Why is the Weil restriction of the multiplicative group a torus? In Serre's book about abelian $\ell$-adic representations, he claims that $\textrm{Res}_{K/\mathbb{Q}} (G_{m})$ is a $d$-dimensional torus if $[K:\mathbb{Q}] = d$. Why is this true?
According to the example in these notes (http://alpha.math.uga.edu/~pete/SC5-AlgebraicGroups.pdf), $\textrm{Res}_{\mathbb{C}/\mathbb{R}} (G_m)$ is isomorphic to $\left\{(Y_1, Y_2) \neq (0, 0)\right\}$. Wouldn't a two-dimensional torus be $\left\{(Y_1, Y_2): Y_1, Y_2 \neq 0\right\}$? Are these two sets actually isomorphic as varieties? Or does it matter that this is about $\mathbb{C}/\mathbb{R}$ instead of $K/\mathbb{Q}$?
 A: Recall the definition of a torus over $k$ is an algebraic group $T$ defined over $k$ so that after base-changing up to the algebraic closure $\overline{k}$, $T_{\overline{k}}$ is isomorphic to a finite product of $\Bbb G_m$ defined over $\overline{k}$.
The central issue here for you is that over fields that are algebraically closed, there is exactly one torus of each dimension up to isomorphism. On the other hand, if you are working over a field which is not algebraically closed, there can be multiple non-isomorphic tori of a given dimension. Your second example shows that you've found two different torii of dimension 2 over $\Bbb R$.
To answer your original question, you may compute that after base changing to $\overline{\Bbb Q}$, the torus $Res(K)$ is isomorphic to $d$ copies of $\Bbb G_m$.
A: Looking at your example over $\Bbb C$ and $\Bbb R$,
it is the affine variety
$$\{(x,y):x^2+y^2\ne0\}$$
which is two dimensional. The group operation is $(x_1,y_1)\cdot(x_2,y_2)
=(x_1x_2-y_1y_2,x_1y_2+x_2y_1)$. Over the ground field $\Bbb C$ it
is isomorphic to $\{(z_1,z_2):z_1z_2\ne0\}\cong G_m\times G_m$ (it splits).
But over $\Bbb R$ it isn't isomorphic to $G_m\times G_m$ (it's a nonsplit
torus).
A: To show that the Weil restriction $\operatorname{Res}_{L/K}(\mathbb{G}_m)$ is a torus, where $L/K$ is a finite (separable) extension of fields, you can show that the base change to some field $L'$ is isomorphic to a product of $\mathbb{G}_m$. In fact the smallest such field is the normal closure $L'$ of $L/K$, so $L'/K$ is Galois. 
To show this you can use a general result on Weil restriction saying that, if $X_K$ is a variety over $K$ (or a general scheme also works), and you base change to $L$, take Weil restriction and base change to $L'$ you get $[L:K]$ copies of $X_{L'}$:  $$\operatorname{Res}_{L/K}(X_K\otimes_K L)\otimes_K L'\cong (X_K\otimes_K L')^{[L:K]}.$$
This result can be proved using the Yoneda lemma and the functorial interpretation of the Weil restriction as the right-adjoint of the base change functor. 
A: Here's a more intuitive proof that $\textrm{Res}_{K/\mathbb{Q}} (G_{m/K}) \otimes_{\mathbb{Q}} K$ is just $d$ copies of $G_{m/\mathbb{Q}}$. 
To restrict scalars for $G_{m/K}$, we have to use the $\mathbb{Q}$-basis for $K$, which we can write as $\alpha_1, \ldots, \alpha_d$. The multiplication takes place by rewriting each $\alpha_i\alpha_j$ with respect to the basis of the $\alpha_i$. (This is all defined over $\mathbb{Q}$.) 
The important point is that $x =c_1\alpha_1 + \ldots + c_d\alpha_d$ lies in  $\textrm{Res}_{K/\mathbb{Q}} (G_{m/K})$, for $c_i \in \mathbb{Q}$, if and only if $N(x) \neq 0$. How do we write $N(x)$ in terms of the $c_i$? If the automorphisms of $K$ are given by $\textrm{id} = \sigma_1, \ldots, \sigma_d$, then $N(x) = \prod\sigma_i(x)$. Since each $\sigma_i$ is given by some $\mathbb{Q}$-linear transformation of $K$, the product is a polynomial in $N(x)$. 
Now, to think about $\textrm{Res}_{K/\mathbb{Q}} (G_{m/K}) \otimes_{\mathbb{Q}} K$, we have to allow the $c_i$ to take values in $K$, not just $\mathbb{Q}$. Even though $N(x)$ is irreducible over $\mathbb{Q}$ when considered as a polynomial in the $c_i$, over $K$ it factors, by definition, in an extremely simple way. In particular, the condition $N(x) \neq 0$ is equivalent to the condition $\sigma_i(c_1\alpha_1 + \ldots + c_d\alpha_d) \neq 0$ for each $i$. Over $K$, this is just a linear condition in the $c_i$; therefore, after a change of basis, $\textrm{Res}_{K/\mathbb{Q}} (G_{m/K}) \otimes_{\mathbb{Q}} K$ is $d$ copies of the torus defined by $x\neq 0$. 
