Two different maximal ideals of a commutative ring

Question

I'm currently working on the following problem:

Let $R$ be a commutative ring, and let $I, J$ be maximal ideals of $R$. Prove that, if $a, b \in R$, there exists some element $c \in R$ such that $c-a \in I$ and $c - b \in J$.

I know that if $I,J$ are maximal ideals of $R$, then for all ideals $K$ of $R$, we have the following:

1) If $I \subseteq K \subseteq R$, either $K = I$ or $K = R$.

2) If $J \subseteq K \subseteq R$, either $K = J$ or $K = R$.

I suppose I'm not seeing the trick to producing this element $c$ such that both $c-a$ and $c-b$ lie in $I$ and $J$, respectively, for two ring elements in $R$. Any help as to producing this element $c$ would be appreciated.

Thanks!

Answer 1

$I+J=R$, since they are maximal, therefore there are $i \in I, j \in J$ such that $i+j=1$.

Let $c=ai+bj$.

$c-a = a(i-1)+bj=a(-j)+bj=(b-a)j \in J\\ c-b = ai+b(j-1)=ai+b(-i)=(a-b)i \in I.$

Answer 2

Hint:

Prove the sequence (without hypotheses on $I$ and $J$) $$0\longrightarrow R/I\cap J\longrightarrow R/I\otimes R/J\longrightarrow R/(I+J)\longrightarrow 0$$ is exact, then use that if $I$ and $J$ are distinct maximal ideals, $I+J=R$.

Answer 3

$I $ is not contained in $J$ thus there exists $d\in I, \not \in J$.

Because $J$ is a maximal ideal $d \in R/J^\times$ and $\exists e\in R$, $de =1\in R/J$.

Similarly find $f\in J, g\in R, fg = 1\in R/I$.

Then for all $a,b\in R$ $$c = afg+bde,\qquad c-a\in I, c-b\in J$$ and hence $(a,b) \mapsto afg+bde$ is an embedding $R/I\times R/J \to R/IJ$ and an isomorphism since its inverse is $u \mapsto (u\bmod I,u\bmod J)$