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 $\pi^{*}(\widetilde{M_*}) = \widetilde{M_* \otimes_{S_*} R_*}$.


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 $$\bigoplus\limits_{p+q=n} M_p\otimes_{R_0} N_q={\mathbb k}[x,y]_n,\quad\text{ while }\quad(M_\bullet\otimes_{R_\bullet} N_\bullet)_n=R_n={\mathbb k}[x]_n.$$

  • $\begingroup$ Does this construction coincide with the pullback of a quasicoherent sheaf on a projective scheme, as in the motivation section of my question? It seems like the answer is yes from your example, $R_* \otimes_{R_*} R_* = R_*$ is what should be expected. $\endgroup$ – Lorenzo Najt Jan 17 '16 at 22:39
  • $\begingroup$ A related question: Is $\operatorname{Proj}(S_* \otimes_{R_*} T_*) = Proj(S_*) \times_{\operatorname{Proj}(R_*)} \operatorname{Proj}(T_*)$ under this definition? $\endgroup$ – Lorenzo Najt Jan 17 '16 at 23:21

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.

  • $\begingroup$ Thanks, this is a good insight. Do you happen to know if this construction agrees with pullbacks of quasicoherent sheaves on projective schemes? It seems that in this case, the additional equivalence relationship does imply that $((S_* \otimes_{R_*} M_*)[r^{-1}])_0 = (S_*[r^{-1}])_0 \otimes_{R_*[r{-1}]} ((M_*)_r)_0$, since the fractions of terms of the form $S_{-p} \otimes M_p$ can be balanced - but I might be making a mistake in this mess of symbols. I am thinking of my question here: math.stackexchange.com/questions/1616965/… $\endgroup$ – Lorenzo Najt Jan 19 '16 at 17:34

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