# Ring Elements are Either Nilpotent or Unital $\implies$ Ring has a Unique Prime Ideal

EDIT: I originally wrote the proof down wrong. Here is the corrected proof and question:

Let $R$ be a commutative ring with identity.

1. Suppose each element of $R$ is either a unit or a nilpotent element.

2. Consider that $N = \cap P_i$, for $\{P_i\}$ the set of prime ideals of $R$.

3. Now suppose, for sake of contradiction, that $R$ had at least two distinct prime ideals $P_1 \ne P_2$. Let $d$ be an element inside either $P_1$ or $P_2$ that is strictly outside of their intersection.

4. Now $d$ is either nilpotent or it is a unit by hypothesis.

5. Suppose $d$ is nilpotent. Then $d \in \cap P_i \subseteq (P_1 \cap P_2)$, contradicting our choice of $d$ as lying strictly outside the intersection of $P_1$ and $P_2$.

6. Then $d$ must strictly be a unit.

7. But if $d$ is strictly a unit, then any ideal $I$ which contains $d$ is $R$ itself since $u^{-1} \in R \implies u^{-1} u = 1 \in I = R$.

8. This last step means that either $P_1$ or $P_2$ is $R$ itself, contradicting it from being a prime ideal.

9. From here we have that there cannot be more than one prime ideal of $R$.

Question: But how do we know that there is at least one prime ideal in $R$? If for example $R$ is a field, then there is no prime ideal in $R$ so that wouldn't this theorem be false?

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In a field the prime ideal is the zero ideal. – LASV Nov 26 '13 at 2:47

Here is a short proof of your original claim: Let $P$ be a prime ideal of $R$. Then $P$ contains all nilpotent elements (because $a^n \in P \Rightarrow a \in P$). Moreover, $P$ does not contain a unit. Therefore, $P$ coincides with the set of nilpotent elements. In particular, there is only one prime.