[proofplan] We prove both adjunctions by writing down the hom-set bijections explicitly and verifying that the formulas are well-defined, linear, inverse to each other, and natural. For extension of scalars, the map is evaluation on pure tensors of the form $1_S \otimes m$. For coextension of scalars, the map is evaluation at $1_S$, and the inverse sends an $R$-[linear map](/page/Linear%20Map) $g:\varphi^*N \to M$ to the $S$-linear map $n \mapsto (s \mapsto g(sn))$. [/proofplan]

[step:Make the extension of scalars module structure explicit] Regard $S$ as an $(S,R)$-bimodule by left multiplication on $S$ and by the right $R$-action \begin{align*} s \cdot r := s\varphi(r) \end{align*} for $s \in S$ and $r \in R$. For a left $R$-module $M$, the [tensor product](/page/Tensor%20Product) $S \otimes_R M$ is therefore defined, and it becomes a left $S$-module by \begin{align*} a \cdot (s \otimes m) := as \otimes m \end{align*} for $a,s \in S$ and $m \in M$. This action is well-defined because it respects the tensor relation: \begin{align*} a \cdot ((s\varphi(r)) \otimes m) &= as\varphi(r) \otimes m \\ &= as \otimes rm \\ &= a \cdot (s \otimes rm). \end{align*} Thus the functor \begin{align*} S \otimes_R - : R\operatorname{-Mod} \to S\operatorname{-Mod} \end{align*} is well-defined on objects, and on morphisms it sends an $R$-linear map $u:M \to M'$ to the $S$-linear map \begin{align*} S \otimes_R u: S \otimes_R M &\to S \otimes_R M' \\ s \otimes m &\mapsto s \otimes u(m). \end{align*} [/step]

[step:Construct the extension of scalars adjunction]Let $M$ be a left $R$-module and let $N$ be a left $S$-module. Define \begin{align*} \Phi_{M,N}: \operatorname{Hom}_S(S \otimes_R M,N) &\to \operatorname{Hom}_R(M,\varphi^*N) \\ F &\mapsto \bigl(m \mapsto F(1_S \otimes m)\bigr). \end{align*} If $F:S \otimes_R M \to N$ is $S$-linear, then $\Phi_{M,N}(F)$ is $R$-linear because, for $r \in R$ and $m \in M$, \begin{align*} \Phi_{M,N}(F)(rm) &= F(1_S \otimes rm) \\ &= F(\varphi(r) \otimes m) \\ &= \varphi(r)F(1_S \otimes m) \\ &= r \cdot \Phi_{M,N}(F)(m) \end{align*} in the restricted $R$-module $\varphi^*N$. Conversely, define \begin{align*} \Psi_{M,N}: \operatorname{Hom}_R(M,\varphi^*N) &\to \operatorname{Hom}_S(S \otimes_R M,N) \end{align*} as follows. For an $R$-linear map $g:M \to \varphi^*N$, let $\Psi_{M,N}(g)$ be the map induced by the $R$-balanced function \begin{align*} B_g:S \times M &\to N \\ (s,m) &\mapsto s g(m). \end{align*} The function $B_g$ is balanced because, for $s \in S$, $r \in R$, and $m \in M$, \begin{align*} B_g(s\varphi(r),m) &= s\varphi(r)g(m) \\ &= s g(rm) \\ &= B_g(s,rm). \end{align*} Thus there is a unique group homomorphism \begin{align*} \Psi_{M,N}(g): S \otimes_R M &\to N \\ s \otimes m &\mapsto s g(m). \end{align*} It is $S$-linear because, for $a,s \in S$ and $m \in M$, \begin{align*} \Psi_{M,N}(g)(a \cdot (s \otimes m)) &= \Psi_{M,N}(g)(as \otimes m) \\ &= asg(m) \\ &= a \cdot \Psi_{M,N}(g)(s \otimes m). \end{align*} Finally, the two maps are inverse. If $F:S \otimes_R M \to N$ is $S$-linear, then for $s \in S$ and $m \in M$, \begin{align*} \Psi_{M,N}(\Phi_{M,N}(F))(s \otimes m) &= s\Phi_{M,N}(F)(m) \\ &= sF(1_S \otimes m) \\ &= F(s \otimes m). \end{align*} If $g:M \to \varphi^*N$ is $R$-linear, then for $m \in M$, \begin{align*} \Phi_{M,N}(\Psi_{M,N}(g))(m) &= \Psi_{M,N}(g)(1_S \otimes m) \\ &= g(m). \end{align*} Hence \begin{align*} \operatorname{Hom}_S(S \otimes_R M,N) \cong \operatorname{Hom}_R(M,\varphi^*N). \end{align*}[/step]

