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Give a ring with identity $R$ and its ideals $I, J$, with $I+J=R$ and $IJ=0$, I was required to prove $R\simeq R/I \times R/J$.

My understanding is $R/I\cap J \simeq R/I \times R/J$ so $I\cap J=0$ must hold if I'm able to do so.

Work done:

Since $1$ belongs to $R$, $1=u+(1-u)$ with $u$ in $I$ and $1-u$ in $J$, with $u(1-u)=0$.

For arbitary $v\in I\cap J$ we have $uv = 0$ and $v(1-u)=0$ i.e. $v=vu$.

If the ring is commutative or just $JI=0$ (we have $vu=0$ and $v=uv$) will force $v=0$. Neither commutativity nor $JI=0$ is given. I am aware of this conterexample but this example uses left ideals $(x)=Rx$, while the ideals defined in my book is always $(x)=Rx+xR+RxR$.

Edit: It seems $R=\mathbb{Z}\langle x,y|2x+2y-1=xy=0\rangle,I=(x),J=(y)$ have $I+J=R$, $IJ=0$ and $I\cap J=JI=(yx)$ (note x and y does not commute). Is this a counterexample or something is wrong?

Micheal Johnson
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  • You might want to check out this: https://math.stackexchange.com/questions/1222474/a-noncommutative-counterexample-to-the-following-property-if-i-j-are-comaxima?rq=1. The answer showed that $IJ+JI = I \cap J$, so I am not sure what is happening in your case. – DuduBob Jan 20 '21 at 13:05
  • So $IJ=0$ means $I\cap J=JI$, but $JI=0$ is precisely what is NOT given... – Micheal Johnson Jan 20 '21 at 13:10
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    Yeah, so you probably should work out the example given in that answer and see if it is also a counterexample for your problem. – DuduBob Jan 20 '21 at 13:13

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Here’s maybe an easier counterexample. Take $R$ to be the ring of lower triangular 2x2 matrices over a field, with basis $e,f,x$ such that $e,f$ are orthogonal idempotents and $x=fxe$. Set $I=ReR$ having basis $e,x$, and $J=RfR$ having basis $f,x$. Then $I+J=R$ and $IJ=0$ but $JI$ has basis $x$.

Andrew Hubery
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