Please, take a look at The Chinese Remainder Theorem for Rings. for the theorem. My text gives an example to show that the theorem is not true for the non-commutative case. I do not understand the example so I hope that you can explain to me.

Example: Consider the ring $R$ of non-commutative real polynomials in $X$ and $Y$. Let $I$ be the principle two-sided ideal generated by $X$ and $J$ be the principle two-sided ideal generated by $XY + 1$. Then $I + J = R$ but $I \cap J \neq IJ$.

Can you explain to me why $I + J = R$ and why $I \cap J \neq IJ$? Thanks.

  • $\begingroup$ The Chinese remainder theorem , as stated at the linked question does hold for noncommutative rings. The only thing that doesn't carry over from the commutative version is the product=intersection assertion. $\endgroup$ – rschwieb Feb 9 '17 at 5:28

\begin{align*} &xy \in I \text{ and } xy + 1 \in J\\[6pt] \implies\; &1 \in I+J\\[6pt] \implies\; &I+J = R \end{align*}

For the other question, let $r = (xy + 1)x$.

Then $r \in I$ and $r \in J$, hence $r \in I\cap J$.

However the lack of commutativity of the variables $x,y$ implies $r \notin IJ$.

Therefore $I\cap J \ne IJ$.

  • $\begingroup$ Thanks. Anyway, for a clarification, by IJ do they mean cartesian product? $\endgroup$ – geniusacamel Feb 9 '17 at 2:31
  • $\begingroup$ No, it's the two-sided ideal generated by the internal element-wise products. $$IJ = \text{the ideal generated by } \{ij \mid i \in I, j \in J\}$$ $\endgroup$ – quasi Feb 9 '17 at 2:41

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