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I have the following definition: Suppose that $g \in \mathrm{GL}_n(k)$ is regular and semisimple. Define $T_g := \mathrm{Cent}_{\mathrm{GL}_n(k)}(g)$ to be the centralizer of $g$ in $\mathrm{GL}_n(k)$. My advisor and I managed to proof that we have $T_g \cong k[g]^\times$ as multiplicative subgroups of $\mathrm{GL}_n(k)$.

Now my advisor claims that $T_g$ is in fact an algebraic torus, i.e. a finite product of copies of $k^\times$, but how so? The multiplication in a product $(k^\times)^r$ is defined entrywise, but the action of $g$ on $k[g]^\times$ is very weird, especially $g \cdot g^{n-1}$.

Gargantuar
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    See if you can work it out explicitly for $n = 2$ first. – Sammy Black Mar 03 '24 at 21:09
  • Are you writing $k[g]$ as shorthand for the group algebra over $k$ of the cyclic subgroup generated by $g$? And is $k$ algebraically closed here, or at least contains all roots of unity? – Alex Wertheim Mar 03 '24 at 21:15
  • @AlexWertheim As far as $k[g]$ is concerned, I think yes, similar to the notation for polynomial rings (note that $\langle g\rangle$ has order $n$). And as far as $k$ itself is concerned: Definitely not. Actually, $k$ will be a $p$-adic local field. – Gargantuar Mar 03 '24 at 21:23
  • @Gargantuar in that case, the unique (surjective) $k$-algebra morphism $k[X] \to k[g]$ sending $X$ to $g$ factors through an isomorphism $k[X]/\langle X^{n}-1\rangle \cong k[g]$. But $k[X]/\langle X^{n}-1\rangle$ is a product of fields, and which fields they are depends on the decomposition of $X^{n}-1$ in $k[X]$. So I don't see how your advisor's claim holds, unless one of the things I mentioned above is true. – Alex Wertheim Mar 03 '24 at 21:31
  • @AlexWertheim Ok, I'm sorry, that was misleading and I made a mistake. $g$ is not simply a group element, but $g$ comes with a relation, namely the one of the characteristic=minimal polynomial $f(X)\in k[X]$. In this case, we have $k[g]\cong k[X]/(f(X))$ as $k$-algebras, and now we look at the units. – Gargantuar Mar 03 '24 at 21:39
  • @Gargantuar ah, I see. In that case, I suppose there's probably a strategy here where you use the properties of $g$ to tell you something about how $f(X)$ splits in $k$, but now we're out of my depth. Sorry I can't help more with that :( – Alex Wertheim Mar 03 '24 at 21:44
  • @SammyBlack Ok, I see at least for the case $n=2$. Suppose that $g=(g_{ij})_{i,j}$ is the matrix, so $k[g]={aI_2+bg:a,b\in k}$. The question is which elements in $k[g]$ are invertible, i.e. when is $\det(aI_2+bg)\ne0$. Solving the normalized $\det(XI_2-g)=\mathrm{char}(g;X)=0$ gives at most two solutions $\lambda_1,\lambda_2\in k$, where $\mathrm{char}$ is the characteristic polynomial. Now map $k[g]^\times\to(k^\times)^2$, $(a,b)\mapsto(a-\lambda_1b,a-\lambda_2b)$ (or to $(k^\times)^1$ if there is only one Eigenvalue). This forces $\mathrm{char}(g;X)$ to never be zero. – Gargantuar Mar 03 '24 at 22:20
  • Being regular semi-simple means that (at least over $\overline{k}$, which you may assume WLOG to check something is a torus) up to conjugation, $g$ is a diagonal matrix with distint entries. Can you then directly check that the centralizer is the diagonal torus? – Alex Youcis Mar 05 '24 at 03:50

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To get it of the unanswered list: the claim that $T_g \cong (k^\times)^r$ is generally wrong. As @Alex Youcis mentioned, one has to pass to an algebraic closure $\bar k$ of $k$, which is also required in the definition of an algebraic torus.

So assume that $T_g \cong k[g]^\times$. Since $g$ is regular semisimple, all its Eigenvalues in $\bar{k}$ are distinct, and so $\mathrm{char}(g;X) \in k[X]$ is already its minimal polynomial. This gives $k[g] \cong k[X] / \mathrm{char}(g;X)$. Moreover, $\mathrm{char}(g;X)$ does not have multiple roots in $\bar k$, whence it has no multiple irreducible factors over $k$. Write $\mathrm{char}(g;X) = f_1(X) \dotsm f_r(X)$ where $f_1, \dots, f_r$ are pairwise distinct and irreducible. Then by the Chinese remainder theorem, $k[g] \cong \prod_{i=1}^r k[X] / f_i(X)$ where $k[X] / f_i(X)$ are fields that are finite over $k$. This is all we can say over $k$.

If we pass to $\bar k$, i.e. if we tensor with $\bar k$, then $\bar k[X] / f_i(X)$ are fields that are finite extensions over $\bar k$. As algebraically closed fields have no proper algebraic (and in particular finite) extensions, we must have $\bar k[X] / f_i(X) = \bar k$ and $r = n$. Thus, $\bar k[g] \cong \bar k^n$.

To obtain $T_g$ in both cases, take the units of the ring $k[g]$, being isomorphic to $\prod_{i=1}^r (k[X] / f_i(X))^\times$ or $(\bar k^\times)^n$ resp.

Gargantuar
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