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

Question

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

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

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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

1. If a hom. $\phi:G\to H$ of diagonalisable linear algebraic groups is injective, then the induced hom. $\phi^*:X^*(H)\to X^*(G)$ is surjective
2. Salvaging Exercise 3.2.10(2) of Springer's, "Linear Algebraic Groups (Second Edition)".

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.

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Please help :)

Answer 1

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.

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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 $H=G$.

2. (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, $H=G$.