# The torsion subgroup of a diagonalisable linear algebraic group $G$ with ${\rm char}(k)=p$ (alg. closed $k$) is dense in $G$

This is Exercise 3.2.10(5b) of Springer's, "Linear Algebraic Groups (Second Edition)".

## The Question:

Let $$p$$ be the characteristic exponent of an algebraically closed field $$k$$. Let $$G$$ be a diagonalisable linear algebraic group over $$k$$ with character group $$X$$. Prove that the subgroup of elements in $$G$$ of finite order is dense in $$G$$.

That is, it is dense with respect to the Zariski topology.

## The Details:

For the definition of linear algebraic groups I work with, see this question of mine: Show that $({\rm id}\otimes \Delta)\circ\Delta=(\Delta\otimes{\rm id})\circ\Delta$ "translates" to associativity of linear algebraic groups

From $$\S$$3.3.1 ibid.:

Let $$G$$ be a linear algebraic group. A homomorphism of algebraic groups $$\chi: G\to \Bbb G_m$$ is called a rational character (or simply a character). The set of rational characters is denoted by $$X^*(G)$$. It has a natural structure of abelian group, which we write additively. The characters are regular functions on $$G$$, so lie in $$k[G]$$. By Dedekind's theorem [La2, Ch. VIII, $$\S$$4] the characters are linearly independent elements of $$k[G]$$.

[. . .]

A linear algebraic group $$G$$ is diagonalisable if it is isomorphic to a closed subgroup of some group $$\Bbb D_n$$ of diagonal matrices.

Let $$H$$ be a closed subgroup of diagonalisable linear algebraic group $$G$$. Define

$$H^\bot=\{\chi\in X^*(G)\mid \chi(H)=\{1\}\}.$$

The previous part of the exercise can be phrased like so:

Denote by $$G_n$$ the subgroup of elements of $$G$$ of order dividing $$n$$ and $$\gcd(n,p)=1$$. Then $$(G_n)^\bot=nX$$.

(I do not know how to prove this part, but my supervisor suggested the part in question can be done without recourse to this one.)

## Context:

I don't know what to do. Therefore, to provide context, I will answer the questions listed here:

• What are you studying?

A postgraduate research degree in linear algebraic groups.

• What text is this drawn from, if any? If not, how did the question arise?

(See above.)

• What kind of approaches (to similar problems) are you familiar with?

For recent questions of mine from the same exercise set in the book, see

The second one salvages the first.

• What kind of answer are you looking for? Basic approach, hint, explanation, something else?

A full answer would be preferable. If someone could describe the basic approach to solving it, that would be great though.

• Is this question something you think you should be able to answer? Why or why not?

No; at least, not yet. My topology skill level is quite low.

It's difficult to know which (equivalent) definition of dense to use. (Springer doesn't specify which.)

I have spent a good few hours on it (spread over a few weeks; I had COVID recently, so was out of action then) and so would like to move on.

• It's fairly trivial to show that the torsion subgroup is indeed a subgroup Commented Jun 27, 2023 at 22:21
• I suppose I could look at $G=k^*$. Intuitively, the result makes sense. Commented Jun 27, 2023 at 22:26

Consider a closed immersion $$f: G \rightarrow \mathbb{G}_m^{\oplus n}$$. By your question 1, this induces a surjection on rational characters. By the structure theorem of abelian groups, there exists a basis $$(\chi_1,\ldots,\chi_n)$$ of $$X^{\ast}(\mathbb{G}_m^{\oplus n})$$ and integers $$d_1\mid \ldots \mid d_n$$ such that $$(\chi_i)_{|G}$$ has order $$d_i$$, and $$(d_1\chi_1,\ldots,d_n\chi_n)$$ is a basis of the kernel of $$X^{\ast}(\mathbb{G}_m^{\oplus n}) \rightarrow X^{\ast}(G)$$.

Then, after considering the closed immersion $$(\chi_1,\ldots,\chi_n): G \rightarrow \mathbb{G}_m^{\oplus n}$$, we can assume that $$\chi_i$$ is simply the $$i$$-th projection.

By Dedekind’s theorem and because the characters are a $$k$$-generating family of $$k[\mathbb{G}_m^{\oplus n}]$$, the characters of $$G$$ are a $$k$$-basis of $$k[G]$$.

It follows that $$k[G]$$ is exactly $$k[\mathbb{G}_m^{\oplus n}]/(\chi_i^{d_i}-1)$$. In other words, $$G \cong \bigoplus_i \mu_{d_i}$$, with $$\mu_0=\mathbb{G}_m$$.

Now, let’s show that every character of $$\bigoplus_i \mu_{d_i}$$ that vanishes (ie evaluates to $$1$$) on points of finite order vanishes at every point.

