Lack of unique factorization of idealsI'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.

Question

Lack of unique factorization of idealsI&#039;m aware of the result that integral domains admit unique factorization of ideals iff they are Dedekind domains.It&#039;s clear that       Z    [      −    3    ] is not a Dedekind domain, as it is not integrally closed. I&#039;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&#039;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.

Maeve Holloway · Accepted Answer

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 &quot;intuition&quot; 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&#039;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 &quot;paradoxical&quot; 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.)

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

Answered question

Answer & Explanation

New Questions in College Statistics