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Convention: rings are unital and commutative.

Let $S=\bigoplus_{d\geq 0}S_d$ be a graded ring. For $n\in\mathbb Z$ let $S(n)$ be the graded $S$-module defined by $S(n)_d=S_{n+d}$. It is claimed in Stacks Project (2.27.2) that the canonical map $$ \phi:S(n)\otimes_S S(m)\to S(n+m),x\otimes y\mapsto xy $$ is an isomorphism. Surjectivity for me is clear(*). Injectivity not. Since the tensor product commutes with direct sums, we have $$ S(n)\otimes_S S(m)=\bigoplus_{i,j}S_{n+i}\otimes_S S_{m+j} $$ while $$ S(n+m)=\bigoplus_d S_{n+m+d}. $$ Let's assume $n=m$. If we take $i\neq j$, then we have the direct summand $$ (S_{n+i}\otimes S_{n+j})\oplus (S_{n+j}\otimes S_{n+i})\subset S(n)\otimes_S S(m). $$ Clearly then $(x\otimes y)\oplus (-y\otimes x)$ is mapped to zero. Where am I making a mistake? (assuming the claim that we have an isomorphism is correct)

(*) Thanks to thomasz pointing out that surjectivity isn't clear at all, here is a counter(**) example:

Take $S=k[x,y,z]$ with $\deg k=0,\deg x=1,\deg y=2,\deg z=3$. Then $S(1)\otimes_S S(2)\to S(3)$ doesn't reach $z\in S(3)$.

(**) This example doesn't work, as pointed out in the comments. I assumed we have an $\mathbb N$-grading, but in Stacks Project (1.10.56) we have the following convention: graded rings have an $\mathbb N$-grading, while graded modules have a $\mathbb Z$-grading. The twist operation is defined for graded modules and yields a graded module. Thus $S(n)$ is $\mathbb Z$-graded (meaning that also for $d<0$ we have $S(n)_d=S_{n+d} $). With this in mind, surjectivity is clear; we can reach $x\in S(n+m)_d$ by $x\otimes 1\in S(n)_{m+d}\otimes S(m)_{-m}$).

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    $\begingroup$ Is surjectivity clear? If multiplication in $S$ is trivial, then I think it pretty obviously isn't surjective, unless $S(n+m)$ is zero. You need some hypotheses about $S$ at least. $\endgroup$
    – tomasz
    Jun 16, 2023 at 21:26
  • $\begingroup$ @tomasz Let $x\in S_{n+m+d}$, then $x\otimes 1$ is mapped to $x$. $\endgroup$
    – Sha Vuklia
    Jun 16, 2023 at 21:27
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    $\begingroup$ Even if your rings are unital, $1$ is $0$-graded, so it doesn't work unless $n=0$ or $m=0$. $\endgroup$
    – tomasz
    Jun 16, 2023 at 21:30
  • $\begingroup$ @tomasz Sorry, you're right. $\endgroup$
    – Sha Vuklia
    Jun 16, 2023 at 21:30
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    $\begingroup$ Perhaps they had $S=A[X_0,\dots,X_n]$ in mind. I'll leave a comment on StacksProject and await their reply. $\endgroup$
    – Sha Vuklia
    Jun 16, 2023 at 21:35

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My counter-example for injectivity doesn't work: I forgot that we are tensoring over $S$ (and not for example over $S_0$), and therefore $$ x\otimes y+ (-y\otimes x)=x\otimes y- x\otimes y=0. $$ Also, I couldn't use that the tensor-product commutes with the direct sum, because $S_n$ is an $S_0$-module and not an $S$-module.

I see now that $\phi$ is in fact an isomorphism; its inverse is given by $$ \phi^{-1}:S(n+m)\to S(n)\otimes_S S(m),x\mapsto x\otimes 1. $$ Note that this is an $S$-module map. It is also a graded map, since it maps $x\in S(n+m)_d=S_{n+m+d}$ to $x\otimes 1\in S(n)_{m+d}\otimes S(m)_{-m}$ which has degree $m+d-m=d$.

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