[guided]We want to prove that $S \otimes_R -$ is left adjoint to restriction of scalars. The universal property should say that giving an $S$-linear map out of $S \otimes_R M$ is the same as giving an $R$-linear map out of $M$. Let $M$ be a left $R$-module and let $N$ be a left $S$-module. Define \begin{align*} \Phi_{M,N}: \operatorname{Hom}_S(S \otimes_R M,N) &\to \operatorname{Hom}_R(M,\varphi^*N) \\ F &\mapsto \bigl(m \mapsto F(1_S \otimes m)\bigr). \end{align*} The only point needing verification is $R$-linearity. For $r \in R$ and $m \in M$, the tensor relation gives \begin{align*} 1_S \otimes rm = \varphi(r) \otimes m. \end{align*} Therefore, using $S$-linearity of $F$, \begin{align*} \Phi_{M,N}(F)(rm) &= F(1_S \otimes rm) \\ &= F(\varphi(r) \otimes m) \\ &= \varphi(r)F(1_S \otimes m) \\ &= r \cdot \Phi_{M,N}(F)(m), \end{align*} where the last equality is precisely the restricted $R$-module structure on $\varphi^*N$. Now start with an $R$-linear map $g:M \to \varphi^*N$. We define an $S$-linear map on the tensor product by the expected formula \begin{align*} s \otimes m \mapsto s g(m). \end{align*} To justify that this descends to the tensor product, define \begin{align*} B_g:S \times M &\to N \\ (s,m) &\mapsto s g(m). \end{align*} For $s \in S$, $r \in R$, and $m \in M$, \begin{align*} B_g(s\varphi(r),m) &= s\varphi(r)g(m) \\ &= s g(rm) \\ &= B_g(s,rm), \end{align*} where the middle equality uses $R$-linearity of $g$ after restricting scalars on $N$. Hence $B_g$ is $R$-balanced and induces a unique map \begin{align*} \Psi_{M,N}(g): S \otimes_R M &\to N \\ s \otimes m &\mapsto s g(m). \end{align*} It is $S$-linear because multiplying the tensor by $a \in S$ on the left gives \begin{align*} \Psi_{M,N}(g)(a \cdot (s \otimes m)) &= \Psi_{M,N}(g)(as \otimes m) \\ &= asg(m) \\ &= a \cdot \Psi_{M,N}(g)(s \otimes m). \end{align*} The formulas are inverse to each other. For $F:S \otimes_R M \to N$ and a pure tensor $s \otimes m$, \begin{align*} \Psi_{M,N}(\Phi_{M,N}(F))(s \otimes m) &= s\Phi_{M,N}(F)(m) \\ &= sF(1_S \otimes m) \\ &= F(s \otimes m), \end{align*} and pure tensors generate $S \otimes_R M$ as an abelian group. Conversely, for $g:M \to \varphi^*N$, \begin{align*} \Phi_{M,N}(\Psi_{M,N}(g))(m) &= \Psi_{M,N}(g)(1_S \otimes m) \\ &= g(m). \end{align*} So the hom-sets are naturally bijective.[/guided]

