$\mathbb Z [x]$ module structure on $\mathbb Z$ For each integer $n \in \mathbb Z$, define the ring homomorphism
$$φ_n :\mathbb Z [x]\to \mathbb Z, \ \ φ_n(f)=f(n).$$
This provides a $\mathbb Z[x]$-module structure on $\mathbb Z$ given by $$f ◦ a = f(n) · a$$ for all $f \in \mathbb Z[x]$ and $a ∈ Z$. Given two integers $m, n \in \mathbb Z$, how can we compute the tensor product $$\mathbb Z ⊗_{\mathbb Z[x]} \mathbb Z?$$ Here the left and right modules are respectively determined by the homomorphisms $φ_n$ and $φ_m$. 
I believe that the answer is related to integers $n$ and $m$ but I cannot constitute a rigorous argument. Any suggestions?
 A: By the isomorphism theorem you have $\Bbb Z_{\phi_m} \cong \Bbb Z[X]/(X-m)$ as $\Bbb Z[X]$-algebras. Hence you have
$$\Bbb Z[X]/(X-m) \otimes_{\Bbb Z[X]} \Bbb Z[X]/(X-n) \cong \Bbb Z[X]/(X-m,X-n)$$ 
as $\Bbb Z[X]$-algebras (and in particular $\Bbb Z[X]-modules).
A: Let $\mathbb Z_n$ denote $\mathbb Z$ with the module structure given by $\phi_n$.
Note that, since $x\cdot(m-n) = m\cdot(m-n) \in (m-n)\mathbb Z$, the submodule $(m-n)\mathbb Z_m$ of $\mathbb Z_m$ is identical to the ideal $(m-n)\mathbb Z$ of $\mathbb Z$.
Regard the following map:
$$\mathbb Z_m\times \mathbb Z_n \to \mathbb Z_m/(m-n)\mathbb Z_m$$
$$(a, b) \mapsto ab+(m-n)\mathbb Z_m$$
This is bilinear with respect to $\mathbb Z[x]$ because for $a\in\mathbb Z_m$ and $b\in\mathbb Z_n$:
$$(xa)b+(m-n)\mathbb Z_m = mab+(m-n)\mathbb Z_m = x(ab)+(m-n)\mathbb Z_m$$
$$a(xb)+(m-n)\mathbb Z_m = nab+(m-n)\mathbb Z_m = mab+(m-n)\mathbb Z_m = x(ab)+(m-n)\mathbb Z_m$$
Therefore, one gets an induced homomorphism of $\mathbb Z[x]$-modules:
$$f:\mathbb Z_m\otimes_{\mathbb Z[x]} \mathbb Z_n \to \mathbb Z_m/(m-n)\mathbb Z_m$$
Obviously, this is surjective. Let $\alpha\in\ker f$. Then, there is $a\in\mathbb Z$ such that $\alpha = a(1\otimes 1)$.
Then, $0 = f(\alpha) = a+(m-n)\mathbb Z_m$, which means that $a = k(m-n)$ for some $k\in\mathbb Z$. Therefore:
$$\alpha = k(m-n)(1\otimes 1) = (km\otimes 1)-(k\otimes n) = ((xk)\otimes 1)-(k\otimes (x1)) = 0$$
Thus $f$ is injective. In conclusion:
$$\mathbb Z_m\otimes_{\mathbb Z[x]} \mathbb Z_n \cong \mathbb Z_m/(m-n)\mathbb Z_m$$
