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I am beginner in algebra. I want to know if every power of a prime ideal is a principal ideal. Is the statement correct or is there a counterexample?

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Hint (because this looks like homework): It would be enough to find a ring with a prime ideal that is not principal. –  Henning Makholm Dec 20 '11 at 15:02
    
@HenningMakholm Why did you delete it, I was thinking the same.. was it something wrong in that? –  AD. Dec 20 '11 at 15:03
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@AD: the last sentence of the question as originally asked sounded to me like a translation from a homework exercise in some non-English language -- so I decided I'd rather not provide a complete solution. –  Henning Makholm Dec 20 '11 at 15:05
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Vahid: Start by looking at a non-PID. –  AD. Dec 20 '11 at 15:09
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Vahid, please try to use a respective language. –  AD. Dec 20 '11 at 15:19

2 Answers 2

Hint $\ $ In a domain D, every power of a prime ideal is principal iff D is a PID. Indeed, if every power of a prime ideal is principal then every prime ideal is principal, so D is a PID, by the proof below. Conversely, in a PID every ideal is principal, hence so is every power of a prime ideal.

Below is a proof of the key step, from my 2008/11/10 Ask an Algebraist post.

In reply to "principal prime ideals imples PID", posted by Student on November 9, 2008:

Let $\rm\:R\:$ be an integral domain. Let every prime ideal in $\rm\:R\:$ be principal.
Prove that $\rm\:R\:$ is a principal ideal domain (PID).

Below I give a simpler, more general way to view the proof, and references. First let's recall a standard proof, e.g. that given by P.L. Clark, paraphrased below.

Proof $\rm\:\! 1\!:\:$ If not then the set of all nonprincipal ideals is nonempty. Let $\rm\:\{I_j\}\:$ be a chain of nonprincipal ideals and put $\rm\:I = \bigcup_j I_j.\:$ If $\rm\:I = (x)\:$ then $\rm\:x \in I_j\:$ for some $\rm\:j,\:$ so $\rm\:I = (x) \subset I_j\:$ implies $\rm\:I = I_j\:$ is principal $\rm\:\Rightarrow\Leftarrow\:$

Thus by Zorn's Lemma there is an ideal $\rm\:I\:$ which is maximal with respect to the property of not being principal. As is so often the case for ideals maximal with respect to some property or other, we can show that $\rm\:I\:$ must be prime. Indeed, suppose that $\rm\:ab \in I\:$ but neither $\rm\:a\:$ nor $\rm\:b\:$ lies in $\rm\:I.\:$ So ideal $\rm\:J = (I,a)\:$ is strictly larger than $\rm\:I,\:$ so principal: $\rm\:J = (c).\:$ $\rm\:(I:a)\ :=\ \{r \in R : ra \in I\}\:$ is an ideal containing $\rm\:I\:$ and $\rm\:b,\:$ so strictly larger than $\rm\:I\:$ and thus principal: say $\rm\:(I:a) = (d).\:$ Let $\rm\:i\in I,\:$ so $\rm\:i = uc.\:$ Now $\rm\:u(c) \subset I\:$ so $\rm\:ua \in I\:$ so $\rm\:u \in (I:a).\:$ Thus we may write $\rm\:u = vd\:$ and $\rm\:i = vcd.\:$ This shows $\rm\:I \subset (cd).\:$ Conversely, $\rm\:d \in (I:a)\:$ implies $\rm\:da \in I\:$ so $\rm\:d(I,a) = dJ \subset I\:$ so $\rm\:cd \in I.\:$ Therefore $\rm\:I = (cd)\:$ is principal $\rm\:\Rightarrow\Leftarrow\quad$ QED

I show that the second part of the proof is just an ideal-theoretic version of a well-known fact about integers. Namely suppose that the integer $\rm\:i>1\:$ isn't prime. Then, by definition, there are integers $\rm\:a,b\:$ such that $\rm\:i\mid ab,\:$ $\rm\:i\nmid a,\:$ $\rm\:i\nmid b.\:$ This immediately yields a proper factorization of $\rm\:i,\:$ namely $\rm\ i = c\:\! (i:c)\:$ where $\rm\:c = (i,a).\:$ Thus: not prime $\Rightarrow$ reducible $ $ (or: irreducible $\Rightarrow$ prime). Below is an analogous constructive proof generalized to certain ideals.

Theorem $\rm\ \:$ Suppose that $\rm\:R\:$ is a ring, and suppose that $\rm\:I\ne 1$ is an ideal of $\rm\:R\:$ which satisfies the property: $\ $ ideal $\rm\:J\supset I\: \Rightarrow\: J\mid I\:.\:$ Then $\rm\:I\:$ not prime $\rm\:\Rightarrow\:$ $\rm\:I\:$ reducible (properly).