[step:Make the coextension of scalars module structure explicit] For a left $R$-module $M$, regard $S$ as a left $R$-module by \begin{align*} r \cdot s := \varphi(r)s \end{align*} for $r \in R$ and $s \in S$. Define a left $S$-module structure on $\operatorname{Hom}_R(S,M)$ by \begin{align*} (s \cdot f)(t) := f(ts) \end{align*} for $s,t \in S$ and $f \in \operatorname{Hom}_R(S,M)$. First, $s \cdot f$ is $R$-linear. For $r \in R$ and $t \in S$, \begin{align*} (s \cdot f)(\varphi(r)t) &= f(\varphi(r)ts) \\ &= r f(ts) \\ &= r(s \cdot f)(t). \end{align*} Second, the assignment is a left $S$-action. For $s_1,s_2,t \in S$ and $f \in \operatorname{Hom}_R(S,M)$, \begin{align*} (s_1 \cdot (s_2 \cdot f))(t) &= (s_2 \cdot f)(ts_1) \\ &= f(ts_1s_2) \\ &= ((s_1s_2)\cdot f)(t), \end{align*} and \begin{align*} (1_S \cdot f)(t) = f(t1_S)=f(t). \end{align*} Thus \begin{align*} \operatorname{Hom}_R(S,-): R\operatorname{-Mod} \to S\operatorname{-Mod} \end{align*} is well-defined on objects, and on morphisms it sends an $R$-linear map $u:M \to M'$ to \begin{align*} \operatorname{Hom}_R(S,u): \operatorname{Hom}_R(S,M) &\to \operatorname{Hom}_R(S,M') \\ f &\mapsto u \circ f. \end{align*} This map is $S$-linear because, for $s,t \in S$, \begin{align*} \operatorname{Hom}_R(S,u)(s \cdot f)(t) &= u(f(ts)) \\ &= (s \cdot (u \circ f))(t). \end{align*} [/step]

[step:Construct the coextension of scalars adjunction]Let $N$ be a left $S$-module and let $M$ be a left $R$-module. Define \begin{align*} A_{N,M}: \operatorname{Hom}_S(N,\operatorname{Hom}_R(S,M)) &\to \operatorname{Hom}_R(\varphi^*N,M) \\ F &\mapsto \bigl(n \mapsto F(n)(1_S)\bigr). \end{align*} If $F:N \to \operatorname{Hom}_R(S,M)$ is $S$-linear, then $A_{N,M}(F)$ is $R$-linear because, for $r \in R$ and $n \in N$, \begin{align*} A_{N,M}(F)(rn) &= F(\varphi(r)n)(1_S) \\ &= (\varphi(r) \cdot F(n))(1_S) \\ &= F(n)(1_S\varphi(r)) \\ &= F(n)(\varphi(r)) \\ &= rF(n)(1_S) \\ &= rA_{N,M}(F)(n). \end{align*} Conversely, define \begin{align*} B_{N,M}: \operatorname{Hom}_R(\varphi^*N,M) &\to \operatorname{Hom}_S(N,\operatorname{Hom}_R(S,M)) \end{align*} as follows. For an $R$-linear map $g:\varphi^*N \to M$, define \begin{align*} B_{N,M}(g):N &\to \operatorname{Hom}_R(S,M) \end{align*} by requiring that, for $n \in N$, \begin{align*} B_{N,M}(g)(n):S &\to M \\ t &\mapsto g(tn). \end{align*} For fixed $n \in N$, this map is $R$-linear because, for $r \in R$ and $t \in S$, \begin{align*} B_{N,M}(g)(n)(\varphi(r)t) &= g(\varphi(r)tn) \\ &= r g(tn) \\ &= rB_{N,M}(g)(n)(t). \end{align*} The map $B_{N,M}(g)$ is $S$-linear because, for $s,t \in S$ and $n \in N$, \begin{align*} B_{N,M}(g)(sn)(t) &= g(t(sn)) \\ &= g((ts)n) \\ &= B_{N,M}(g)(n)(ts) \\ &= (s \cdot B_{N,M}(g)(n))(t). \end{align*} Finally, the two maps are inverse. If $F:N \to \operatorname{Hom}_R(S,M)$ is $S$-linear, then for $n \in N$ and $t \in S$, \begin{align*} B_{N,M}(A_{N,M}(F))(n)(t) &= A_{N,M}(F)(tn) \\ &= F(tn)(1_S) \\ &= (t \cdot F(n))(1_S) \\ &= F(n)(1_St) \\ &= F(n)(t). \end{align*} If $g:\varphi^*N \to M$ is $R$-linear, then for $n \in N$, \begin{align*} A_{N,M}(B_{N,M}(g))(n) &= B_{N,M}(g)(n)(1_S) \\ &= g(1_S n) \\ &= g(n). \end{align*} Hence \begin{align*} \operatorname{Hom}_S(N,\operatorname{Hom}_R(S,M)) \cong \operatorname{Hom}_R(\varphi^*N,M). \end{align*}[/step]

