Let $G$ be an affine algebraic group over an algebraically closed field $k$ whose characteristic is $p>0$. Can $\mathcal{U}(G)$, the set of unipotent elements of $G$, be characterized as all elements $g\in G$ such that $g^{p^t}=1$ for some $t\in\mathbb{N}$? If not, what is $\mathcal{U}(G)$? If so, what are equivalent definitions of $\mathcal{U}(G)$ and why are they equivalent?
I'm just trying to grow in my understanding of the definition of unipotence. Thanks for your help!
On
The set of unipotent elements are indeed the elements whose orders are some power of the characteristic of the field.
To see this, let $g\in G$ and embed $G$ in some general linear group $H$ over the field $k$. Let $p$ be the characteristic of $k$.
Since the order of the image of $g$ in $H$ is a power of $p$ if and only if the order of $g$ is a power of $p$, we just need to show that a matrix over a field of characteristic $p$ is unipotent if and only if it has order a power of $p$.
Let $A$ be a unipotent matrix, so $(A-I)^m = 0$ for some $m$. Let $n$ be given such that $p^n\geq m$. Then we have $0 = (A - I)^{p^n} = A^{p^n} - I$ so $A$ has order a power of $p$.
On the other hand, if we assume that $A^{p^n} = I$ for some $n$ then we have $(A - I)^{p^n} = A^{p^n} - I = 0$ so $A$ is unipotent.
More "intrinsically", although you still need an embedding, is to embed $G \to GL(k[G])$. (We assume $k$ is algebraically closed). Here $k[G]$ is the ring of regular functions on $G$, and $G$ acts on $k[G]$ by $$(g\cdot f)(x) = f(g^{-1}x)$$
Of course, $k[G]$ is an infinite dimension vector space, but you can show that it is the direct limit of finite dimensional $G$-stable subspaces. An unipotent element in $GL(k[G])$ is then defined to be an element unipotent on any restriction to the finite dimensional $G$-stable pieces.
But then for this definition to be used, you would need to check that under a morphism of algebraic groups, unipotent element gets sent to a unipotent one. Also for $GL_n$, unipotence in the usual sense is the same as this sense.