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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?

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The latter product does not give a $\mathbb{Z}_2$-graded ring. – Jyrki Lahtonen Dec 12 '12 at 9:18
I suggest you to start the sentences by capital letters, as the grammar rules. – user26857 Dec 12 '12 at 10:13
Ok YACP, thank you for the remark!! – palio Dec 12 '12 at 10:16

When you see $R$ as a $\mathbb Z_2$-graded ring, then the direct sum $R=R_0\oplus R_1$ is a direct sum of abelian groups, not a direct sum of rings. Hence, the product on $R$ is not given by component-wise multiplication, but rather by what you describe first: $$(r_0+r_1)\cdot(s_0+s_1) = (r_0s_0 + r_1s_1) + (r_1s_0+r_0s_1).$$ Note, in particular, that $R_1$ is not a ring! It has no $1$ and isn't closed under multiplication.

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oh i think i see what you mean.. is that why we choose multiplication in $\mathbb C$ by $(x,y)*(x',y')=(xx'-yy',xy'+x'y)$ rather than $(xx',yy')$ is it for respect of gradation reasons? – palio Dec 12 '12 at 9:26
I think originally we chose the multiplication in $\mathbb C$ for reasons that are quite different: Namely, having a root of $-1$ available. It was probably observed later that $\mathbb C$ can be interpreted as a $\mathbb Z_2$-graded algebra over $\mathbb R$. – Jesko Hüttenhain Dec 12 '12 at 9:32
and also this is a decomposition of $\mathbb C=\mathbb R\oplus \mathbb R$ rather then $\mathbb C=\mathbb R\oplus \mathbb R i$ – palio Dec 12 '12 at 9:33

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