TITLE: Are the unipotent and nilpotent varieties isomorphic in bad characteristics?
QUESTION [17 upvotes]: In characteristic 0 or good prime characteristic, there are standard ways to relate the unipotent variety $\mathcal{U}$ of a simple algebraic group $G$ and the nilpotent variety $\mathcal{N}$ of its Lie algebra $\mathfrak{g}$.  Recall that $p$ is good just when it fails to divide any coefficient of the highest root (the root system being irreducible), bad otherwise.  The only possible bad primes are $2,3,5$.  In characteristic 0, algebraic versions of the exponential and logarithm maps  provide Ad $G$-equivariant isomorphisms in both directions, whereas in good characteristic $p>0$, the less direct arguments of Springer in The unipotent variety of a semi-simple group yield similar results.    [The isogeny type of $G$ adds some complications here.]
There is scattered literature on the varieties $\mathcal{U}$ and $\mathcal{N}$ when $p$ is bad, often treated case-by-case, e.g., four papers by Lusztig posted on arXiv starting with Unipotent elements in small characteristic, along with papers by his student T. Xue.  A serious challenge when $p$ is bad is to find a uniform explanation for the failure of the numbers of unipotent classes and nilpotent orbits to agree in some cases: the details were worked out by Holt–Spaltenstein and others.   In spite of this breakdown in $G$-equivariance, a natural question can be raised:

Are the two varieties $\mathcal{U}$ and $\mathcal{N}$ isomorphic in all characteristics, for example when $G$ is simply connected?

The answer does not seem to be written down explicitly (?), but for example one can see indirectly that both varieties have the same dimension in all characteristics: the number of roots.  Existence of regular nilpotent elements in the Lie algebra of a simple algebraic group in bad characteristics by S. Keny, a former student of Steinberg, showed case-by-case that regular nilpotent elements always exist and form a dense orbit in $\mathcal{N}$.
By definition, the isotropy group in $G$ of such an element has dimension equal to the rank of $G$.

REPLY [5 votes]: Update: Paul Levy points out in the comments that a reasonable way of defining the nullcone in $\operatorname{Lie}(G) = \mathfrak{g}$ is as the zero set of the homogeneous invariants of positive degree — i.e. of $(k[\mathfrak{g}]_+)^G = (S^+\mathfrak{g}^*)^G$.
But with this definition, my original comment isn't valid. Indeed, if $G = \operatorname{PGL}_2$ then the co-adjoint representation of $G$ on $\mathfrak{g}^*$ has a fixed vector, which is a linear invariant in $(k[\mathfrak{g}]_+)^G$ whose zero locus is — as in Dave Stewart's original answer — the span of the root vectors.
I suppose all my original objection (below) really amounted to was that if $X$ is the affine scheme defined by the $\mathbf{Z}$-algebra $R=\mathbf{Z}[A,B,C]/\langle A^2 + 4BC\rangle$, then for any field $k$ of char. not 2, $X_k$ identifies with the nilpotent variety of $\mathfrak{pgl}_{2,k}$, but if $k$ has char. 2, $X_k$ is not reduced.

Original post: I'm going to write this as an "answer", though I think it mostly amounts to a comment on Dave Stewart's answer.
It is not completely clear to me that the statement "the nilpotent variety of $\mathfrak{pgl}_2$ is reduced" is correct when $p=2$. Well, I suppose that more precisely I mean: it isn't clear that the scheme of nilpotent elements should be viewed as reduced.
Taking a basis $x,y,h$ of $\mathfrak{pgl}_2$ (say, in its 3-dimensional representation), one finds that $ah + bx + cy$ is nilpotent just in case
$a^2+4bc=0$. Of course, in char. 2 this amounts to $a^2=0$, which suggests that the scheme of nilpotent elements shouldn't be viewed as a reduced subscheme. (If you don't want to write down the matrices, see e.g. Jantzen "Nilpotent Orbits in Representation Theory" §2.7.)
I do doubt (?) that this nilpotent scheme is isomorphic to the scheme of unipotent elements of $\operatorname{PGL}_2$, but (assuming that doubt is correct — I didn't think too carefully about it) I think the reason is more complicated than the statement "one is reduced and the other isn't".