Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.

It's clear that $\mathbb{Z}[\sqrt{-3}]$ is not a Dedekind domain, as it is not integrally closed. I'm having difficulty demonstrating directly that $\mathbb{Z}[\sqrt{-3}]$ does not admit unique factorization of ideals.

Obviously we only need one example of an ideal that doesn't have a unique factorization into prime ideals, but a good answer would provide either a process or some intuition in finding such an example.

share|improve this question
    
My number theory is rusty, but I had some vague idea that prime ideals were invertible, and the product of invertible ideals was invertible. So maybe there is an ideal that is not a product of primes? I think factorizations into prime ideals are likely to be unique by definition of prime. –  Jack Schmidt May 18 '11 at 1:11
    
@Jack: This may be more along the lines of what I should be showing. I'll edit my question so I'm not precluding a possibly correct answer. –  Hans Parshall May 18 '11 at 1:20
    
Cool. I added an answer for a ring I know better, Z[i], except the not-dekekind-domain version. Some people think it is a silly ring, but I claim it is much better for understanding orders in number rings than the deceptive charlatan known as Z[√−3]! –  Jack Schmidt May 18 '11 at 1:22
1  
But the fact that the ring is not integrally closed proves the ring doesn't have unique factorization of ideals, which is a good thing to know: failing to satisfy a logical consequence of unique factorization of ideals implies there is no unique factorization of ideals without having to supply a direct example. I think that is itself rather nice! –  KCd May 18 '11 at 1:38
add comment

4 Answers 4

up vote 11 down vote accepted

In the ring ${\mathbf Z}[\sqrt{-3}]$, the ideal $P = (2,1+\sqrt{-3})$ is prime since it has index 2 in the ring. Note $P^2 = (4,2+2\sqrt{-3},-2 + 2\sqrt{-3}) = (4,2+2\sqrt{-3}) = (2)(2,1+\sqrt{-3}) = (2)P$, where $(2)$ is the principal ideal generated by 2 in ${\mathbf Z}[\sqrt{-3}]$. If there were unique factorization of ideals into products of prime ideals in ${\mathbf Z}[\sqrt{-3}]$ then the equation $P^2 = (2)P$ would imply $P = (2)$, which is false since $1+\sqrt{-3}$ is in $P$ but it is not in $(2)$.

In fact we have $P^2 \subset (2) \subset P$ and this can be used to prove the ideal (2) does not even admit a factorization into prime ideals, as follows. If $Q$ is a prime ideal factor of (2) then $(2) \subset Q$, so $PP = P^2 \subset Q$, which implies $P \subset Q$ (from $Q$ being prime), so $Q = P$ since $P$ is a maximal ideal in ${\mathbf Z}[\sqrt{-3}]$. If (2) is a product of prime ideals then it must be a power of $P$, and $P^n \subset P^2$ for $n \geq 2$, so the strict inclusions $P^2 \subset (2) \subset P$ imply (2) is not $P^n$ for any $n \geq 0$.

The "intuition" that the ideal $P = (2,1+\sqrt{-3})$ is the key ideal to look at here is that $P$ is the conductor ideal of the order ${\mathbf Z}[\sqrt{-3}]$. The problems with unique factorization of ideals in an order are in some sense encoded in the conductor ideal of the order. So you want to learn what a conductor ideal is and look at it in several examples. For example, the ideals $I$ in ${\mathbf Z}[\sqrt{-3}]$ which are relatively prime to $P$ (meaning $I + P$ is the unit ideal (1)) do admit unique factorization into products of prime ideals relatively prime to $P$. That illustrates why problems with unique factorization of ideals in ${\mathbf Z}[\sqrt{-3}]$ are closely tied to the ideal $P$.

If you look at the ideal notation $P' = (2,1+\sqrt{-3})$ in the larger ring ${\mathbf Z}[(1+\sqrt{-3})/2]$, which we know has unique factorization of ideals, then we don't run into any problem like above because $P' = 2(1,(1+\sqrt{-3})/2) = (2)$ in ${\mathbf Z}[(1+\sqrt{-3})/2]$, so $P'$ is actually a principal ideal and the "paradoxical" equation $PP = (2)P$ in ${\mathbf Z}[\sqrt{-3}]$ corresponds in ${\mathbf Z}[(1+\sqrt{-3})/2]$ to the dumb equation $P'P' = P'P'$. (The ideal $P'$ in ${\mathbf Z}[(1+\sqrt{-3})/2]$ is prime since the quotient ring mod $P'$ is a field of size 4: ${\mathbf Z}[(1+\sqrt{-3})/2]$ is isom. as a ring to ${\mathbf Z}[x]/(x^2+x+1)$, so ${\mathbf Z}[(1+\sqrt{-3})/2]/P' = {\mathbf Z}[(1+\sqrt{-3})/2]/(2)$ is isom. to $({\mathbf Z}/2{\mathbf Z})[x]/(x^2+x+1)$, which is a field of size 4.)

