A $\mathbb Z_2$-gradation on a ring.

Question

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?

Answer 1

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.