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 $$\left(\sum_{k \neq n} \bigoplus_{i + j = k} M_i\otimes N_j\right)\cap\bigoplus_{i + j = n} M_i\otimes N_j = \{0\}.$$


2 Answers 2


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.

  • 1
    $\begingroup$ 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 $\endgroup$ Aug 28, 2019 at 18:22
  • $\begingroup$ @Max: I'm aware of that, but the modules $M_i\otimes N_j$ are not $R$-modules, so your argument doesn't apply? $\endgroup$ Aug 28, 2019 at 18:27
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    $\begingroup$ @withoutfeather They are $R$-modules since the ring in question has the trivial grading. $\endgroup$
    – Jxt921
    Aug 28, 2019 at 18:30
  • $\begingroup$ @Jxt921 Yes, you're right. I didn't read the question very carefully. $\endgroup$ Aug 28, 2019 at 18:35

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