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Let $R$ be a ring with a single proper ideal $I$. I need to prove that $I$ is prime, but all I have at my disposal to do it with are the definitions of prime and maximal ideals as stated below:

  • An ideal $I$ of a ring $R$ is prime if $I \neq R$ and the following condition is satisfied: for all ideals $J_{1}$, $J_{2}$ of $R$, if $J_{1}J_{2} \subseteq I$, then $J_{1} \subseteq I$ or $J_{2} \subseteq I$.
  • An ideal $I$ of $R$ is maximal if $I \neq R$ and there is no proper ideal of $R$ containing $I$ (i.e., if $J$ is an ideal of $R$ properly containing $I$ then $J = R$).

$R$ is not necessarily commutative, it does not necessarily have unity, so I can't assume any of those things in my proof.

This is my attempt:

Suppose $R$ is a ring with a single proper ideal $I$. Then, since there is no proper ideal $J$ of $R$ containing $I$, $I$ is maximal.

Now, the only ideals contained in $I$ are the zero ideal $\{0\}$ and $I$ itself. So, for any $J_{1}$ of $R$ such that if $J_{2} = \{0\}$, we have that $J_{1}J_{2} = J_{1}\{0\} = \{0\} \subseteq I$, then $J_{2} = \{0\} \subseteq I$. Moreover, for any $J_{1}$ of $R$ such that if $J_{2}=I$, we have that $J_{1}J_{2}=J_{1}I \subseteq I$, then $J_{2}=I \subseteq I$. (However, the only choices for $J_{1}$ in either case are $\{0\}$ or $I$ in order for $J_{1}J_{2} \subseteq I$). Therefore, $I$ is prime.

Is this all there is to it? It just seems so silly...

If not, how should I prove it?

Thank you,

  • Your argument boils down to "if $J_2=0$ then $J_2\subseteq I$ and if $J_2=I$ then $J_2\subseteq I$." Is that what you're supposed to prove? – anon Mar 20 '17 at 07:08
  • @arctictern I think so. I'm supposed to prove that $I$ is prime, and that's what it means to be prime, isn't it? –  Mar 20 '17 at 07:11
  • You said $I$ is prime if $J_1J_2\subseteq I$ implies $J_1\subseteq I$ or $J_2\subseteq I$. Is that what you did? – anon Mar 20 '17 at 07:13
  • @arctictern The only other possibilities for what $J_{1}$ could be are ${0}$ and $I$, so I thought so, but perhaps I didn't. –  Mar 20 '17 at 07:14
  • The more I reread your argument the less I understand the English. Call "if $J_2=0$ then $J_1J_2\subseteq I$" condition X. You're saying that if $J_1$ is such that condition X holds, then $J_2=0$. But that's not true. Just because $J_1J_2\subseteq I$ would be true if $J_2=0$ doesn't mean that $J_2$ is $0$. Moreoever, condition X is always true, no matter what $J_1$ is. And finally, I don't see how that goes about proving the implication in the definition of prime ideal. – anon Mar 20 '17 at 07:20
  • @arctictern I'm trying so hard to stay awake, I'm not surprised it didn't make sense. I'm going to try to sleep some and then work on the rest of this in the morning. –  Mar 20 '17 at 07:21
  • One widespread naming scheme says that proper means "not the whole ring" and nontrivial means "not the whole ring and not the zero ideal." Just to make sure, you definitely want the latter meaning, right? – rschwieb Mar 20 '17 at 10:48
  • @rschwieb yup. For both of those. –  Mar 20 '17 at 12:00

2 Answers2

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I think the property does not hold if $R$ is not assumed to have a unit.

Indeed, take $p$ to be your favorite prime number, and consider the ideal $R = p\mathbb{Z}/p^3\mathbb{Z}$, which is a ring without a unit. Its ideals are precisely the ideals of $\mathbb{Z}/p^3\mathbb{Z}$ that are contained in $p\mathbb{Z}/p^3\mathbb{Z}$. But any ideal of $\mathbb{Z}/p^3\mathbb{Z}$ is of the form $k\mathbb{Z}/p^3\mathbb{Z}$ where $k\mid p^3$. So any ideal of $R$ will be of the form $k\mathbb{Z}/p^3\mathbb{Z}$ where $p\mid k\mid p^3$.

Now for it to be proper we must have $p<k<p^3$, and thus $k=p^2$. So $R$ verifies the assumption : it has only one proper ideal $I$. However this proper ideal is not prime, indeed denoting $\overline{x}$ the class of $x$ mod $p^3$, one has $\overline{p}\times \overline{p} \in I$, but $\overline{p}\notin I$.

So arctic tern's proof works when $R$ has a unit, but the property isn't true if $R$ doesn't have a unit.

user26857
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Maxime Ramzi
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  • interesting observation. –  Mar 20 '17 at 12:17
  • so is what you're saying that if $R$ is a ring without a unit, then $R$ having a single proper ideal $I$ does not imply that $I$ is prime, or just that the way arctictern did it is not how you prove it in that case? –  Mar 20 '17 at 12:34
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    Well I have found an example of a ring without unit that contradicts what you want to prove: so you can't prove it. My comment under arctic tern's answer was just to say that his proof did not work if $R$ didn't have a unit, but here I've proved that no proof can work, because it's not true – Maxime Ramzi Mar 20 '17 at 13:13
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    So "single ideal" does not imply "prime ideal" in a non unital ring – Maxime Ramzi Mar 20 '17 at 13:20
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I am not sure what it is you're trying to do in your argument. You seem to be showing that if $J_2$ is contained in $I$, then it is contained in $I$, which is both tautologically true and not what you're supposed to be showing.

You want to prove

$$ J_1J_2\subseteq I \implies (J_1\subseteq I ~\textrm{ or }~ J_2\subseteq I).$$

This is equivalent to the contrapositive

$$ (J_1\not\subseteq I ~\textrm { and }~ J_2\not\subseteq I)\implies J_1J_2\not\subseteq I.$$

In your case, the only ideal not contained in $I$ is $R$ itself, in which case $J_1\not\subseteq I$ is equivalent to $J_1=R$, and similarly $J_2\not\subseteq I$ is equivalent to $J_2=R$, and so $J_1J_2=R\not\subseteq I$, as desired.

anon
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  • thank you. My brain is sleep-deprived. –  Mar 20 '17 at 07:17
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    How does $J_1 J_2 = R$ if $R$ is not supposed to have a unit (according to OP, it's not assumed)? For instance, $R=2\mathbb{Z}$ is a ring without a unit (indeed it is an ideal of $\mathbb{Z}$), and so $RR = 4\mathbb{Z} \neq R$. So the claim that $J_1J_2 = R$ has to be proved in more detail, if it is true (at least I think so). – Maxime Ramzi Mar 20 '17 at 10:16
  • @Max Ah, you're right. Good catch. This is why my ring's always have an identity... – anon Mar 20 '17 at 15:53
  • @arctic tern : Yes, I'm myself used to the convention that "a ring has a unit", but since OP precisely said that he didn't assume that his ring had an identity, I mentioned it – Maxime Ramzi Mar 20 '17 at 16:04
  • @Max Yeah. (Also, she.) – anon Mar 20 '17 at 16:07