5

As is well-known, $Z[\sqrt{-5}]$ is not a ufd because $6$ has more than one prime factorization in this ring: $6=2\cdot 3$ and $6=(1+\sqrt{-5})(1-\sqrt{-5})$. But both of these prime factorizations have the same number $(2)$ of prime factors...Am I correct that in $Z[\sqrt{-29}], 30=2\cdot 3\cdot 5$ and $30=(1-\sqrt{-29})(1+\sqrt{-29})$ are prime factorizings of $30$ that have different numbers of factors?

Also would $Z[\sqrt{-2309}]$ give as distinct prime factorizations $2310=2\cdot 3 \cdot 5\cdot 7\cdot 11=(1+\sqrt{-2309})(1-\sqrt{-2309})$? ($2309$ is a prime number).

What about $Z[\sqrt{-30029}]$, would that give $30030=2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13=(1+\sqrt{-30029})(1-\sqrt{-30029})$ as distinct prime factorizations?($30029$ is prime)...Does this show the number of primes in distinct prime factorizings be different? I'm worried that the norms will cause some of my "primes" to be nonprimes. Thanks.

Adam Hughes
  • 36,777
Richard Peterson
  • 619
  • You should look into improving the format of this question and its readability. http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Vincent Jul 30 '14 at 01:56
  • 1
    One of the things you have to check is that the factorizations are into irreducibles and are different by more than a unit. Take $6=3\cdot 2=(-1)(1+\sqrt{-2})(1-\sqrt{-2})\cdot\sqrt{-2}^2$ these are two factorizations, but only the one on the right is into irreducibles. – Adam Hughes Jul 30 '14 at 02:08
  • 1
    It is my understanding, you will abundantly let me know if this understanding is the slightest bit wrong, that $\mathbb{Z}[\sqrt{-2}]$ is UFD and thus the irreducibles are primes after all. $\mathbb{Z}[\sqrt{-29}]$, on the other hand, has class number 6, so the irreducibles are not primes, and from Bill's answer it follows that there can be multiple factorizations of differing lengths. I would still warn you to not be led astray by incomplete factorizations and/or multiplication by units, especially in real rings. – Robert Soupe Jul 30 '14 at 04:01

1 Answers1

5

It is a classical result of Carlitz (1960) that a number ring has an element with different length factorizations into irreducibles iff it has class number $> 2.\,$ For a proof, and a survey of related results see Half-factorial domains, a survey by Scott T. Chapman and Jim Coykendall.

Bill Dubuque
  • 272,048