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.

\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 $I\cap J \ne IJ$.
