This might sound like a silly question, but I seem to be getting different definitions from different texts.

Let $k$ be some field and $G$ some linear algebraic group over $k$. Now, the standard definition I have seen, for instance in

Milne, J. S., Algebraic groups. The theory of group schemes of finite type over a field, Cambridge Studies in Advanced Mathematics 170. Cambridge: Cambridge University Press (ISBN 978-1-107-16748-3/hbk; 978-1-316-71173-6/ebook). xvi, 644 p. (2017). ZBL1390.14004.

is that a character is a $k$-group homomorphism $\chi:G\rightarrow \mathbb{G}_m$. But, there is this result in

Platonov, Vladimir; Rapinchuk, Andrei; Rapinchuk, Igor, Algebraic groups and number theory. Volume 1 (to appear), ZBL07671765.

which says that if $G$ is torus, then the following two conditions are equivalent:

  1. $G$ is split
  2. all its characters are defined over $k$.

Yes, Platonov and Rapinchuk define their characters over a much larger algebriacally closed field $\Omega$, but wouldn't this mean that the two definitions of a character and hence the structure of the character group are different? Am I missing something? Are the definitions somehow similar?

  • 1
    $\begingroup$ The field of algebraic groups has had some issues with relying on texts written before schemes, where in order to work over a non-algebraically-closed field, one has to use some workarounds. One of those workarounds is doing everything over an algebraically closed field, but only considering morphisms/subobjects defined over a smaller field. I think there's a good chance that's what's happening in the second definition. $\endgroup$
    – KReiser
    Commented May 3, 2023 at 14:39
  • 1
    $\begingroup$ I can the logic in what you are saying, but the latter book was writter in 1994 and they do alude to schemes in the book, but mention that they will mostly work with classical varieties and that things would match up with you were to work with schemes. $\endgroup$ Commented May 3, 2023 at 14:59
  • 3
    $\begingroup$ Right, bearing in mind that I'm not an expert on this area and would be happy to be corrected by one, the impression I've got from my colleagues who have more experience in algebraic groups is that some fundamental results were written back during the transition to schemes and have not been fully translated to the more modern language. If a book written in '94 relies more heavily on those results and doesn't update them, then you're in a spot where you're still using Weil's foundations (this is heavily suggested by your reference to a much larger acf $\Omega$). Best of luck! $\endgroup$
    – KReiser
    Commented May 3, 2023 at 15:09

1 Answer 1


Yes, please don't feel bad about this confusion. It is for semi-historical reasons as KReiser rightly pointed out above, and can be quite confusing.

Namely, let us fix a field $K$ and an (affine) algebraic group $G$ over $K$. Then, for any separable$\color{red}{^{(\ast)}}$ algebraic extension $L/K$ one can form the group of $L$-rational characters


where these are homomorphisms of group $L$-schemes. There is somewhat varying terminology in this regard, but often people call $X_{K^\mathrm{sep}}$ the group of characters of $G$, and denote it just $X^\ast(G)$ (or $X(G)$). They then might call the elements of $X_K^\ast(G)$ the rational characters of $G$. Although, again, this is author dependent, and so you have to sort of feel out the definition in each case.

To understand the relationship between the rational characters over differing fields, it's useful to introduce some extra structure. If $L/K$ is Galois, with Galois group $\Gamma_{L/K}$, then there is a natural action of $\Gamma_{L/K}$ on $X^\ast_L(G)$ given as follows:

$$\sigma\cdot \chi:= \sigma_G\circ \chi\circ\sigma_{\mathbb{G}_{m,L}}^{-1}.$$

Here for a $K$-scheme $X$ and $\sigma$ in $\Gamma_{L/K}$, I am denoting by $\sigma_X$ the natural morphism $X_L\to X_L$ which is the identity on the $X$-factor and the induced map on $\mathrm{Spec}(L)$ given by $\sigma^{-1}\colon L\to L$ (you can remove the inverse here if you are OK with it being a right, opposed to left, action). It is a good exercise to check that this is a well-defined action (e.g. that $\sigma\cdot\chi$ really can still be made sense of as an element of $X^\ast_L(G)$).

From Galois descent for affine schemes (e.g. see [Poonen,§4.4]) there is a natural identification


for any tower of extensions $L/M/K$ (with $L/K$ Galois). Here I am being somewhat imprecise and identifying $X^\ast_M(G)$ with a subset of $X_L^\ast(G)$ via the (injective) group map

$$X^\ast_M(G)\to X^\ast_L(G),\qquad \chi\mapsto \chi_L,$$

(the base change map).

In particular, one sees that the natural map

$$X_K^\ast(G)\to X^\ast_{K^\mathrm{sep}}(G)=X^\ast(G),$$

is a bijection if and only if $\Gamma_{K^\mathrm{sep}/K}$ acts trivially on $X^\ast(G)$. As you mentioned, if $G$ is a torus, then this is equivalent to $G\cong \mathbb{G}_{m,K}^n$, i.e. that it is split (and in fact, you can use closely related ideas to define the maximal split subtorus -- see [Springer, Proposition 13.2.4]).

I hope this clarifies things!

$\color{red}{(\ast):}$ This is a little too technical to mention above, but you can really have $L/K$ be any algebraic extension. But, if $L/K$ is purely inseparable then the natural map $X^\ast_K(G)\to X_L^\ast(G)$ is a bijection.


[Poonen] Poonen, B., 2017. Rational points on varieties (Vol. 186). American Mathematical Soc..

[Springer] Springer, T.A. and Springer, T.A., 1998. Linear algebraic groups (Vol. 9). Boston: Birkhäuser.

  • $\begingroup$ This is incredibly helpful! Thank you. $\endgroup$ Commented May 4, 2023 at 10:09

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