# Chinese Remainder Theorem for rings.

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.

• 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. – rschwieb Feb 9 '17 at 5:28

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 $I\cap J \ne IJ$.
• No, it's the two-sided ideal generated by the internal element-wise products. $$IJ = \text{the ideal generated by } \{ij \mid i \in I, j \in J\}$$ – quasi Feb 9 '17 at 2:41