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Let $R$ be a ring. Prove that each element of $R$ is either a unit or a nilpotent element iff the ring $R$ has a unique prime ideal.

Help me some hints.

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    $\begingroup$ Are you assuming commutativity? And $F[x]/(x^3)$ is a counterexample to the question as currently phrased. Did you mean to ask if there is a unique prime ideal? $\endgroup$
    – rschwieb
    Dec 6, 2013 at 14:43

3 Answers 3

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$F[x]/(x^3)$ consists of units and nilpotent elements, but has four ideals, so this suggests you meant something more like unique prime ideal.

This is indeed true for commutative rings. The hypothesis that nonunits are nilpotent means that the nilradical is a maximal ideal. But considering that all prime ideals contain the nilradical, the nilradical is precisely the one prime ideal in the ring.

Conversely, if you assume the ring has one prime ideal, then there is clearly only one maximal ideal, and everything inside it is a nonunit, hence nilpotent.

The statement is false for noncommutative rings. $M_2(R)$ has exactly one prime ideal: $\{0\}$. Needless to say there are non-nilpotent non-units in this ring (for example $\begin{bmatrix}1&0\\0&0\end{bmatrix}$.)

It might be interesting though to follow up and see if any of the new one-sided prime ideal definitions makes this work in noncommutative rings.

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  • $\begingroup$ I don't understand your proof of the converse implication. Every maximal ideal is prime, so there is only one maximal ideal; and no maximal ideal is equal to the entire ring, which means it cannot contain any units. How do you conclude from this that every element is nilpotent? $\endgroup$
    – user193319
    Feb 12, 2018 at 22:11
  • $\begingroup$ @user193319 In any commutative ring, the intersection of all prime ideals is the set of nilpotent elements. If there is one prime ideal, then it is made of nilpotent elements. $\endgroup$
    – rschwieb
    Feb 12, 2018 at 22:36
  • $\begingroup$ How might one solve this without knowing that fact? I am working on this very problem in Dummit and Foote, but that fact hasn't been proven yet. $\endgroup$
    – user193319
    Feb 12, 2018 at 23:23
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    $\begingroup$ @Alearner when there is one prime ideal, it must be the unique maximal ideal of the entire ring. In such a ring, there nonunits are precisely the things in the maximal ideal. Since there is only one prime, everything inside it is nilpotent. Thus you have units and nilpotents and nothing else. $\endgroup$
    – rschwieb
    Sep 23, 2020 at 16:55
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    $\begingroup$ @Alearner That's part of it, yes. $\endgroup$
    – rschwieb
    Sep 23, 2020 at 17:11
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http://am-solutions.wikispaces.com/Solutions+to+Chapter+1

"Let $A$ be a ring, $R$ its nilradical. Show that the following are equivalent:

1) $A$ has exactly one prime ideal;

2) every element of $A$ is either a unit or nilpotent;

3) $A/R$ is a field.

Proof. 1) ⇒ 2). Observe that $R$, which is the intersection of the prime ideals, is equal to the given prime ideal; and that $A$ is a local ring. Thus $A−R=A^∗$ and by definition $R$ consists of all nilpotent elements.

2) ⇒ 3). The quotient map $A→A/R$ is surjective. Since ring homomorphisms map units to units, $x∈A/R$ is either $0$ or a unit.

3) ⇒ 1). All prime ideals contain $R$, and $R$ is a maximal ideal: hence there is one prime ideal. "

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Hint for $\Leftarrow$:

Every ring with identiy has a prime ideal. Let $P$ be a prime ideal of $R$. Then it contains all the nilpotent elements (why?). It does not contain a unit (why?), so it is in fact the set of nilpotent elements of $R$ and hence because of the condition the unique prime ideal.

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