Eilenberg-Watt's theorem reference. I'm looking for a reference (with proof!) for the following result:
Let $A$ and $B$ be unital, associative rings.

(Eilenberg-Watt) Let $F: {}_{A}\mathrm{Mod} \to {}_{B}\mathrm{Mod}$ be a functor that is
additive and preserves direct sums and quotients. Then there is a
natural isomorphism $$\alpha: P \otimes_A \bullet \to F$$ where
$P=F(A)$ has the natural right $A$-module structure.

In short, an additive functor that preserves direct sums and quotients coincides with the functor $P \otimes_A \bullet$.
Please check if all assumptions match, because I have seen plenty of variations of this theorem but not this exact one.
 A: If we require that for any map $\varphi : X \to Y$ then $F(coker \varphi) \simeq coker F(\varphi) $, then the result is true.
Let $M$ be a module, and let $\{m_i\}_{i\in I}$ be a set of generators. This defines a map $\psi: A^{\oplus I} \to M$ sending the 1 in the $i$-th component to $m_i$. Now for any relation $\sum a_{ij} m_i =0$ in $M$, let $r_j\in A^{\oplus I}$ be the element $\sum a_{ij}$. This defines a map $\varphi: A^{\oplus J}\to A^{\oplus I}$, and you can check that $M \simeq coker \varphi $. Let me drop the $\oplus$ above the $A$'s. We can use the cokernel commutativity to write
$$ F(M) \simeq F(coker \varphi ) \simeq coker F(\varphi) $$
Note also that
$$F(A^I) \simeq F(A) ^I \simeq F(A) \otimes A^I$$
and that $F$ being additive implies also
$$F(\varphi) = F(A) \otimes \varphi$$
The latter formula means to tensor with the identity on $F(A) $. Also, recall that tensoring with a module is a left adjoint functor, thus preserve cokernels. In particular we have
$$ coker F(\varphi) \simeq coker F(A) \otimes \varphi \simeq F(A) \otimes coker \varphi \simeq F(A) \otimes M $$
And voilà. Maybe this was the result you were looking for?
A: See this question and answer for a proof. The idea is exactly the one described in the second part of Andrea's answer: for any $A$-module, you have a presentation $A^{(J)}\to A^{(I)}\to M\to 0$, which you can then use together with your assumptions to deduce that $F(M)\cong F(A)\otimes_A M$.
Note that $F(A)$ has a $B$-module structure by definition, and also a right $A$-module structure given by $F(r_a), a\in A$ where $r_a : A\to A$ is given by $x\mapsto xa$ (this is a morphism of left $A$-modules).
To get naturality of the isomorphism $F(M)\cong F(A)\otimes_A M$, the trick is to define, for any $F$ and any $M$ (no assumptions so far) a natural morphism $F(A)\otimes_A M\to F(M)$. One then uses the assumptions to prove that this natural morphism is an isomorphism.
This natural morphism is defined by the following composite: $M\to \hom(A,M) \to \hom(F(A),F(M))$ and then the universal property of the tensor product. There is one thing to note here : $\hom(A,M)\to \hom(F(A),F(M))$ is a morphism of $A$-modules because of the definition of the right $A$-module structure on $F(A)$.
This gives you a natural morphism, which is obviously an isomorphism when $M=A$, and since both sides preserve direct sums and cokernels, it is an isomorphism for any $M$.
Let me mention two additional things with respect to some points in Andrea's answer :
1- In the example he gives in the first part of the question, there is a mistake : we never consider $P\otimes_\mathbb C X$, but $P\otimes_A X$ - in particular the dimension is not the product of the dimensions.
2- There is no difference between preserving cokernels of general maps and preserving cokernels of injections: indeed if you preserve the latter, then you preserve epimorphisms, and then for a general cokernel sequence $M\to N\to C\to 0$, you can separate it as $M\to \mathrm{im}(M\to N)\to 0$ and $0\to \mathrm{im}(M\to N)\to N\to C\to 0$.
This gives two a priori different definitions of right exact functors which end up agreeing.
