Lack of unique factorization of ideals I'm aware of the result that integral domains admit unique f

Amappyaccon22j7e 2022-05-07 Answered
Lack of unique factorization of ideals
I'm aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.
It's clear that Z [ 3 ] is not a Dedekind domain, as it is not integrally closed. I'm having difficulty demonstrating directly that Z [ 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.
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Answers (1)

Maeve Holloway
Answered 2022-05-08 Author has 23 answers
In the ring Z [ 3 ], the ideal P = ( 2 , 1 + 3 ) is prime since it has index 2 in the ring. Note P 2 = ( 4 , 2 + 2 3 , 2 + 2 3 ) = ( 4 , 2 + 2 3 ) = ( 2 ) ( 2 , 1 + 3 ) = ( 2 ) P the principal ideal generated by 2 in Z [ 3 ]
If there were unique factorization of ideals into products of prime ideals in Z [ 3 ] then the equation P 2 = ( 2 ) P would imply P=(2), which is false since 1 + 3 is in P but it is not in (2).
In fact we have P 2 ( 2 ) 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 ) Q, so P P = P 2 Q, which implies P Q (from Q being prime), so Q=P since P is a maximal ideal in Z [ 3 ].
If (2) is a product of prime ideals then it must be a power of P, and P n P 2 for n 2, so the strict inclusions P 2 ( 2 ) P imply (2) is not P n for any n 0.
The "intuition" that the ideal P = ( 2 , 1 + 3 ) is the key ideal to look at here is that P is the conductor ideal of the order Z [ 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 Z [ 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 Z [ 3 ] are closely tied to the ideal P.
If you look at the ideal notation P P = ( 2 , 1 + 3 ) in the larger ring Z [ ( 1 + 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 + 3 ) / 2 ) = ( 2 ) in Z [ ( 1 + 3 ) / 2 ], so P′ is actually a principal ideal and the "paradoxical" equation P P = ( 2 ) P in Z [ 3 ] corresponds in Z [ ( 1 + 3 ) / 2 ] to the dumb equation P P = P P . (The ideal P′ in Z [ ( 1 + 3 ) / 2 ] is prime since the quotient ring mod P′ is a field of size 4: Z [ ( 1 + 3 ) / 2 ] is isom. as a ring to Z [ x ] / ( x 2 + x + 1 ), so Z Z [ ( 1 + 3 ) / 2 ] / P = Z [ ( 1 + 3 ) / 2 ] / ( 2 ) is isom. to ( Z / 2 Z ) [ x ] / ( x 2 + x + 1 ), which is a field of size 4.)
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