Question:
For each $n\in\mathbb{Z}$, define the ring homomorphism
$$\phi_n:\mathbb{Z}[x]\to\mathbb{Z}$$
by $\phi_n(f)=f(n)$. This gives a $\mathbb{Z}[x]$-module structure on $\mathbb{Z}$, i.e,
$$f\circ a=f(n)\cdot a$$
for all $f\in\mathbb{Z}[x]$ and $a\in \mathbb{Z}$. Now given two integers $n,m\in\mathbb{Z}$, compute the tensor product $\mathbb{Z}\otimes_{\mathbb{Z}[x]}\mathbb{Z}$ where the left-hand copy of $\mathbb{Z}$ uses the module structure from $\phi_n$ and the right-hand copy of $\mathbb{Z}$ uses the module structure from $\phi_m$. (Note: the answer depends on the numbers $n$ and $m$)
Answer:
I first analyze the case where $n=m$. First observe that as constant polynomials do exist in $\mathbb{Z}[x]$, any basic tensor $a\otimes b=ab\otimes 1$. Then I write the map
$$\rho:\mathbb{Z}\otimes_{\mathbb{Z}[x]}\mathbb{Z}\to\mathbb{Z}$$
defined by $\rho(a\otimes 1)=a$. This map in fact yields a $\mathbb{Z}[x]$-module isomorphism, i.e, $\mathbb{Z}\otimes_{\mathbb{Z}[x]}\mathbb{Z}\cong \mathbb{Z}$.
Now without loss of generality suppose $m>n$. Again by the same observation, any basic tensor $a\otimes b=1\otimes ab$. Now consider $1\otimes (m-n)$ and the polynomial $f(x)=x-n$. This would yield $1\otimes m-n=1\otimes f(m)\cdot 1=f(n)\cdot1\otimes 1=0\otimes 1=0$. However, $1\otimes(m-n)=(m-n)(1\otimes1)$ and since $m-n\neq0$, $1\otimes1=0$. Therefore, $\mathbb{Z}\otimes_{\mathbb{Z}[x]}\mathbb{Z}=0$.
I want to ask if you spot any mistake in my proof. Thanks in advance…
Best Answer
Here’s an alternative approach. Let us write $\mathbb Z_{[m]}$ for the $\mathbb Z[x]$-module where $x$ acts as multiplication by $m$.
There is an exact sequence (free resolution) of $\mathbb Z[x]$-modules $$ \mathbb Z[x]\to\mathbb Z[x]\to\mathbb Z_{[m]}\to0 $$ where the first map sends $1\mapsto m-x$ and the second map sends $1\to1$.
As tensoring is right exact, we can tensor this with $\mathbb Z_{[n]}$ to obtain another exact sequence. We now use that $\mathbb Z[x]\otimes\mathbb Z_{[n]}\cong\mathbb Z_{[n]}$ to write this exact sequence as $$\mathbb Z_{[n]}\to\mathbb Z_{[n]}\to\mathbb Z_{[m]}\otimes\mathbb Z_{[n]}\to0 $$ As an endomorphism of $\mathbb Z[x]\otimes\mathbb Z_{[n]} $ the first map sends $1\otimes1$ to $(m-x)\otimes1$. Using the isomorphism with $\mathbb Z_{[n]}$ we obtain the endomorphism $1\mapsto m-n$.
So the tensor product $\mathbb Z_{[m]}\otimes\mathbb Z_{[n]}$ is the abelian group $\mathbb Z/(m-n)$ with $\mathbb Z[x]$-module structure where $x$ acts as multiplication by $n$ (equivalently $m$).
This approach of computing the tensor product by first replacing one module by a projective (or free) resolution is both common and useful.