Let $R$ be a ring for which we have a $\mathbb Z_2$-gradation meaning that $R=R_0\oplus R_1$, for example when $R=\mathbb C$ we have that $R_0=\mathbb R$ and $R_1=\mathbb R i$. I'm having trouble seeing that an element $r\in R$ is uniquely described as $r=r_0+r_1$ or as a unique couple $(r_0,r_1)$. If we multiply two elements $r$ and $r'$ in $R$ we get different results. In the first case $r=r_0+r_1$ and $r'=r_0'+r_1'$ we have a product $(r_0+r_1)*(r_0'+r_1')=(r_0r_0'+r_1r_1')+(r_0r_1'+r_1r_0')$ which is a product that respects that gradation since $(r_0r_0'+r_1r_1')\in R_0$ and $(r_0r_1'+r_1r_0')\in R_1$ but if we take $r=(r_0,r_1)$ and $r'=(r_0',r_1')$ then the product $(r_0r_0',r_1r_1')$ does not respect the gradation since both $r_0r_0'$ and $r_1r_1'\in R_0$ is there any explanation for this?
A $\mathbb Z_2$-gradation on a ring.
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abstract-algebra
ring-theory
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1The latter product does not give a $\mathbb{Z}_2$-graded ring. – 2012-12-12
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1I suggest you to start the sentences by capital letters, as the grammar rules. – 2012-12-12
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0Ok YACP, thank you for the remark!! – 2012-12-12