Let $$R$$ be a ring and $$M = \bigoplus_{k \geq 0} M_k, N = \bigoplus_{k \geq 0} N_k$$ graded $$R$$-modules. I'm having trouble seeing why $$M\otimes N = \bigoplus_{k \geq 0} \bigoplus_{i + j = k} M_i\otimes N_j$$ would be grading of $$M\otimes N$$. In particular, I don't understand why we need to have $$\left(\sum_{k \neq n} \bigoplus_{i + j = k} M_i\otimes N_j\right)\cap\bigoplus_{i + j = n} M_i\otimes N_j = \{0\}.$$

It follows from the distributivity of tensor product over direct sums :

$$A\otimes (\bigoplus_i B_i) \cong \bigoplus_i (A\otimes B_i)$$naturally in all variables, with the canonical maps (i.e. $$(\sum_j a_{ij}\otimes b_{ij})_{i\in I} \mapsto \sum_{i,j}a_{ij}\otimes b_{ij}$$ and $$a\otimes (b_i)_{i\in I} \mapsto (a\otimes b_i)_{i\in I}$$ respectively - a funny note : if we write elements of $$\bigoplus_i$$ as $$\sum_i$$, then the first map is $$\sum_i a_i\otimes b_i \mapsto \sum_i a_i\otimes b_i$$, which could lead to some confusions, it doesn't precisely because this is an isomorphism)

In particular, the canonical map is an isomorphism $$\bigoplus_n \bigoplus_{p+q=n} M_p\otimes N_q \overset{\simeq}\to M\otimes N$$ which explains why it's a grading

Take $$P=\bigoplus_{k\geq 0}\bigoplus_{i+j=k}M_i\otimes N_j$$. $$P$$ is naturally a graded $$R$$-module, and we will prove that $$P\cong M\otimes_R N$$.

The map $$f: M\otimes_R N \to P, m\otimes n\mapsto \sum m_i\otimes n_j$$ (where $$m=\sum m_i,n=\sum n_j$$, $$m_i,n_j$$ are homogeneous elements) is well-defined by the universal property of $$M\otimes_R N$$. Again, the map $$M_i\otimes N_j\to M\otimes_R N, m_i\otimes n_j\to m_i\otimes n_j$$ is well-defined as homomorphism of abelian groups. Taking direct sum, one obtains a map $$g: P\to M\otimes_R N$$ that maps $$m_i\otimes n_j$$ with $$m_i,n_j$$ homogeneous to $$m_i\otimes n_j\in M\otimes_R N$$.

It is easy to see that $$f$$ and $$g$$ are inverse of each other, hence they are isomorphisms of abelian groups. Notice that $$f$$ is a homomorphism of $$R$$-module, so it is an isomorphism of $$R$$-modules. Since $$M\otimes_R N$$ is isomorphic as $$R$$-modules to a graded $$R$$-module, it is naturally graded.

P.S. While I wrote this answer, there is another answer which used the distributivity of tensor product over direct sums. But I don't think my answer is redundant, because the distributivity only implies that $$P$$ and $$M\otimes N$$ isomorphic as abelian groups. To show that they are isomorphic as $$R$$-modules, one has to write down the map explicitly.

• For your PS : your answer may not be redundant, but not for the reason you mention. Indeed distributivity of tensor products works equally well for $R$-modules as for abelian groups Aug 28, 2019 at 18:22
• @Max: I'm aware of that, but the modules $M_i\otimes N_j$ are not $R$-modules, so your argument doesn't apply? Aug 28, 2019 at 18:27
• @withoutfeather They are $R$-modules since the ring in question has the trivial grading. Aug 28, 2019 at 18:30
• @Jxt921 Yes, you're right. I didn't read the question very carefully. Aug 28, 2019 at 18:35