(Note: here, I’m using “vanish” only in the sense of “is $$1$$ when evaluated at every point”. In positive characteristic, groups aren’t always reduced, so characters can be unipotent, so a “vanishing” character need not be the constant character equal to $$1$$. A typical example of a “vanishing” character would be the inclusion $$\mu_p \subset \mathbb{G}_m$$ over $$\overline{\mathbb{F}_p}$$.)

Indeed, let $$\chi$$ be such a character.

Consider $$\chi_j: \mu_{d_j} \rightarrow \bigoplus_i \mu_{d_i} \overset{\chi}{\rightarrow}\mathbb{G}_m$$: this character also vanishes on points of finite order – but points of finite order are Zariski-dense in $$\mu_{d_i}$$, so this character vanishes at every point of $$\mu_{d_i}$$.

Since $$\chi$$ is the product of $$\chi_j \circ p_j$$, where $$p_j$$ is the projection on the $$j$$-th factor, $$\chi$$ vanishes at every point.

Finally, let $$f \in k[G]$$ vanish (ie evaluates to zero, from now on) at every point of finite order: we need to show that $$f$$ is nilpotent. We know that $$f$$ is a $$k$$-linear combination of characters.

If two characters $$\alpha,\beta$$ evaluate to the same value at every point of finite order, their quotient $$q$$ evaluates to $$1$$ at every point of finite order, hence at every point, and $$q-1$$ is nilpotent, hence $$\alpha-\beta$$ is nilpotent.

So up to replacing $$f$$ with $$f+\nu$$ for some nilpotent function $$\nu$$, we can assume that the characters restrict to pairwise distinct group homomorphisms $$G(k)_{tors} \rightarrow k^{\times}$$; moreover, $$f$$ is a linear combination of them that vanishes at every point of $$G(k)_{tors}$$. By independence of characters (on a “classical” aka non-algebraic group), the linear combination is zero, so $$f$$ is zero and we are done.

Edit: here are other ways to deduce the result from the “decomposition” of $$G$$ established earlier.

1. (simplified version of the above)

Let $$H$$ be the Zariski-closure of the torsion points of $$G$$. It’s a subgroup of $$G$$.

Consider the pull-back of $$H$$ through the map $$\mu_{d_j} \rightarrow \bigoplus_i{\mu_{d_i}}=G$$. It’s a closed subgroup of $$\mu_{d_j}$$ containing all its torsion points, so it’s all of $$\mu_{d_j}$$: hence $$H$$ contains the $$\mu_{d_j}$$ summand.

Since $$H$$ is a subgroup, it contains the sum of the $$\mu_{d_j}$$, which is $$G$$, hence $$H=G$$.

1. (a counting argument)

The normal form shows that, for any linear diagonalizable group $$G$$, there is an integer $$N_G$$ coprime to $$p$$ such that for every multiple $$n$$ of $$N_G$$ coprime to $$p$$, $$G$$ has exactly $$|\pi_0(G)|n^{\dim{G}}$$ $$n$$-torsion points.

Applying this result to $$H$$ and $$G$$, there are infinitely many $$n$$ such that $$|\pi_0(H)|n^{\dim{H}}=|\pi_0(G)|n^{\dim{G}}$$. It follows that $$H$$ and $$G$$ have the same dimension and the same number of connected components.

Let $$H^0$$ be the intersection of $$H$$ and the connected component of unity $$G^0$$ of $$G$$. Then $$H^0$$ is a closed subgroup of $$G^0$$ and an open subgroup of $$H$$.

Moreover, $$\dim{H_0}=\dim{H}=\dim{G}=\dim{G_0}$$. Since $$G$$ is a connected group, it is irreducible, hence $$H_0=G_0$$.

Then $$H$$ is a reunion of (disjoint) cosets of $$G^0$$. Since the same holds for $$G$$, and since $$H$$ and $$G$$ have the same number of connected components, $$H=G$$.

• Thank you! However, I don't yet know what the $\mu_{d_i}$ are. I've looked in Borel, Humphreys, Springer, and de Graaf so far, albeit briefly, yet found nothing on them; is it standard notation and what does it mean? Commented Jul 3, 2023 at 22:25
• I just meant the $d_i$-th roots of unity (as in, the kernel of the $d_i$-th power, from $\mathbb{G}_m$ to itself). I thought that was standard, but I’m not working in algebraic groups, so I don’t know too well what’s standard. Commented Jul 3, 2023 at 22:30
• My supervisor verified this answer for me, though I still don't understand it. He's not surprised, given that it uses ideas I haven't encountered before. I cannot, therefore, accept it just yet. I might give you the bounty in the grace period though. Commented Jul 6, 2023 at 0:14
• I see. I’m not sure if it helps, but I added in a simplified version of the argument and a different proof once you have the decomposition as a sum of $\mu_{d_i}$. Commented Jul 6, 2023 at 14:16