This is Exercise 3.2.10(2) of Springer's book, "Linear Algebraic Groups (Second Edition)". According to Approach0, it is new to MSE.

The Question:

Let $\phi:G\to H$ be a homomorphism of diagonalisable (linear algebraic) groups (over an algebraically closed field $k$). Denote by $\phi^*$ the induced homomorphism $X^*(H)\to X^*(G)$. If $\phi$ is injective, then $\phi^*$ is surjective.

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.

We can say

$$\begin{align} \phi^*:X^*(H)&\to X^*(G),\\ f &\mapsto f\circ \phi. \end{align}$$


One way to view the initial setup is via the commutative diagram

$\require{AMScd}$ $$\begin{CD} k[H] @> f\mapsto f\circ\phi >> k[G]\\ @A i AA @AA j A\\ X^*(H) @>>\chi \mapsto \chi\circ\phi > X^*(G), \end{CD}$$

where $i,j$ are inclusions.

My supervisor suggested I try examples, a couple of which being:

Example: Let $G=k^*$, $H=k^*\times k^*$, and consider either

$$\begin{align} \phi:=i_1: G&\to H,\\ x &\mapsto (x,1), \end{align}$$


$$\begin{align} \phi:=i_2: G&\to H,\\ x &\mapsto (1,x). \end{align}$$

I didn't get very far. I have a rough idea that we have

Example: Let $G=k^*$, $H=k^*\times k^*$, so that

$$\begin{align} t: G&\to H,\\ x &\mapsto (x^m,x^n) \end{align}$$

is injective if and only if $\gcd(m,n)=1$.

This is also a suggestion from my supervisor.

Another hint from my supervisor is that we can use part of Theorem 3.2.3:

If linear algebraic group $G$ is diagonalisable, then the elements of $X^*(G)$ form a $k$-basis of $k[G]$.

I'm not sure how to use this, though.

Further Context:

I've been stuck on this for weeks, on & off. It has me questioning my aptitude for linear algebraic groups because I have made little progress.

For an idea of my experience, see the following questions of mine:

I hope this is enough context.

Even More Context:

Since I have little to add in terms of attempts, I will answer the questions here:

  • What are you studying?

A PhD in linear algebraic groups, first year.

  • 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?

See above.

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

A full explanation is preferred, please.

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

No. Given the amount of time I have spent on it so far, to no avail, I don't think it's something I can do alone. My supervisor has been helping as well.

Please help :)

  • 3
    $\begingroup$ It's completely ridiculous that this was downvoted $\endgroup$
    – FShrike
    May 26 at 13:49
  • $\begingroup$ @FShrike It's Not! Hater are gonna hate! $\endgroup$ May 26 at 15:51
  • $\begingroup$ This question has a follow-up. $\endgroup$
    – Shaun
    Jun 8 at 21:50

1 Answer 1


Unless I'm missing something here, the exercise is wrong as stated.

Let $k$ of characteristic $p>0$ and let $G=H=\Bbb G_m$. Let $\phi:G \to G$ be the map that raises everything to the $p$-th power. Then $\phi$ is injective (as $p$-th roots in char $p$ are unique), but on $X(G)=\Bbb Z$, $\phi$ induces the "multiplication by $p$"-map, which is not surjective.

So what is going on here?

I think this behaviour is a good example of why in finite characteristic, algebraic groups via classical algebraic geometry (even over an algebraically base closed field) is not the right approach. You just lose too much when you don't allow nilpotents.

Let me sketch you, even if you don't know schemes, why group schemes are "better" in this case. The thing is that the homomorphism I gave above is not "injective" as a morphism of group schemes: it has a kernel. Now describing this kernel might be tricky if you don't know schemes, but you can think of it as the "$p$-th roots of unity", let's call it $\mu_p$.

I think it should be clear that the kernel of raising to the $p$-th power has something to do with $p$-th roots of unity. If you want to think in terms of Hopf algebras, then we have a Hopf algebra $k[T]/(T^p-1)$ (with the comultiplication defined uniquely by $\Delta(T)=T \otimes T$). This is non-reduced, so it won't give rise to an algebraic group. It still gives rise to a perfectly fine group scheme, though. The classical approach can't distinguish this group scheme from the trivial group scheme, because there are no non-trivial $p$-th roots of unity in $k$, but $\mu_p$ is still nontrivial as a group scheme.

One way one can "see" that $\mu_p$ is non-trivial is to use the "functor of points" approach. A group scheme can be identified with (basically) a representable functor on $k$-algebras with values in groups. Now $\mu_p$ represents the functor $A \mapsto \{a \in A \mid a^p=1\}$. If we allow non-reduced $A$ this is clearly distinct from the terminal functor that sends everything to the one-point set.

  • $\begingroup$ This question has a follow-up. $\endgroup$
    – Shaun
    Jun 8 at 21:50

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