## Unique Factorisation Domains/Fields..

In my notes/online I have that for a set to be a field is a stronger condition than for a set to be a unique factorisation domain (obviously), but I'm confused about the concept of irreducibility in a field. For example in R\{0} there are no irreducible elements as far as I can tell? I'm trying to show whether a different ring is a UFD or not but again I find the concept of irreducibility difficult as here again the ring seems to have no irreducible elements.

 PhysOrg.com science news on PhysOrg.com >> Ants and carnivorous plants conspire for mutualistic feeding>> Forecast for Titan: Wild weather could be ahead>> Researchers stitch defects into the world's thinnest semiconductor
 Recognitions: Homework Help Science Advisor in any ring there are three classes of non zero elements: units, zero divisors, non unit non zero divisors. irreducibles are always, by definition, a subset of the non units. in a field, all non zero elements are units, hence there is no possibility of irreducibles.

Recognitions:
 Quote by Zoe-b In my notes/online I have that for a set to be a field is a stronger condition than for a set to be a unique factorisation domain (obviously), but I'm confused about the concept of irreducibility in a field. For example in R\{0} there are no irreducible elements as far as I can tell? I'm trying to show whether a different ring is a UFD or not but again I find the concept of irreducibility difficult as here again the ring seems to have no irreducible elements. Thanks in advance.
perhaps you should tells us which ring you are trying to prove is a UFD.

## Unique Factorisation Domains/Fields..

Ok that makes sense, so is the statement that all fields are UFDs correct? Or just nonsense because the concept of factorisation only makes sense with regards to irreducibles?

Let 2^(1/n) be the real positive nth root of 2. If we consider Z [ 2^(1/n) ] as the image of Z[X] under the evaluation map f(X) -> f(2^(1/n)), then let R= union from n=1 to infinity of (Z [ 2^(1/n) ]). I have shown already that R is a subring of the reals but as far as I can see it has no irreducible elements-
If a is a member of Z[2^(1/p)] for some natural number p, say a = g(2^(1/p)) for some polynomial g in Z[X]. Then we can always take out a factor of 2^(1/2p) to give

a = 2^(1/2p) * h(2^(1/2p))

where h is some polynomial in Z[X] (with the same integer coefficients as g but different powers). So a is reducible.

Am I on the completely wrong track?

Recognitions: