I have a pair of related questions about Weil Restriction. Let $E/F$ be a field extension, and let $A$ be an $E-$algebra. Assume that all relevant restrictions of scalars exist. We have a norm map $n: A \rightarrow RA$. I wish to show that for $f \in A$ we have $R(A_f) \simeq (RA)_{n(f)}$ (this is problem 11.4.7(6) in Springer's Linear Algebraic Groups).
Using functoriality, given the map $A \rightarrow A_f$, we have an induced map $RA \rightarrow R(A_f)$ and since $n(f) \in RA$ is mapped to something invertible in $R(A_f)$, by universal properties, the morphism $RA \rightarrow R(A_f)$ factors through $(RA)_{n(f)} \rightarrow R(A_f)$. However, I'm not sure how to proceed here.
A more unified perspective here is probably given by the functors of points and thinking relatively. So here, working over a base scheme $S' \rightarrow S$, for an $S$-scheme $Y$ and an $S'$-scheme $X$ we have
$Hom_{S'}(Y \times_S S', X)=Hom_S(Y, \Pi_{S'/S}X).$
Moreover in this setting, we know that taking restrictions of scalars commutes with base change. So my question is; what is the schematic formulation of the above question? Let $X=Spec(A)$ be an $S'$ scheme. Then my impression is that the base change diagram for localization would be $Spec(A_f) \times_{S'} Spec(A)$.
Then by an easy argument, we have
$\Pi_{S'/S}(Spec(A_f) \times_{S'} Spec(A)) \simeq \Pi(Spec(A_f)) \times_{\Pi(S')} \Pi(Spec(A))$.
Though it seems close, I cannot quite figure out if this is the correct analogue of the statement above. Moreover here it seems I don't even introduce the norm map at all, or use any universal properties about localization, so I am unsure of myself.
Please let me know if you have any comments or if I should clarify anything.
$\newcommand{\Res}{\mathsf{Res}}$ $\newcommand{\Spec}{\mathrm{Spec}}$$\newcommand{\Hom}{\mathrm{Hom}}$
I'd like to tell you a way to think about this that I think is instructive. I hope you feel similarly even though, most likely, it's outside the purview of what Springer is actually discussing.
For the sake of simplicity, let us assume that $E/F$ is Galois and let me denote by $\Gamma$ the group $\mathrm{Gal}(E/F)$ and we denote a general element of $\Gamma$ by $\sigma$ or $\tau$.
Let us begin by defining for all $\sigma$ in $\Gamma$ the $E$-algebra
$$A^\sigma:= A\otimes_{E,\sigma}E$$
where this notation means that we're taking the tensor product of $E$-algebras where $E\to A$ is the structure map and $E\to E$ is the map given by $\sigma$. We consider this an $E$-algebra by defining $e(a\otimes b):=a\otimes (eb)$.
Let us note that we have a map of $F$-algebras
$$\sigma:A\to A^\sigma:a\mapsto a\otimes 1$$
Note though that this map is not $E$-linear. In fact,
$$\sigma(ea)=(ea)\otimes 1 = a\otimes \sigma(e)=\sigma(e)(a\otimes 1)$$
so $\sigma:A\to A^\sigma$ is $\sigma$-linear.
We now consider the $E$-algebra
$$A^{\otimes \Gamma}:=\bigotimes_{\sigma\in\Gamma_F}A^\sigma$$
where the tensor product on the right hand side is a tensor product of $E$-algebras. We shall denote a general simple tensor in $A^{\otimes\Gamma}$ by $\displaystyle \bigotimes a_\sigma$ (i.e. the $\sigma^\text{th}$-coordinate in the simple tensor is $a_\sigma$). Note that $A^{\otimes\Gamma}$ carries a natural $\Gamma$-action by permuting the coordinates or, more explicitly,
$$\tau\left(\bigotimes a_\sigma\right)=\bigotimes b_\sigma,\qquad b_{\tau\sigma}=a_\sigma$$
Note that the action of $\Gamma$ is not $E$ linear, but is $F$-linear.
Let us now consider the $F$-algebra $(A^{\otimes \Gamma})^{\Gamma}$--the $\Gamma$-fixed points of $A^{\otimes \Gamma}$. We have an obvious inclusion of $F$-algebras
$$\iota:(A^{\otimes\Gamma})^{\Gamma}\hookrightarrow A^{\otimes \Gamma}$$
Less obviously is the fact that the induced map of $E$-algebras
$$(A^{\otimes\Gamma})^{\Gamma}\otimes_F E\to A^{\otimes\Gamma}:x\otimes e\mapsto ex$$
is an isomorphism of $E$-algebras. In fact, it's actually an isomorphism of $E$-algebras with an action of $\Gamma$ where $\Gamma$ acts on the source by its action on $E$!
Why is $A^{\otimes\Gamma}$ or $(A^{\otimes{\Gamma}})^{\Gamma}$-important? Well, note that for any $F$-algebra $R$ the obvious map
$$\Hom_F\left((A^{\otimes \Gamma})^\Gamma,R\right)\to \Hom_E\left ((A^{\otimes \Gamma})^\Gamma\otimes_F E,R\otimes_F E\right)^{\Gamma}$$
is a bijection where the $\Gamma$ action on
$$\Hom_E\left ((A^{\otimes \Gamma})^\Gamma\otimes_F E,R\otimes_F E\right)$$
takes a homomorphism $\alpha$ to $\sigma\circ \alpha\circ \sigma^{-1}$ where $\sigma^{-1}$ is acting on $(A^{\otimes \Gamma})^\Gamma\otimes_F E$ by its action on $E$ and the action of $\sigma$ is on $R\otimes_F E$ also acting by $E$.
But, we've already noted that we have an isomorphism
$$(A^{\otimes\Gamma})^{\Gamma}\otimes_F E\to A^{\otimes\Gamma}:x\otimes e\mapsto ex$$
of $E$-algebras with $\Gamma$-action. Thus, we see that
$$\Hom_E\left ((A^{\otimes \Gamma})^\Gamma\otimes_F E,R\otimes_F E\right)^{\Gamma}=\Hom_E\left (A^{\otimes\Gamma},R\otimes_F E\right)^{\Gamma}$$
but what is a $\Gamma$-equivariant map of $E$-algebras $A^{\otimes \Gamma}\to R\otimes_F E$? Well, by the definition of tensor product over $E$, its a collection of maps of $E$-algebras
$$f_\sigma:A^\sigma\to R\otimes_F E$$
where we abbreviate $f_{\mathrm{id}}$ to $f$ such that for any collection $a_\sigma\in A^\sigma$ you have
$$f_\sigma(a_\sigma)=\sigma(f(a))$$
In other words, you see that such data is an entirely determined by $f$.
In other words, summing everything up, there is a natural series of bijections
$$\begin{align}\Hom_F\left((A^{\otimes\Gamma})^\Gamma,R\right) &= \Hom_E\left((A^{\otimes\Gamma})^\Gamma\otimes_F E,R\otimes_F E\right)^\Gamma\\ &= \Hom_E(A^{\otimes\Gamma},R\otimes_F E)^\Gamma\\ &= \Hom_E(A,R\otimes_F E)\end{align}$$
or, in other words, we have shown that
$$\Res_{E/F}\Spec(A)=\Spec\left((A^{\otimes\Gamma})^\Gamma\right)$$
More explicitly we have a bijection
$$J:\Hom_F\left((A^{\otimes\Gamma})^\Gamma,R\right)\xrightarrow{\approx}\Hom_E(A,R\otimes_F E)$$
given by taking $f$ to $(f\otimes 1)\mid_A$.
Now, what does this have to do with the norm map?Note we have a multiplicative map
$$N:A\to A^{\otimes \Gamma}:a\mapsto \bigotimes \sigma(a)$$
which we call the norm map. Note that this map is not additive but does have image in $(A^{\otimes\Gamma})^\Gamma$.
Thus, if $a\in A$ then $N(a)\in (A^{\otimes\Gamma})^\Gamma$. Given our above discussion it's now easy to verify that
$$\Res_{E/F}(D(a))=D(N(a))$$
Indeed, what is a map of $F$-schemes
$$\Spec(R)\to D(N(a))$$
but a map of $F$-algebras
$$(A^{\otimes\Gamma})^\Gamma\to R$$
such that $N(a)$ maps to a unit. What is a map of $E$-schemes
$$\Spec(R\otimes_F E)\to\Spec(A)$$
but a map of $E$-algebras
$$A\to R\otimes_F E$$
Note then that under our above bijection
$$\Hom_F\left((A^{\otimes\Gamma})^\Gamma,R\right)\xrightarrow{\approx}\Hom_E(A,R\otimes_F E)$$
one sees that
$$f(N(a))=J(f)(a)$$
and so $f(N(a))$ is a unit iff $J(f)(a)$ is. This is precisely what we want.
Let me say one last word on what the relationship is between the norm map
$$N:A\to (A^{\otimes\Gamma})^\Gamma$$
and more familiar versions of the norm map. Namely, let us suppose that there is some $F$-algebra $B$ such that $A=B\otimes_F E$. Then, we note that we have an isomorphism of $E$-algebras
$$A^\sigma=B\otimes_F E\otimes_{E,\sigma}E\to A=B\otimes_F E:(b\otimes e_1)\otimes e_2\mapsto b\sigma(e_1)e_2$$
With this, one can show that there is an isomorphism of $E$-algebras
$$A^{\otimes \Gamma}\cong B\otimes_E E^{\otimes\Gamma}\cong A\otimes_E (E^{|\Gamma|})\cong B^{|\Gamma|}$$
Moreover, it's not hard to check then that the natural map
$$B\to (A^{\otimes\Gamma})^\Gamma$$
is an isomorphism. Thus, we see that the norm map is a map
$$N:B\otimes_F E=A\to (A^{\otimes\Gamma})^\Gamma=B$$
Now, let $x\in B\otimes_F E$ act on the left of $B\otimes_F E$ by left multiplication denote this by $\ell_x$. Since $B\otimes_F E$ is a free $B$-module of finite rank, one can consider $\det(\ell_x)\in B$. Then, one has , under the above identifications, that $N(x)=\det(\ell_x)$.