If $M_*$ and $N_*$ are graded modules over the *graded* ring $R_*$, what is the definition of $M_* \otimes_{R_*} N_*$?

Question

Quick question (hopefully): What is the correct definition of a tensor product of two graded $R_*$-modules and/or graded $R_*$-algebras $M_*$ and $N_*$ over the graded ring $R_*$?

$M_* \otimes_{R_*} N_* = ?$

If R is not graded I know how to do this, but when $R_*$ is graded the usual construction doesn't work ($N_i$ isn't a $R_*$ module).

The motivation is that I want to understand what happens when the quasicoherent sheaf associated to a graded module is pulled back along $\pi: $ Proj($R_*$) $\to$ Proj($S_*$), provided this map is defined by some map of graded rings $S_* \to R_*$. I am guessing that a correct algebraic definition for this construction will show that $\pi^{*}(\widetilde{M_*}) = \widetilde{M_* \otimes_{S_*} R_*}$.

Answer 1

Usually the tensor product of two graded $R_\bullet$-modules $M_\bullet, N_\bullet$ is defined as having $n$-th component $$\left(M_{\bullet}\otimes_{R_\bullet} N_\bullet\right)_n\ :=\ \left(\bigoplus\limits_{p+q=n} M_p\otimes_{\mathbb Z} N_q\right)\left/\left(m r\otimes n - m\otimes r n\ |\ a+b+c=n, m\in M_a, r\in R_b, n\in N_c\right)\right..$$ In particular, this is a quotient of $\bigoplus\limits_{p+q=n} M_p\otimes_{R_0} N_q$, but usually much smaller.

For example, take $R_\bullet=M_\bullet=N_\bullet:={\mathbb k}[x]$ for some commutative ring $\mathbb k$, with $\text{deg}(x)=1$. Then $$\bigoplus\limits_{p+q=n} M_p\otimes_{R_0} N_q={\mathbb k}[x,y]_n,\quad\text{ while }\quad(M_\bullet\otimes_{R_\bullet} N_\bullet)_n=R_n={\mathbb k}[x]_n.$$

Answer 2

Just a comment -- an indirect but intuitive way to remember this is the following:

First of all, as a module $M_* \otimes_{R_*} N_*$ is just the ordinary tensor product of $R_*$-modules. This is the only reasonable construction (we can always apply the operation of "forgetting the grading", which reduces the construction to this ordinary tensor product).

So the module structure is unique. That leaves the grading: observe that $M_* \otimes_{R_*} N_*$ is generated by homogeneous elements of the form $m \otimes n$. Clearly, the degree of such an element is $\deg(m) + \deg(n)$.

Moreover, the defining relations of $R_*$-bilinearity respect this rule -- multiplying by scalars raises the degree in the `intuitive' way -- which means it's well-defined to grade elements of $M_* \otimes_{R_*} N_*$ by (a) finding any way to represent them as sums of simple tensors, then (b) using the above rule.

Following this reasoning a few more steps leads to the nice computation done by Hanno.