share|improve this answer
2  
This is great, thank you. I particularly appreciate the comparison with the $\mathbb{Z}[(1 + \sqrt{-3})/2]$, showing how the "failure" is "resolved". –  Hans Parshall May 18 '11 at 2:09
add comment

Let me take an example with no fractions, Z[2i]=Z[x]/(x^2+4). Now (2+2i)*(2-2i) = 8, but 2+2i is irreducible. So that is a little crazy.

What about the ideal (2)? In the quotient ring we have Z[x]/(2,x^2+4) = Z[x]/(2,x^2) is not a domain, so (2) is not prime. How does it factor into prime ideals?

I claim it doesn't. It ought to factor into (1+i)*(1-i), but there are no such elements in Z[2i]. How to prove it? Well if J = P*Q, then at least J ≤ P, so let's look for primes above (2).

Z[x]/(2,x^2) is a nice local ring with a unique maximal ideal (2,x)/(2,x^2), which pulls back to (2,2i). Since there is only one prime above J=(2), we must have J is some power of P=(2,2i), or else J has no such prime factorization.

Well P*P = (4,4i,-4) = (4,4i) does not contain 2, so that is done. J has no prime factorization.


In case you want to stick with the evil Dr. √−3 and his house of lies, Z[√−3] = Z[x]/(x^2+3), here is an example there:

(1+√−3)(1-√−3) = 4, but 2, 1+√−3, and 1-√−3 are all irreducible. Evil.

What about the ideal (2)? In the quotient ring we have Z[x]/(2,x^2+3) = Z[x]/(2,(x+1)^2) is again not a domain, so (2) is not prime. How does it factor into prime ideals?

I claim it doesn't. (I am little unclear on what it should factor into, since (1+√−3)/2 is a unit, rather than a prime). At any rate, we look for prime ideals above it.

Since Z[x]/(2,(x+1)^2) is a nice local ring with unique maximal ideal (2,x+1)/(2,(x+1)^2) that pulls back to (2,1+√−3), we see that if J=(2) has a factorization into prime ideals, it must be a power of P=(2,1+√−3). However, P*P = (4,2+2√−3,−2+2√−3) = (4,2+2√−3) does not contain 2. Hence J has no such factorization into prime ideals.

share|improve this answer
    
Ooh, Keith Conrad points a nice twist for both examples: P*P = 2*P, but 2 ≠ P, so unique factorization explicitly fails too (well, 2 is not prime but still, that is a failure to factorize if I have ever seen one!). –  Jack Schmidt May 18 '11 at 2:01
    
@Jack: Thanks again for the insight that the problem is not uniqueness, but factoring into prime ideals in the first place. I like the $\mathbb{Z}[2i]$ example. –  Hans Parshall May 18 '11 at 2:11
    
Jack: you should write (2) rather than 2 when you are referring to ideals. –  KCd May 18 '11 at 2:12
    
In the ring ${\mathbf Z}[2i]$, the conductor ideal is $P = (2,2i)$, which Jack identifies as a source of problems. Since $P^2 = (4,4i,-4) = (4)$, we have $P^2 \subset (2) \subset P$ with both inclusions being strict. Therefore a proof that the ideal (2) in ${\mathbf Z}[2i]$ is not a product of prime ideals proceeds exactly the same way as in the answer I gave. –  KCd May 18 '11 at 2:15
    
In the ring ${\mathbf Z}[i]$, the ideal notation $(2,2i)$ is $2(1,i) = (2)(1) = (2)$, a principal (though not prime) ideal and the inclusion $(2) \subset (2,2i)$ is not strict as it was with ideals in ${\mathbf Z}[2i]$. That is consistent with why the strange behavior in ${\mathbf Z}[2i]$ can not occur in ${\mathbf Z}[i]$, which we know has unique factorization of ideals. –  KCd May 18 '11 at 2:19
show 6 more comments

The arithmetic of nonmaximal quadratic orders comes up naturally when studying binary quadratic forms, so in particular Cox's book contains some discussion of this.