[guided]The right adjoint is built from the idea that an $S$-linear map into $\operatorname{Hom}_R(S,M)$ is determined by what its associated functions do at $1_S$. Let $N$ be a left $S$-module and let $M$ be a left $R$-module. Define \begin{align*} A_{N,M}: \operatorname{Hom}_S(N,\operatorname{Hom}_R(S,M)) &\to \operatorname{Hom}_R(\varphi^*N,M) \\ F &\mapsto \bigl(n \mapsto F(n)(1_S)\bigr). \end{align*} We verify $R$-linearity. For $r \in R$ and $n \in N$, the restricted $R$-action on $N$ is $rn=\varphi(r)n$. Thus \begin{align*} A_{N,M}(F)(rn) &= F(\varphi(r)n)(1_S) \\ &= (\varphi(r) \cdot F(n))(1_S) \\ &= F(n)(1_S\varphi(r)) \\ &= F(n)(\varphi(r)) \\ &= rF(n)(1_S) \\ &= rA_{N,M}(F)(n). \end{align*} The key point is that $F(n):S \to M$ is $R$-linear, and $S$ is viewed as a left $R$-module through $\varphi$. Now start with an $R$-linear map $g:\varphi^*N \to M$. Define \begin{align*} B_{N,M}(g):N &\to \operatorname{Hom}_R(S,M) \end{align*} by setting, for $n \in N$, \begin{align*} B_{N,M}(g)(n):S &\to M \\ t &\mapsto g(tn). \end{align*} For this to be meaningful, we first check that $B_{N,M}(g)(n)$ is $R$-linear as a map from $S$ to $M$. For $r \in R$ and $t \in S$, \begin{align*} B_{N,M}(g)(n)(\varphi(r)t) &= g(\varphi(r)tn) \\ &= r g(tn) \\ &= rB_{N,M}(g)(n)(t). \end{align*} So $B_{N,M}(g)(n)$ lies in $\operatorname{Hom}_R(S,M)$. Next we check $S$-linearity of $B_{N,M}(g)$. For $s,t \in S$ and $n \in N$, \begin{align*} B_{N,M}(g)(sn)(t) &= g(t(sn)) \\ &= g((ts)n) \\ &= B_{N,M}(g)(n)(ts) \\ &= (s \cdot B_{N,M}(g)(n))(t). \end{align*} Since this equality holds for every $t \in S$, it proves \begin{align*} B_{N,M}(g)(sn)=s \cdot B_{N,M}(g)(n). \end{align*} Finally, we prove that evaluation at $1_S$ and the construction $n \mapsto (t \mapsto g(tn))$ undo each other. Starting with $F:N \to \operatorname{Hom}_R(S,M)$, for $n \in N$ and $t \in S$, \begin{align*} B_{N,M}(A_{N,M}(F))(n)(t) &= A_{N,M}(F)(tn) \\ &= F(tn)(1_S) \\ &= (t \cdot F(n))(1_S) \\ &= F(n)(1_St) \\ &= F(n)(t). \end{align*} Starting with $g:\varphi^*N \to M$, for $n \in N$, \begin{align*} A_{N,M}(B_{N,M}(g))(n) &= B_{N,M}(g)(n)(1_S) \\ &= g(1_Sn) \\ &= g(n). \end{align*} Thus the second natural bijection is established.[/guided]

[step:Verify naturality and conclude the two adjunctions] The constructions above are natural in all module variables because they are defined by precomposition, postcomposition, tensoring morphisms, and evaluation at $1_S$. For the left adjunction, if $u:M' \to M$ is $R$-linear, $v:N \to N'$ is $S$-linear, and $F:S \otimes_R M \to N$ is $S$-linear, then both naturality composites send $m' \in M'$ to \begin{align*} vF(1_S \otimes u(m')). \end{align*} For the right adjunction, if $v:N' \to N$ is $S$-linear, $u:M \to M'$ is $R$-linear, and $F:N \to \operatorname{Hom}_R(S,M)$ is $S$-linear, then both naturality composites send $n' \in N'$ to \begin{align*} u(F(v(n'))(1_S)). \end{align*} Therefore the bijections are natural. Hence $S \otimes_R -$ is left adjoint to $\varphi^*$, and $\operatorname{Hom}_R(S,-)$ is right adjoint to $\varphi^*$. [/step]