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I am run into the following in an Algebra text:

"Let $R_0=\mathbb Z/2\mathbb Z⊕\mathbb Z/2\mathbb Z⊕\cdots$ viewed as a ring without identity, with addition and multiplication defined componentwise. Let $R=\mathbb Z⊕R_0$ be the ring obtained by "adjoining" an identity $1\in \mathbb Z$ to $R_0$."

My question is:

What is the identity of $R$?

If it is the pair $(1,0)$ so for any nonzero element $e\in R_0$ we would have $(1,0)(0,e)=(0,e)$ so $(0,0)=(0,e)$ and therefore $e=0$, a contradiction.

Any leading answer would be thanked.

user26857
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karparvar
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1 Answers1

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Multiplication is not defined coordinatewise, rather $(n,r)(m,s)=(nm,rs+rm+ns)$. In particular $(1,0)(n,s)=(n,s)$. Writing $(n,r)$ as $n+r$ should help in understanding why we have defined multiplication in such a way.

Pedro
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  • Thanks for the answer. Now, if $P$ is a prime ideal containing $R_0$ and $e\in R_0$ we know that $e$ is idempotent so $(1-e)e=0$. How could we get that $e=0$? And how could we deduce that the element $(2,0)\in R$ is not a zero divisor in $R_P$? – karparvar Aug 19 '15 at 06:08
  • @karparvar Under this operation (which is not the only candidate, but I admit is probably the most likely candidate) $e^2=e$ for everything in $R_0$. Why would there be any expectation for $e=0$? – rschwieb Aug 19 '15 at 10:23
  • @rschwieb I meant $e=0\in R_P$. I thought a little and realized that: it is because $e\in P$ whence $1-e$ is not in $P$, so, the equality $(1-e)e=0$ gives $e/1=0$ as an element of $R_P$. – karparvar Aug 19 '15 at 10:32
  • @karparvar Under this multiplication, $(2,0)(0,x)=0$ for any $x$ in the $R_0$, and hence in any localization. Unless $(0,R_0)$ collapses to zero in the localization, it looks like (2,0) is a zero divisor. – rschwieb Aug 19 '15 at 10:34
  • @karparvar Ah ok, e would definitely be zero in the localization, yes. – rschwieb Aug 19 '15 at 10:38
  • @karparvar that also explains why $(0, R_0)$ does collapse to zero when localizing at any prime containing $R_0$ :) – rschwieb Aug 19 '15 at 10:40