As it happens, when I taught a course based (sometimes closely, sometimes not) on Cox's book, I included a "case study" of factorization in the ring $\mathbb{Z}[\sqrt{-3}]$: see Section 3 of these notes.

share|improve this answer
add comment

HINT $\ $ Given any fraction $\rm\:w\:$ witnessing that the nonmaximal order $\rm\:D\:$ is not integrally closed, i.e. a proper fraction over $\rm\:D\:$ that's a root of a monic $\rm\:f(x)\in D[x]\:,\:$ the proof below shows, for the fractional ideal $\rm\ I = (w,1),\ \ I^{\:n} =\: I^{\:n-1}\:,\:$ but $\rm\ I\ne 1\:$ (by $\rm\:w\not\in D$), $\:$ contra unique factorization.

Here $\rm\: n=2,\ w = (-1 + \sqrt{-3})/2\ $ is integral over $\rm\:D = \mathbb Z[\sqrt{-3}]\:,\:$ being a root of $\rm\:x^2+x+1\:.$ Alternatively, clearing denominators from the fractional ideals to obtain integral ideals, we obtain $\rm\ J^2 =\ 2\ J\:,\:$ but $\rm\ J \ne (2)\:,\:$ for $\rm\ J = 2\ (w,1) = (-1+\sqrt{-3},2)\:,\: $ also contra unique factorization. For further remarks, including relations with Dedekind's conductor ideal and irrationality proofs, see my discussion with Gerry Myerson in this sci.math thread circa May 20, 2009.

THEOREM $\ $ Dedekind domains $\rm\:D\:$ (so PIDs) are integrally closed.

Proof $\ $ Suppose a fraction $\rm\:w\:$ over $\rm\:D\:$ is integral over $\rm\:D\:,$ i.e. in terms of fractional ideals over $\rm\:D\:$

suppose $\rm\:\ \ w^n\:\ \in\:\ (w^{n-1},\ldots,w,1)\quad\quad\ \ [*]$

Now $\rm\: (w,1)^n\ =\ (w^n,w^{n-1},\ldots,w,1) $

$\rm\quad\quad\quad\quad\quad\quad\quad\quad =\ (w^{n-1},...,w,1)\ $ by $\ [*] $

Thus $\rm\ (w,1)^n\: =\ (w,1)^{n-1}\ $

Thus $\rm\ (w,1)\ =\ (1)\ $ by cancelling the invertible ideal $\rm\:(w,1)^{n-1}\:.$

Hence $\rm\ w \in (1) = D\:.\ $ Therefore $\rm\:D\:$ is integrally closed. $\quad$ QED

REMARK $\ $ A common similar proof uses $\rm\ I^2 =\: I,\ \ I = (w^n,...,w,1)\:.$

NOTE $\ $ If fractional ideals are unfamiliar then one may clear denominators from the prior proof by scaling it by $\rm\ b^n\:,\:$ where $\rm\ w = a/b\:,\:$ i.e.

$\rm\quad\quad\quad\quad (w,1)^n\ =\ (w,1)^{n-1}\ $ which, upon scaling through by $\rm\ b^n\ $

$\rm\quad \Rightarrow\quad (a,b)^n\: =\: b\ (a,b)^{n-1}$

$\rm\quad \Rightarrow\quad (a,b)\ =\ (b)\ \ $ by cancelling the invertible ideal $\rm\ (a,b)^{n-1}\: $ from above

$\rm\quad \Rightarrow\quad b\: |\: a\ $ in $\rm\: D\ $ i.e. $\rm\ a/b\in D\:.$

share|improve this answer
1  
Bill: please clarify in your post what you mean by "ideal (w,1)" when w is not in D. Also, the theorem you state only requires that D satisfies cancellation of (nonzero) ideals, not that it is Dedekind. That is, any domain satisfying cancellation of ideals is integrally closed. –  KCd May 18 '11 at 4:33
    
@KCd I've edited the proof to emphasize that the first version works with fractional (vs. integral) ideals. The proof mentions Dedekind domains due to the context it was excerpted from - see the linked sci.math post. –  Bill Dubuque May 18 '11 at 5:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.