# Is it true that $S(n)\otimes_S S(m)\cong S(n+m)$?

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 $$x\otimes 1\in S(n)_{m+d}\otimes S(m)_{-m}$$).

• 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. Jun 16, 2023 at 21:26
• @tomasz Let $x\in S_{n+m+d}$, then $x\otimes 1$ is mapped to $x$. Jun 16, 2023 at 21:27
• Even if your rings are unital, $1$ is $0$-graded, so it doesn't work unless $n=0$ or $m=0$. Jun 16, 2023 at 21:30
• @tomasz Sorry, you're right. Jun 16, 2023 at 21:30
• Perhaps they had $S=A[X_0,\dots,X_n]$ in mind. I'll leave a comment on StacksProject and await their reply. Jun 16, 2023 at 21:35

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 $$m+d-m=d$$.