Consider the ring $R$ of non commutative real polynomials in $X$ and $Y$. Denote by $I$ the principal two-sided ideal generated by $X$ and $J$ the principal two-sided ideal generated by $XY+1$. Then $I+J=R$, but $I∩J≠IJ$. I understand the first part since $1\in I+J$, but I could not show $I∩J≠IJ$. Any example to show? Thanks
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$R$ is not clear, what do you mean by "non-commutative real polynomials in $X$ and $Y$" ? For me, "real polynomials in $X$ and $Y$" sounds like $R=\mathbb R[X,Y]$ which is commutative. Maybe you mean $R$ non commutative, and consider $R[X,Y]$ ? – Surb Oct 01 '19 at 08:39
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This is the exact wording taken from a lecture note. I think you are right in interpretation. Any example to show. Thanks – Fellow InstituteOfMathophile Oct 01 '19 at 08:43
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I searched online, there do exist non commutative polynomials in which variables do not commute. I think the author means that R<x,y> – Fellow InstituteOfMathophile Oct 01 '19 at 08:50
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I think non commutative here means that $XY \neq YX$. The base ring is just the reals. – quarague Oct 01 '19 at 08:50
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The key idea here is that if the ring doesn't commute, we don't necessarily have $IJ=JI$ whereas $I \cap J= J \cap I$. So consider the element $(XY+1)\cdot X$. It is in $I \cap J$ but not in $IJ$ hence $I \cap J \neq IJ$.
quarague
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I feel myself so dumb. You made it so simple. Thanks. – Fellow InstituteOfMathophile Oct 01 '19 at 09:04
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Ah, I see. We can quotient further by $X\left(XY+1\right)$ and $Y\left(XY+1\right)$; this will kill every element of $IJ$. The quotient has a confluent rewriting system $XXY \to -X$ and $YXY \to -Y$, but the element $\left(XY+1\right)X$ is reduced modulo this quotient. – darij grinberg Jan 22 '21 at 03:13