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


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*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,

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.
