Non-representability of Weil Restriction Let $f \colon R \to S$ be a ring homomorphism and $X$ an $S$-scheme. We define the functor $R_{S/R}(X)$ on $R$-algebras $T$ by
$$
R_{S/R}(X)(T) = X(T \otimes_R S).
$$
If $S/R$ is a field extension (more generally, finite and locally free), and $X$ is quasi-projective, then this functor is known to be representable.
Is there an example where this is not representable? Can one find an example that is smooth and finite type over $S$? What about if $X$ is a simple non-separated scheme, such as the affine line with two origins?
 A: Let $R =\mathbb R$, $S = \mathbb C$ (any quadratic field extension works) and let $X$ be the doubled line. 
We have a map of functions from the Weil restriction of $X$ to the Weil restriction of $\mathbb A^1$, which is $\mathbb A^2$. So if the representing scheme exists then it must map to $\mathbb A^2$.
Now take $T = \mathbb C[x,y]$ so $T \otimes_R S = \mathbb C[x,y] \times \mathbb C[x,y]$. Consider the map from $T \otimes_R S$ to $X$ where the coordinate on the affine line is $ (x+iy, x-iy)$ but on the first component we map to the affine line with the first origin and on the second we map to the affine line with the second origin. 
This defines a map from $T$ to the Weil restriction. If the Weil restriction is a scheme then there must be an open affine $\operatorname{Spec} T'$ in $\operatorname{Spec}  T$, that maps to an open affine $\operatorname{Spec} U$ in the Weil restriction, which maps to the Weil restriction, which maps to $\mathbb A^2$. Dualizing we have maps of rings $\mathbb R[x,y] \to \mathbb C[x,y] = T \to T'$ and $\mathbb R[x,y] \to U \to T'$ forming a commutative diagram. 
The image of $U$ inside $T'$ either is contained in $\mathbb R(x,y)$ or not. In either case we will derive a contradiction.
If the image is contained in $\mathbb R[x,y]$, then because $\operatorname{Spec} 
 T'$ contains the origin, the image must be a subring of $\mathbb R(x,y)$ that the map $\mathbb R[x,y] \to \mathbb C$ sending $x$,$y$ to $0$ factors through. Thus it must be $\mathbb R[x,y]$ adjoining some rational functions that are well-defined at $0$. So in fact we have a factorization $\mathbb R[x,y] \to U \to \mathbb R \to \mathbb C$ sending $x,y$ to $0$. But this implies that the $T$-point of the Weil restriction we wrote down earlier, restricted to the origin, descends to $\mathbb R$, which would make it Galois-invariant. But it is not Galois-invariant, because the two components, exchanged by Galois, map to the two different origins.
If the image is not contained in $\mathbb R$, then it contains some rational function $f$ in $\mathbb C(x,y)$. Choose some $u,v \in \mathbb R$, not both $0$, such that the point $(u,v) \in \mathbb A^2$ lies in $\operatorname{Spec}  T'$, and for which $f(u,v)$ is well-defined but not an element of $\mathbb R$. This is easy because these are all generic conditions. For such an $x,y$, the induced map  $\mathbb R[x,y] \to U \to T' \to \mathbb C$ sending $x$ to $u$ and $y$ to $v$, sends $f$ to $f(x,y)\notin \mathbb R$ and thus has image $\mathbb C$.
Now if the map from the ring of functions of any affine neighborhood of a $\mathbb C$-point of a scheme to $\mathbb C$ has image $\mathbb C$, that point does not descend to $\mathbb R$ (because that can be checked affine-locally). However, our chosen point $(u,v)$ does descend to $\mathbb R$, because $u, v \in \mathbb R$ and $u,v$ are not both $0$ so $u+iv, u-iv \neq 0$, and thus we lie in the affine line with doubled origin. So this contradicts our conclusion that the image of $U$ in $\mathbb C$ is $\mathbb C$.
