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 $g \cdot g^{n-1}$.
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.