This uses the tag because there might be more than one valid answer, and it's a matter of guesswork to some extent; but there is a right answer (in theory).

Thoughts and Motivation:

As a follow-up to

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

(whose notation I use here) in which I asked for a proof of Springer's exercise . . .

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.

. . . I asked my supervisor for clarification on what the intended exercise might be.

We spoke about distinctions in category theory between "injective homomorphism" and "monomorphism" and he suggested the amendment that went something like,

"Injective" should be "closed embedding onto its image."

But this seemed like a guess. Here "closed" is with respect to the Zariski topology, of course.

I thought about this for a while and it makes sense, although, as before, I cannot prove the adjusted version.

My suspicion is that it holds only for $\operatorname{char}(k)=0$. That's independent of my supervisor's stipulation. But, again, I cannot prove this.

So . . .

The Question:

What might the intended exercise be? That is, how might we adjust the exercise as stated in the book so that it's correct?

NB: I'm aware that those two questions (below The Question) could have different answers. I just thought they complement each other nicely and may indeed end up equivalent; if not, I'm interested in the latter, given its potential to be more general and therefore of more use to the community.

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

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


Diagonalisable groups are key to the theory of linear algebraic groups, and I am embarking on research into the latter; my PhD studies so far are mostly in learning the area and I'm still finding my feet.

Please help :)


1 Answer 1


Replacing "injective" with closed immersion works. In fact, one could take any of the following, because they're all equivalent for a morphism $\phi:G \to H$ of linear algebraic groups:

  • $\phi$ is a closed immersion.
  • $\phi$ is "injective as a group functor", meaning that for every $k$-algebra $R$, we have that $G(R) \to H(R)$ is injective.
  • $\phi$ is a monomorphism in the category of algebraic groups over $k$.
  • $\phi$ is a monomorphism in the category of varieties over $k$.
  • $\phi$ is a monomorphism in the category of $k$-group schemes.
  • The kernel of $\phi$ as a group scheme is trivial.

For proofs, see Milne - Algebraic Groups, Section 5d

Also let me note that in characteristic $0$, the kernel on the group scheme level is actually reduced and hence an algebraic group, so in characteristic $0$ this is also equivalent to simply injective.

  • $\begingroup$ Thank you. I would like to construct/see a proof someday of the intended exercise. I wonder whether these conditions could be weakened in some way . . . Perhaps I'll ask questions along these lines soon. $\endgroup$
    – Shaun
    Commented Jun 9, 2023 at 12:52
  • 1
    $\begingroup$ @Shaun here's a proof idea: For example in characteristic $0$, the category of diagonalizable algebraic groups is dual to the category of finitely generated abelian groups via the functor $G \mapsto X^*(G)$, meaning that the $\mathbf{DiagGrps}_{k}^{op}$ is equivalent to $\mathbf{ab}$ via this functor (the categories being those that I mentioned). Now if $\phi:G \to H$ is a monomorphism, then $X^*(H) \to X^*(G)$ will be an epimorphism of f.g. abelian groups by abstract nonsense. But epimorphisms of (f.g.) abelian groups are the surjective homomorphisms. For char p this can be modified. $\endgroup$ Commented Jun 9, 2023 at 12:57
  • $\begingroup$ Excellent! Thank you, again. Assuming nobody else answers, the bounty is as good as yours! $\endgroup$
    – Shaun
    Commented Jun 9, 2023 at 14:29
  • $\begingroup$ Please let me know where I can find a proof of this duality, @LukasHegar :) $\endgroup$
    – Shaun
    Commented Jun 9, 2023 at 14:54
  • 1
    $\begingroup$ @Shaun the functor in the other direction takes a f.g. abelian group $A$ and constructs the group algebra $k[A]$ as a Hopf algebra with comultiplication given on basis elements by $a \mapsto a \otimes a$. One can find a version of this in Milne's Algebraic Groups. In char $0$ it reduces to what I said, in char $p$ it involves some non-reduced group schemes, but he tells you which f.g. abelian groups correspond to reduced group schemes, so one can set up a duality between certain subcategories. $\endgroup$ Commented Jun 9, 2023 at 15:09

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