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I try to unterstand, why the Integer factorization is only working for rings and not for fields. My first idea was, that you don't have a uniqueness quantification for prime "numbers" in fields. Is that correct? It would be nice if someone would explain it to me.


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There are no prime numbers in a field. – André Nicolas Oct 26 '12 at 7:07
up vote 4 down vote accepted

There are no prime numbers in a field. Every non-zero element of a field is a unit, where a unit is defined as a number $a$ which has a multiplicative inverse. A prime $p$ is in general defined as an object which is neither $0$ nor a unit, and such that whenever $p$ divides $ab$, $p$ divides $a$ or $p$ divides $b$. An object $p$ is irreducible if $p$ is neither $0$ nor a unit, and whenever $p=ab$, one of $a$ or $b$ is a unit. So by definition there are neither primes nor irreducibles in a field.

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Thank you, that sounds quite clear – ulead86 Oct 26 '12 at 11:30

Prime elements, irreducible elements etc. are defined in an arbitrary integral domain. An integral domain is called factorial (or UFD) if every non-zero non-unit is a product of prime elements. This decomposition is essentially unique. So this captures the notion of unique prime factorization in ring theory. Now it is trivial to verify that fields are factorial: There is non non-zero non-unit at all, so nothing has to be checked. In particular, the answer to "Why doesn't work Integer factorization for fields?" is: No, it works! The answers so far indicate that it doesn't work because of the abscence of prime elements, but this doesn't matter.

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