0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

\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 $I\cap J \ne IJ$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.