Proof $\rm\:\ \ I\:$ not prime $\rm\:\Rightarrow\:$ $\rm\:\exists\: a,b \notin I\:$ with $\rm\:ab \in I.\:$ $\rm\:A\: :=\: (I,a)\supset I\: \Rightarrow\: A\mid I,\:$ say $\rm\:I = AB\:$; wlog we may assume that $\rm\:b \in B\:$ since $\rm\:A(B,b) = AB\:$ via $\rm\:Ab = (I,a)b \subset I = AB.\:$ Finally the factors $\rm\:A,B\:$ in $\rm\:I = AB\:$ are proper: $\rm\:A = (I,a),\:$ $\rm a \notin I\:$; $\rm\:\ B \supseteq (I,b),\:$ $\rm b \notin I.\quad$ QED

The contains $\Rightarrow$ divides hypothesis: $\rm\:J\supset I\: \Rightarrow\: J\mid I\:$ is true for principal ideals $\rm\:J\:$ (hence proof 1), and also holds true for all ideals in a Dedekind domain. Generally such ideals $\rm\:J\:$ are called multiplication ideals. Rings whose ideals satisfy this property are known as multiplication rings. Their study dates back to Krull.

The OP's problem is Exercise $1\!-\!1\!-\!10,\: p. 8\:$ in Kaplansky: Commutative Rings, viz.

$10.\:$ (M. Isaacs) In a ring $\rm\:R\:$ let $\rm\:I\:$ be maximal among non-principal ideals. Prove that $\rm\:I\:$ is prime. (Hint: adapt the proof of Theorem $7$. We have $\rm\:(I,a) = (c).$ This time take $\rm\:J = \{x\in R\mid xc \in I\}.$ Since $\rm\:J \supset (I,b),\:$ $\rm\:J\:$ is principal. Argue that $\rm\:I = Jc\:$ and so is principal.)

See also this AaA post and the following AMS Math Review.


Mott, Joe Leonard. $ $ Equivalent conditions for a ring to be a multiplication ring.
Canad. J. Math. 16 1964 429--434. $ $ MR 29:119 13.20 (16.00)

If "ring" is taken to mean a commutative ring with identity and a multiplication ring is a "ring" in which, when $\rm\:A\:$ and $\rm\:B\:$ are ideals with $\rm\:A \subset B,\:$ there is an ideal $\rm\:C\:$ such that $\rm\:A = BC,\:$ then it is shown that the following statements are equivalent.

  1. $\rm\:R\:$ is a multiplication ring;
  2. if $\rm\:P\:$ is a prime ideal of $\rm\:R\:$ containing ideal $\rm\:A\:$ then there is an ideal $\rm\:C\:$ such that $\rm\:A = PC\!\:;$
  3. $\rm\:R\:$ is a ring in which the following three conditions are valid:
    a. $\ $ every ideal is equal to the intersection of its isolated primary components;
    b. $\ $ every primary ideal is a power of its radical;
    c. $\ $ if $\rm\:P\:$ is a minimal prime of $\rm\:B\:$ and $\rm\:n\:$ is the least positive integer such that $\rm\:P^n\:$ is an isolated primary component of $\rm\:B,\:$ and if $\rm\:P^n \neq P^{n+1},\:$ then $\rm\:P\:$ does not contain the intersection of the remaining isolated primary components of $\rm\:B.\:$ (Here an isolated $\rm\:P\!\!-\!primary$ component of $\rm\:A\:$ is the intersection of all $\rm\:P$-primary ideals that contain $\rm\:A.)$
    Reviewed by H. T. Muhly
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@Bill: Hi Bill, I know this question was a long time ago, but I just happen to read it. First, I found this proof and statement very interesting, thanks for posting it :). I would like to understand better your sentence "As is so often the case for ideals maximal with respect to some property or other, we can show that I must be prime". Would you have some other examples of this phenomenon ? Or other proofs using this same idea ? Thanks ! –  Bogdan Oct 15 '12 at 0:15
    
@redfiloux Actually those are P. L. Clark's words (from said paraphrased proof). For some further examples see the references here. and see also this answer. –  Bill Dubuque Oct 15 '12 at 0:19
    
@Bill: Great ! Interesting. Thanks for the references. –  Bogdan Oct 15 '12 at 0:32
    
@Bill: That's another good example, which perfectly fits in my building database of such small examples and tricks. Thanks some more. –  Bogdan Oct 15 '12 at 1:09

If you consider the ring $\mathbb R[X,Y]$ , its ideal ${\frak m}=(X,Y)\subset \mathbb R[X,Y]$ is prime (do you see why?). However none of its powers ${\frak m}^n \quad (n\geq 1)$ is principal.

To get you started assume $n=1$ and suppose ${\frak m}$ is principal, that is ${\frak m}=(X,Y)=(f(X,Y))$ for some $f(X,Y)\in \mathbb R[X,Y]$
We would deduce that $(f(X,Y)$ divides both $X$ and $Y$ so that $f(X,Y)$ would be a non-zero constant $r\in \mathbb R$ .
But then the assumption ${\frak m}=(X,Y)=(f(X,Y))=(r)$ is absurd (why?).

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