$\DeclareMathOperator{\Hom}{Hom}$ $\DeclareMathOperator{\Mod}{Mod}$ $\DeclareMathOperator{\End}{End}$ $\DeclareMathOperator{\coker}{coker}$
When talking about tensor product of abelian groups, or modules over some ring, I can prove that it is right exact, for instance by showing it is a left adjoint. But I'm currently dealing with slight issues dealing with the tensor product in a slightly more general setting, first let me state the theorem I'm talking about.
Let $k$ be a commutative ring (with unity). Let $C$ be a $k$-abelian category and $R$ a $k$-algebra. Assume that $C$ admits small inductive limits. Then for any $X\in \Mod(R,C)$ and $N\in \Mod(R^{op})$ the functor $Y\mapsto \Hom_{R^{op}}(N,\Hom_C(X,Y))$ is representable. Furthermore, denoting $N\otimes_R X$ its representative, the functor $$\bullet \otimes_R \bullet : \Mod(R^{op})\times \Mod(R,C)\to C $$ is additive and right exact in each variable.
To avoid any confusion by what I mean here, $k$-Abelian category is the same as an abelian category but $\Hom_C(X,Y)\in \Mod(k)$ (instead of $\Mod(\mathbb{Z}))$, with composition being $k$-bilinear. Also, $\Mod(R,C)$, the category of left $R$ object in $C$, is defined as follows :
- The objects are pairs $(X,\alpha_X)$ where $X$ is an object in $C$ and $\alpha_X:R\to \End_C(X)$ is a $k$-algebra morphism.
- A morphism $(X,\alpha_X)\to(Y,\alpha_Y)$ is a morphism $f:X\to Y$ in $C$ such that $\alpha_Y\circ f = f\circ\alpha_X$
When $C$ is $k$-abelian, $\Mod(R,C)$ is as well, and the forgetful functor $\Mod(R,C)\to C$ is faithful and exact. I have been able, with some help, to prove that the functor $Y\to \Hom_{R^{op}}(N,\Hom_C{(X,Y}))$ is representable. This ends up giving an isomorphism $\Hom_C(N\otimes_R X,Y)\simeq \Hom_R(N,\Hom_C (X,Y))$, hence $\bullet\otimes_R X$ is a left adjoint to $\Hom_C(X,\bullet)$, hence $\otimes_R X$ is right exact. But then I'm not sure how to prove that $\bullet\otimes_R\bullet$ is exact in its second variable, at first I thought the isomorphism above would also allow me to conclude that $N\otimes_R\bullet$ is a left adjoint to some other functor, but I haven't been able to do so. I have similar issues with the additivity, it feels like it should be an easy consequence of the construction, but I don't actually know how to show it's additive in each variable.
So you are interested in the bifunctor $\otimes_R$ defined by the following natural isomorphism: $$\textrm{Hom}_\mathcal{C} (N \otimes_R X, Y) \cong \textrm{Hom}_{R^\textrm{op}} (N, \textrm{Hom}_\mathcal{C} (X, Y))$$ You seek a right adjoint for $N \otimes_R {-} : \textrm{Mod} (R, \mathcal{C}) \to \mathcal{C}$.
For this, we should assume $\mathcal{C}$ is complete. Then $\mathcal{C}^\textrm{op}$ is a cocomplete $k$-linear abelian category and we can apply the dual of the theorem constructing $\otimes_R$ to obtain a bifunctor $\pitchfork_S$ defined by the following natural isomorphism: $$\textrm{Hom}_\mathcal{C} (Y, P \pitchfork_S Z) \cong \textrm{Hom}_S (P, \textrm{Hom}_\mathcal{C} (Y, Z))$$ Here, $S$ is a $k$-algebra, $P$ is a left $S$-module, and $Z$ is an object in $\textrm{Mod} (S, \mathcal{C})$.
We can apply this in particular when $S = k$ to obtain a bifunctor ${\pitchfork_k} : \textrm{Mod} (k)^\textrm{op} \times \mathcal{C} \to \mathcal{C}$. Contravariant functoriality in the first variable implies that any right action on $P$ induces a left action on $P \pitchfork_k Z$. Hence, $N \pitchfork_k Y$ is an object in $\textrm{Mod} (R, \mathcal{C})$. It can then be shown that we have the following natural isomorphism: $$\textrm{Hom}_{R^\textrm{op}} (N, \textrm{Hom}_\mathcal{C} (X, Y)) \cong \textrm{Hom}_{\textrm{Mod} (R, \mathcal{C})} (X, N \pitchfork_k Y)$$ Thus, $N \pitchfork_k {-}$ is the required right adjoint for $N \otimes_R {-}$.
(Incidentally, $N \otimes_R X$ is bilinear in $N$ and $X$ simply because the RHS of the defining isomorphism is bilinear in $N$ and $X$. This is a standard transport-of-structure argument.)