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Given a ring $R$, show that there exists a ring $R'$ with unity such that $R$ is a subring (up to isomorphism) of it.

I am not getting the meaning of the question.

Peter Woolfitt
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Wolverine
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1 Answers1

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Let $\mathbb Z\times R$ be the underlying set of $R'$.

Define addition by: $\langle n,r\rangle+\langle m,s\rangle=\langle n+m,r+s\rangle$.

Define multiplication by $\langle n,r\rangle\langle m,s\rangle=\langle nm,mr+ns+rs\rangle$.

It can be shown that $R'$ is a ring having $\langle1,0\rangle$ as unit.

The map $\phi: R'\rightarrow R$ prescribed by $r\mapsto\langle0,r\rangle$ is an injective ringhomomorphism so that its image is a subring of $R'$ isomorphic with $R$.

drhab
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    This is the Dorroh adjunction of $1$. It is not always the best way to adjoin $1$ since it may not preserve crucial properties. – Bill Dubuque Mar 04 '15 at 15:50
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    @BillDubuque Thank you for the link. New stuff for me. Often when I see comments/answers of you the title of a book - written by someone named J.R.R Tolkien - comes into my mind. – drhab Mar 04 '15 at 16:33