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I am trying to solve Problem 7.4.40 from Dummit and Foote, a part of which states:

Let $R$ be a commutative ring with $1\neq 0$ such that $R$ has exactly one prime ideal. Then every element of $R$ is either nilpotent or a unit.

ATTEMPT: Let $P$ be the unique prime ideal of $R$. We will show that every element of $R\setminus P$ is a unit. Let $r\in R\setminus P$. Then the ideal $(r)$ is either equal to $R$ or is a proper ideal of $R$. If $(r)=R$ then $r$ is a unit. So assume $(r)\subsetneq R$. But then $(r)$ is contained in a maximal ideal $M$ of $R$. Clearly $M\neq P$. But $M$ is also a prime ideal of $R$ since it is a maximal ideal of $R$, giving a contradiction.

So we have shown that $x\in R\setminus P\Rightarrow r$ is a unit.

Also, since $P$ is prime, we have $\mathfrak N(R)\subseteq P$.

What I am struggling with is showing that $\mathfrak N(R)=P$.

NOTE: The result that $\mathfrak N(R)$ is the intersection of all the prime ideal of $R$, where $R$ is a commutative ring with $1\neq 0$, is not yet discussed in the textbook (See the comment in Problem 7.4.26 in D&F). So I think there must be an elementary argument which shows $\mathfrak N(R)=P$ in the question at hand.

Thanks.

user26857
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caffeinemachine
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1 Answers1

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Let $\mathfrak{p}$ be the unique prime ideal of $R$, and $f \in \mathfrak{p}$ any element. To show that $f$ is nilpotent is equivalent to show that $R_f$ is the zero ring. What do you know about the prime ideals of $R_f$? Note that a ring is the zero ring if and only if it has no prime ideals.

Added later: Since the OP is not aware of localizations, let us just define $R_f := R[X]/(fX-1)$ and show that $R_f$ is the zero ring for all $f \in \mathfrak{p}$.

If there is any prime ideal in $R_f$ we also have a prime ideal $\mathfrak{q}$ in $R[X]$ containing $(fX-1)$. Since $\mathfrak{p}$ is the unique prime of $R$ we have $\mathfrak{q} \cap R = \mathfrak{p}$, so $\mathfrak{q}$ contains $f$. But this is a contradiction since the invertible element $\overline{f} \in R_f$ cannot lie in any prime ideal. So $R_f = 0$.

From this it follows easily that $f$ is nilpotent: Since $R_f = 0$ we have $1 \in (fX-1)$, so $fX-1$ is invertible in $R[X]$. Now we know (I hope you know) that a polynomial is invertible if and only if the lowest coefficient is invertible and all other coefficients are nilpotent.

Dune
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  • @Dune Tank you for your response. I am sorry I am not aware of the notation $R_f$ and probably not aware of the concept underlying the notation. I want to do this problem 'by the book'. – caffeinemachine Sep 11 '14 at 08:13
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    @caffeinemachine: $R_f$ is the ring in which $f$ was made invertible. You could equivalently replace $R_f$ by $R[X]/(fX-1)$ and try to follow my hint. But I don't know whether it could be as easily done with this definition of $R_f$ as with the usual. – Dune Sep 11 '14 at 08:16
  • Thank you Dune for the explanation. The link you provided in the comment to the original post is very helpful. – caffeinemachine Sep 11 '14 at 08:26
  • @user26857: I agree - but it's also quite nice (particularly the last sentence) – zcn Sep 11 '14 at 08:54
  • @user26857: It is all cheating. I want to see any specialized proof that does not additionally show that the nilradical is the intersection of all primes in any ring (since the OP wanted to avoid this statement). – Dune Sep 11 '14 at 09:00