Is this another way of stating the Chinese Remainder Theorem?

Question

Assume that $I + J = R$. Let $a,b \in R$. Find an element $u$ of $R$ satisfying $a + I = u + I$ and $b + J = u + J.$

I want to work on this, but I feel there's some issue of a missing theorem I don't understand. Does this call for the chinese remainder theorem or is it the theorem itself? I don't think I understand.

Anyway, here is my sloppy attempt at trying to understand the question, by plugging in examples of ideals and rings I am familiar with:

First, I think choose $I = 2 \mathbb Z$ and $J = 3\mathbb Z$. So $I+J = \mathbb Z$ (because these ideals are relatively prime) and so I'm taking $R= \mathbb Z$ so $I+J = \{2a + 3b; a,b \in R \}$.

To find $u$ I begin a clumsy set of calculations:

$$ \begin{matrix} a + I=u+I & b+J =u+J\\ a+I = (2a+3b)+I & b+J=(2a+3b) +J\\ I=(a+3b)+I & J=(2a+2b)+J\\ \end{matrix} $$

so $$ u = 2a+3b$$ or $$u=0$$ or something?

but what I'm getting at is $u \in I+J$ which is clearly the case. I'm just not sure what magic figure I'm supposed to get.

Answer 1

By hypothesis, there are elements $i\in I$, $j\in J$ such that $i+j=1$. Observe that $j\equiv 1\mod I$ and $i\equiv 1\mod J$. Now take $u=ib+ja$. We have: $$u\equiv ja\equiv 1 a=a\mod I,\quad\text{and similarly}\quad u\equiv ib \equiv 1 b=b\mod J.$$

Answer 2

You can use the CRT to prove this.

Recall that for co-prime ideals $I,J$, that is $I+J=R$, one has $R/IJ$ is isomorphic to $R/I \times R/J$ (as already mentioned in a comment) under the usual isomorphism $u+IJ \mapsto (u+ I , u+ J)$ (this is a slight abuse of notation).

So you have some class modulo $IJ$, say $u+IJ$ that maps to $(a+ I, b+J)$. Takes that $u$.