1

Let $\Bbbk$ be an algebraically closed field of characteristic $p>0$. Let $G$ be a connected affine algebraic group over $\Bbbk$, and write $\Bbbk[G]$ for the coordinate ring of $G$.

Of course $G$ need not be smooth, so let $I\subseteq\Bbbk[G]$ denote the nilradical of $\Bbbk[G]$. My question is: if $f\in\Bbbk[G]\setminus I$ ($\textbf{this is set-minus!}$), then is it correct that $f$ is not a zero divisor in $\Bbbk[G]$?

My intuition is that something stronger is true: namely that there is a splitting of the natural surjection $\Bbbk[G]\to\Bbbk[G]/I$, and that under this splitting $\Bbbk[G]$ is a projective module over $\Bbbk[G]/I$. But perhaps my intuition is totally wrong :)

freeRmodule
  • 1,852
  • See 2. of https://math.stackexchange.com/questions/2523078/irreducible-components-and-connected-components-of-an-algebraic-group#2741374 – Aphelli May 16 '23 at 08:40
  • Hi Aphelli, I don't see why this is enough, but perhaps I am confused. What if $fg=0$ for some $g\in I$, the nilpotent ideal? It would not violate irreducibility. An example might be the algebra $A=\Bbbk[x,y]/(x^p,yx)$. Then $\operatorname{Spec}A$ is irreducible but $y$ is a zero divisor. – freeRmodule May 17 '23 at 00:07
  • 1
    Oops, you’re right. My comment was too hasty, thanks for pointing it out. I think your argument is correct. I believe that one can also argue like this: we know that $G$ is irreducible (if anything, because $G_{red}$ is smooth and connected). So the closure of the set of its embedded points is a proper closed subset of $G$. But it’s also clearly invariant under left translation! So it must be empty, so that $G$ has no embedded points, and its only associated prime ideal is the nilradical. Thus non-nilpotent elements are not contained in any associated prime ideal, so they’re not zero divisors. – Aphelli May 17 '23 at 06:52

1 Answers1

1

I think I answered my question; please confirm if the argument is correct!

Suppose we have nonzero $g\in I$ such that $fg=0$. Then there exists $k$ such that $g\in I^k$ but $g\notin I^{k+1}$. Thus $g$ defines an element of $I^{k}/I^{k+1}$, which is a $G_{red}=Spec(\Bbbk[G]/I)$ equivariant coherent sheaf on $G_{red}$, under left translation say. But this implies it is a vector bundle on $G_{red}$, and thus is torsion free. However, because $fg=0$, we would have that $g$ defines a nonzero torsion element of $I^k/I^{k+1}$, an impossibility.

freeRmodule
  • 1,852