4

So I was thinking about this the other day, say we have three comaximal ideal, $I,J,K$ in a commutative ring. (this means that $I+J=J+K=K+I=R$), does it have to follow that $I+JK=R$?

I was trying to argue that if $I+JK\neq R$ then there is a maximal ideal $M$ containing it, so it also contains $I$, if I manage to show that $M$ contains $K$ or $J$ then I would be done since it would contain their sum, but it would contradict the maximality assumption.

So far I haven't done it (and it might also be false, its just something I wondered when I was reading)

Daniel Montealegre
  • 6,883

2 Answers2

5

Your start is OK:

A maximal ideal containing $I+JK$ would be a prime ideal containing $JK$ and $I$, and so one of $J$ or $K$ are contained in the maximal ideal. Can you see why this spells a contradiction?

rschwieb
  • 160,592
  • I dont see how we get $JK\subset M$ then $J\subset M$ or $K\subset M$. Primality of $M$ gives us that for elements, but how does it follow for actual ideals? – Daniel Montealegre Oct 25 '13 at 20:24
  • Like a priori I could have $j_1k_1\in M$ and $k_1\in M$ and $i_1\notin M$ and $i_2k_2\in M$ with $i_2\in M$ and $k_2\notin M$. – Daniel Montealegre Oct 25 '13 at 20:28
  • 2
    @DanielMontealegre The property is true for prime ideals. Suppose $JK\subseteq M$ and $J\not\subseteq M$; then there is $a\in J$, $a\notin M$. Now take $b\in K$ and do $ab$; what can you say? – egreg Oct 25 '13 at 20:44
  • @egreg thanks!! – Daniel Montealegre Oct 26 '13 at 19:11
  • @DanielMontealegre The "ideal-wise" definition of primeness generalizes the "element-wise" definition to noncommutative rings. Thanks, egreg, for helping explain that :) – rschwieb Oct 28 '13 at 12:54
3

We only need $I+J=R$ and $I+K=R$ to prove that $I+JK=R$ assuming that your commutative ring $R$ has a $1$.

$R\\=R^2\\=(I+J)(I+K)\\=\underbrace{I^2+IK+JI}_{\subseteq \,I}+JK \\\subseteq I+JK $

Thus, $R=I+JK$.

Note: In order to justify $R=R^2$, assuming that $R$ has a $1$ is essential. The other argument also assumes that $R$ has a $1$. It uses the fact that every proper ideal is contained in some maximal ideal. This is a consequence of Zorn's lemma wherein this argument is used: Any ascending chain $I_1\subsetneq I_2\subsetneq\ldots$ of proper ideals is bounded by their union $\bigcup I_k$. What guarantees that this union is a proper ideal? It's the fact that $1$ is not in the union.

Nothing special
  • 3,690
  • 1
  • 7
  • 27