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In Matsumura's commutative ring theory book he says that if R is ring graded by an abelian semigroup then $R_0$ is a subring. In particular he states that $1\in R_0$. But this seems wrong if we just take the product of rings. For example $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$ is $\mathbb{Z}/2\mathbb{Z}$ graded (as a multiplicative semigroup) but the identity {1,1} is in both components of the grading. So my question is in a grading do we require that $R_gR_h\neq 0$?

alex pacun
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    What is your $\mathbb{Z}/2\mathbb{Z}$ grading on $\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$? – Matthew Towers Oct 25 '20 at 17:16
  • Sorry I edited it to be more clear that I was considering the multiplicative structure on $\mathbb{Z}/2\mathbb{Z}$. Let $R_1$ and $R_0$ be either of the two components of the direct sum. So since it is a direct sum of rings $R_1R_1\subset R_1$ and $R_0R_0\subset R_0$ and $R_0R_1=0\in R_0$ so this respects grading. – alex pacun Oct 25 '20 at 17:24
  • It's as a multiplicative semigroup so 1*1 not 1+1 – alex pacun Oct 25 '20 at 17:27

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