# What goes wrong in a ring that does not have unique factorization?

Whenever I have seen unique factorization discussed, it is always with respect to the solution of diophantine equations; the equations are solved by splitting the equation into linear functions over a ring and then invoking unique factorization. But the discussions always give the impression that the failure of unique factorization causes all sorts of problems. Apart from the applications to diophantine equations, why is unique factorization such a desirable property?

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Thanks for getting rid of that terrible pun, Pete. –  Zev Chonoles Jun 1 '10 at 5:20

I think a glance at any elementary number theory textbook will show you numerous topics which rely on uniqueness of factorization. The formulas for the number of divisors, the sum of the divisors, the Euler phi-function, all depend on unique factorization. Going up a level, the Euler product for the Riemann zeta function depends on unique factorization.

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Your last statement seems dubious to me: Dedekind zeta functions also have Euler products. The key property for this is unique factorization of ideals, i.e., that the ring of integers of a number field is a Dedekind domain. –  Pete L. Clark Jun 1 '10 at 5:16
@Pete, I would say unique factorization into ideals is the solution of the problem caused by the failure of unique factorization into elements; I interpreted the question to be about problems caused by the failure of unique factorization into elements. If OP wants to know why unique factorization into ideals is such a desirable property, I'd nominate the Dedekind zeta functions as part of the answer. –  Gerry Myerson Jun 1 '10 at 5:54
There are other characterizations of UFDs which show that a non-UFD lacks certain nice properties. For instance, a Noetherian domain is a UFD iff every height one prime ideal is principal. Applied to the polynomial ring $k[t_1,\ldots,t_n]$, this shows that every irreducible codimension one subvariety is cut out by a single equation.