For a ring $R$ with a single proper ideal $I$, show that $I$ is prime

Question

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,

Answer 1

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.

Answer 2

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.