Unique Factorization Domains

Hello!

This question may seem obvious but I cant seem to figure it out.

In the definition of an irreducible element (when developing the theory for a UFD for *any* general Integral domain), it is required that we exclude all units from being irreducible. Why is this? In other words, what would happen if we did not exclude irreducibles from being units? Would this make every non zero element of the Integral domain uniquely factorizable into irreducible? I cant seem to trace the exact connection between preventing units from being irreducible and the unique factorization thereof...

Re: Unique Factorization Domains

A special case of this is the convention that 1 is not prime. If it were, then it would not be true that every positive integer can be *uniquely* factored as a product of primes, since .

It's not that much more complicated to say that "a positive integer is prime if its only divisors are itself and 1". But if we wanted to extend our definition to all integers, it becomes "p is prime if its only divisors are p, -p, 1, and -1". It doesn't get worse than that for integers, because 1 and -1 are the only units. But if we have more units, then we get a *whole lot* more divisors for any element, just by multiplying it by some combination of units.

The fact that these divisors exist for any element means they're not interesting, and we're motivated to throw them out. But it's not feasible anymore just to list them as exceptions, like we did above--there are too many of them. We could add something to our definition of irreducible to say that divisors which are units or unit multiples of the element don't count, and then weaken our definition of "unique factorization" to say that the units don't count there, either.

Or, we can just say that units aren't irreducible, and not have to change anything else. That way seems much simpler, so that's what we do.

Re: Unique Factorization Domains

also, think of polynomials over a field: x-4 = 1/4(4x-16) = 1/2(2x-8), etc. there are obviously many, many ways to "factor" x-4, but usually, we want to think of linear polynomials as irreducible (up to a unit).

Re: Unique Factorization Domains

Hello! Thanks for the feedback. I hear what you're saying, and it makes quite good sense with polynomial rings and the integers. But I wanted to trace the *exact* reason for not including units as irreducibles: after all an irreducible is an element which has only trivial factorizations. And units only ever factorize trivially. So it seems strange not to include them as irreducibles. But I think I managed to figure it out when looking at a proof of every Principle Ideal Domain having UFD1 (that every nonzero nonunit is a product of irreducibles). Perhaps you could check my reasoning in this regard; it may be totally wrong!

The main reason we dont include them, is because the process of factorization would never end otherwise (this seemed clear to me initially, but I wanted to see where exactly it happens).

In that proof, we begin with a general element nonunit (and non zero since this will never have a unique factorization (0=0*a for any a in the ring)). Then we have two options: Either it is irreducible (in which case we are done), or it is reducible. Then we can write it as a=c*d, where c and d are both not units. Then considering c, it has the same two options. And so the process continues. However, we cannot keep supposing the latter can occur, since using the ascending chain argument leads to a point at which if we have supposed that is reducible, then is an associate of , thus = * ,(u being a unit) and so is irreducible. A contradiction. Thus we must take the former route and assume then that is reducible. Then a (the element we started with in the beginning) has an irreducible factor. That is = * , with being irreducible.

Now as for , it must have an irreducible factor, because of the above. So it can be written as = * , being irreducible. Then either is an irreducible or a unit, in which case we are done. Or else it has an irreducible factor (due to the above). This is repeated again and again. If we keep assuming the latter

, then again using the ascending chain argument, we must get that is an associate of (i.e = * , where u is a unit) . But we assumed the latter, which means = * , giving that = . Now preventing irreducibles from being units means this is a contradiction. And thus we have to assume that is a unit or an irreducible. And thus we are done.

So to this end, without this definition, the process of factorization continues indefinitely (which is what happens if we let 1 be a prime).

Sorry for the bad Tex, not all that familiar wit it!

Re: Unique Factorization Domains

in UFD's we want to have an analogue of prime numbers (and these should be the irreducible elements). if units were allowed to be irreducible, the principal ideals generated by irreducible elements would not satisfy the ascending chain condition (which is what i think you are trying to say).

in practice, when we are actually decomposing elements into their "prime" factors, we use euclid's lemma that for a prime element p|ab --> p|a or p|b. this is trivially true for any unit (since we have u^-1ab, u^-1a and u^-1b), which doesn't help much (yeah 1 divides everything, which tells us very little). what we want to do is split ab into some things genuinely "simpler", in a way that terminates.

i'm sure you can see, that if we take (u) for any unit u, we just get R (since 1 = u^-1u must be in (u)). so allowing (u) as a prime ideal, is counter-productive, in the sense that R/(u) is never interesting. we'd like for R/(p) to be a little bigger, so we can use these quotients as a way of learning about R.

a lot of rings arise as some form of polynomial ring (or their big brothers, power series). in other words, what we're really interested in, is solving equations in one or more variables. the general case may be rather hard, but if we can break it down into a combination of simpler equations, that's major progress. factorization is one method, but if factorization is not unique (such as in Z[i√5]) its not as useful as when it is unique. we can, of course, define whatever terms we like, in whatever way we like, but some definitions seem to work better than others. we want to keep as many "nice" properties of integers as possible (although what constitutes a "nice" property is up for debate).