(\zeta^{p^2},\zeta,\zeta^p)$ all are in our subgroup, and are all on the hyperplane $X+Y+Z=\xi$ that doesn’t contain the origin, and so isn’t a subgroup. In particular, our one-dimensional subgroup is not contained in any plane through the origin.

Answer 2

If you have a subspace of $K^3$ that is also a subgroup of the additive group $\langle K,+\rangle$, then that subspace must be a plane or a line. In this case choose any line that does not lie in the plane and the intersection will be just the origin.

Answer 3

Well, as Gregory Grant noticed, if $K$ has characteristic $\bf 0$, then $G$ must be a $K$-vector space, hence a line, so the answer is obvious.

If $K$ has characteristic $\bf p$, I'd consider the morphism of varieties $\varphi:K\times G\rightarrow K^3,\ (k,g)\mapsto kg$ and say that if $Im\varphi$ is a variety, it is of dimension no more than $dim(K\times G)=2$, hence one can take any $y\notin Im\varphi$. The line $L=Ky$ will do the job.

I am trying to use as little machinery as possible, so any other answer with fewer technicalities is welcome !