What is the extension of the Unique Factorization Theorem to Gaussian Integers?

[huba]

I am not sure I fully understand the extension of the Unique Factorization Theorem (UFT) to Gaussian Integers (GI), by saying that the representation of a GI as a product of primes is unique except for the order of factors and the presence of units.

Is there a similar problem when the UFT is extended to integers?
For example, -6 can be represented as -2*3 or 2*-3 or -1*2*3, or -1*-2*-3.

 

[HallsofIvy]

I am very confused by your question! The "Unique Factorization Theorem" extended to Gaussian integers? The "Unique Factorization Theorem" does not hold for the Gaussian integers: there exist distinct factorizations of some Gaussian integers. I can't think of an example offhand but I don't think it is terribly difficult.

 

[huba]

I was reading W.J. LeVeque's Elementary Theory of Numbers. Theorem 6-8 says what I said.

 

[CRGreathouse]

A nonzero (rational) integer has a unique factorization up to order and the presence of units (1 and -1).

Gaussian integers similarly have unique factorization up to order and units (1, i, -1, -i). Gaussian integers factor 'further' then rational integers, though: 2 is a prime rational integer, but 2 = (1 + i)(1 - i) in the Gaussian integers.

Most Z[sqrt(n)] do not have unique factorization, though.

 

[LukeD]

I am very confused by your question! The "Unique Factorization Theorem" extended to Gaussian integers? The "Unique Factorization Theorem" does not hold for the Gaussian integers: there exist distinct factorizations of some Gaussian integers. I can't think of an example offhand but I don't think it is terribly difficult.

Eh?? The Gaussian integers are a Euclidean domain, so of course they're a ufd

 

[CRGreathouse]

Eh??

Hush, Halls was thinking of \mathbb{Z}[\sqrt{-5}].