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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.

user26857
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Truong
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    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? – rschwieb Dec 06 '13 at 14:43

3 Answers3

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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.

rschwieb
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  • 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? – user193319 Feb 12 '18 at 22:11
  • @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. – rschwieb Feb 12 '18 at 22:36
  • 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. – user193319 Feb 12 '18 at 23:23
  • @user193319 It is basically necessary to prove the lemma that given a multiplicatively closed set, there exists an ideal maximal with respect to being disjoint from that set, and it is necessarily prime. Then, if $x$ is not nilpotent, you can say there is a prime ideal disjoint from the powers of $x$, but because there is only one prime (which is maximal), it must be a unit – rschwieb Feb 13 '18 at 02:16
  • Could you explain more the proof of $\Rightarrow$ direction? – RFZ Jan 30 '19 at 01:58
  • @ZFR If each unit is a unit or nilpotent, the nilradical is a maximal ideal. Every prime ideal must contain the nilradical, so there is exactly one prime ideal. – rschwieb Jan 30 '19 at 02:24
  • @rschwieb for the converse implication , there are elements in a ring other than nilpotent or unit, like zero divisors which are not nilpotent and elements which are not unit nor zero divisors. What about these elements? Why didn't they belong to the prime ideal P, can you tell me please? I just wanna know, as I recently studied about these. – A learner Sep 23 '20 at 16:42
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    @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. – rschwieb Sep 23 '20 at 16:55
  • @rschwieb you are saying, maximal ideal refused to take any zero divisors which are not nilpotent, and elements neither zero divisors nor unit. Why? Is there any proof about this? – A learner Sep 23 '20 at 16:59
  • Ooh! I think you are saying, as nilredical is intersection of all primes, but as there is only one prime, hence nilredical becomes the prime. Is it? – A learner Sep 23 '20 at 17:02
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    @Alearner That's part of it, yes. – rschwieb Sep 23 '20 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. "

Truong
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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.

BIS HD
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