Does there exist a non-quasi-split torus? In a homework, I was asked to prove that any torus is isomorphic to a quotient of a finitely many product of Weil restrictions $Res_{L/k}\mathbb{G}_m$. While solving this, I got an impression that almost all torus is actually isomorphic to direct product of Weil restrictions, so the word "quotient" is unnecessary. I think it reduces to showing that every integral representation $\mathbb{Z}^n$ of a finite group $G$ can be decomposed into representation generated by a single element.
I would appreciate if someone can answer this question positively or negatively.
Thank you!
 A: Sorry for a false statement. Not every cyclic module $M$ over $\mathbb{Z}[Gal(k^{sep}/k)]$ occurs as a Weil restriction $Res_{L/k}\mathbb{G}_m$ for some $L/k$. For example, we have $Spec \mathbb{R}[x,y]/(x^2+y^2-1)$, which corresponds to the sign representation of $Gal(\mathbb{C}/\mathbb{R})=\mathbb{Z}/2\mathbb{Z}$. This is a torus because its base change over $\mathbb{C}$ is $\mathbb{G}_m$, but neither $Res_{\mathbb{C}/\mathbb{R}} \mathbb{G}_m$ nor $\mathbb{G}_{m,\mathbb{R}}$ is isomorphic to this. 
Also, there is a $\mathbb{Z}[G]$-module for a finite group $G$ with an underlying abelian group $\mathbb{Z}^n$, which is indecomposable, but not cyclic. Let $\mathbb{Z}[S_3]$ acts on $\mathbb{Z}^3$ by $$(1,2)(a,b,z)=(b,a,z),\quad(1,2,3)(a,b,z)=(b-a-z,-a-2z,z).$$ This is an indecomposable representation. This follows from the classification of indecomposable representation of $\mathbb{Z}[S_3]$, and this is done in Lee, Myrna Pike. "Integral representations of dihedral groups of order 2p." Trans. Amer. Math. Soc. 110 (1964) 213–231. MR 156896 doi:10.2307/1993702
This cannot be cyclic, since any elements of the form $(X,Y,0)$ in $\mathbb{Z}[S_3][(a,b,z)]$ has a fixed $X+Y\pmod{3}$.
