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$\def\truesubset{\subsetneq}$I took grad school algebra many eons ago and recently decided to re-learn it just because. I have been unable to figure out the following question that is the key to a different problem I was working on (that problem described at the bottom).

Let $R$ be a commutative ring with unity. Let $I_1, I_2$ be two ideals. Are the following two equivalent?

  1. If $I_1, I_2$ are co-prime (e.g. $I_1+I_2=R$), then there is no prime ideal $P$ such that $I_1+I_2 \truesubset P \truesubset R$.

  2. If there is no prime ideal $P$ such that $I_1+I_2 \truesubset P \truesubset R$, then $I_1, I_2$ are co-prime.

In other words, being co-prime and the sum not being contained in any prime ideal are equivalent characterizations.

1 above is obvious from the definition of co-prime. Since $I_1 + I_2=R$, it is impossible to squeeze any SET between $I_1+I_2$ and $R$.

I can see why 2 would be true in a principal ideal ring, but I cannot prove it as stated. I cannot disprove either because I don't have fresh in my mind enough examples of non-principal ideal rings.

FYI, the characterization in 2 was used in this website some years ago to show that if two ideals are co-prime, so are the sums of any powers of the two ideals (e.g. $I_1+I_2=R$ implies $\forall n \in \mathbb N^+, I^n_1+I^n_2=R$).

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  • $\begingroup$ equivalence is by definition, maybe notice that R/I+J is isomorphic to the tensor product of R/I and R/J... $\endgroup$
    – Mars
    Mar 24, 2017 at 9:18
  • $\begingroup$ 2. is basically equivalent to the statement that a ring has no maximal ideals. This would turn the whole algebra upside down :) $\endgroup$
    – MooS
    Mar 24, 2017 at 9:39

2 Answers 2

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Your property 2 is not true -- a counterexample would be to let $I_1$ be some maximal ideal and $I_2$ be $\{0\}$.

Your property 1 is true, but for the trivial reason that you're assuming that $I_1+I_2=R$, then the conclusion is just that there is no prime ideal $P$ such that $R\subsetneq P\subsetneq R$. But of course there is nothing (prime or not, ideal or not) that sits strictly between $R$ and itself.

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  • $\begingroup$ Henning, are you referring to 2 in rschwieb or 2 in my problem? It seems you are giving a counter example to 2 in his, not mine. I still do not have a counterexample to #2 above. I know the example cannot come from a principal ideal ring like the integers. $\endgroup$
    – Pablo A Perez-Fernandez
    Mar 24, 2017 at 14:54
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    $\begingroup$ @PabloAPerez-Fernandez: It's a counterexample to your number 2. With my counterexample $I_1+I_2$ is a maximal ideal (namely, it equals $I_1$), so there is no ideal (prime or not) that is strictly between $I_1+I_2$ and $R$. But $I_1$ and $I_2$ are not coprime. $\endgroup$
    – hmakholm left over Monica
    Mar 24, 2017 at 15:17
  • $\begingroup$ @HenningMakholm---I understand your counter example for #2 in my question. It even works for the integers. Let $I_1 = 2 \mathbb Z$. Then, $I_1$ is maximal. Letting $I_2 = \{0\}$, we see that there is no proper prime ideal properly containing $I_1+I_2$. Yet, $I_1+I_2 \neq \mathbb Z$. Thanks much for the help. $\endgroup$
    – Pablo A Perez-Fernandez
    Mar 24, 2017 at 16:58
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The whole problem stems from the $\subsetneq P$ in your proposed property $2$, as subsequent counterexamples in comments and solutions demonstrated.

But relaxing that to $\subseteq$, the following are obviously equivalent for any $I\lhd R$ (granted Zorn's lemma etc etc):

  1. There is no prime ideal $P$ such that $I\subseteq P$.
  2. There is no maximal ideal $M$ such that $I\subseteq M$.
  3. $I=R$

Applying this to $I=I_1+I_2$, you would indeed be able to conclude that $I_1+I_2=R$.

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