Under What Conditions is an ideal $I \subset R$ in the given ring $R$ prime?

Question

I am working on the following problem:

Let $R$ be an additive abelian group. Turn $R$ into a ring by defining $xy = 0$ for all $x$, $y$ $\in$ $R$. Under which conditions is an ideal $I$ of $R$ prime?

I have already shown that if $I$ is prime, then $R - I$ is at most a singleton by showing the contrapositive:

Suppose $I \subset R$ is an ideal of $R$ such that $R - I$ has at least two elements, $x$ and $y$. Then, $x$, $y$ $\notin I$, but because of how multiplication is defined in the ring $R$, $xy=0$, so $xy \in I$, since $I$ is an ideal and thus an additive subgroup of $R$ (and therefore must contain the additive identity element $0$).

However, since $xy \in I$, but $x \notin I$, $y \notin I$, $I$ is not prime.

What I did here though is just show something that has to be true if we already know that $I$ is prime, NOT a condition that $I$ must meet if we want to conclude that it is prime.

I was therefore wondering if this condition is sufficient as well - i.e., that $R - I$ at most a singleton $\implies$ that $I$ is prime, and if so, how do I show it? From the way multiplication is defined in $R$, for any $v, w \in R$, $vw =0$. Thus, this must hold for any two $a,b \in I$ as well. If it helps, the determination of conditions under which $I$ is prime is part (b) of the question. Part (a) asked me to describe all ideals of $R$, and for that part I answered by proving that all ideals of $R$ are additive subgroups. So, if $R - I$ is a singleton, then its additive inverse can't be in $I$, and so I'm very confused.

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UPDATE: It has been brought to my attention that this condition is NOT also sufficient. Therefore, I am at an absolute loss as what would be a sufficient condition. Could somebody please help me figure out what is? Thank you in advance.

Answer 1

Since $I$ is an ideal of $R$, it must contain $0$.

We will show that $I$ can never be prime.

For the sake of a contradiction, assume that $I$ is prime. Then, there is an element $a$ which does not belong to $I$. Then $aa=0 \in I$, but $a \notin I$. Therefore, $I$ is not prime, and a contradiction has been reached.

Thus, we have proved that there are no prime ideals of $R$.

Answer 2

You can strengthen your result to showing that $I$ is either $(0)$ or $R$ as follows: S'pose for contradiction $R\setminus I=\{x\}$ and $I$ is not $(0)$. Then $\#(I)\geq 2$, so $\#(x+I)\geq 2$ as well. By counting, $x+I\not\subseteq R\setminus I$, i.e. $(x+I)\cap I\neq\emptyset$. Subtracting, $x\in I$ contrary to assumption.

Now, presumably you can show that those cases for $I$ are indeed Edit: not prime?