Unique Factorisation Domains/Fields

[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.

 

[mathwonk]

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.

 

[Deveno]

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.

 

[Zoe-b]

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?

 

[lavinia]

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?

Isn't the number 3 irreducible in this ring?

 

[Zoe-b]

Ok I completely missed that the polynomial can have a constant term (being slow) so yes I think you're right, 3 is irreducible. But I still can't see whether or not it is a unique factorisation domain as any element where the polynomial has a zero constant term is reducible. For example there definitely is not a unique factorisation of the element 2 into irreducibles- infact there isn't any factorisation of 2 into irreducibles as far as I can see, nor (do I think!) is it a unit.

(I've just found a bit on wikipedia saying 'any field is trivially a UFD' which now makes sense to me but obviously my example is not a field).

 

[lavinia]

Ok I completely missed that the polynomial can have a constant term (being slow) so yes I think you're right, 3 is irreducible. But I still can't see whether or not it is a unique factorisation domain as any element where the polynomial has a zero constant term is reducible. For example there definitely is not a unique factorisation of the element 2 into irreducibles- infact there isn't any factorisation of 2 into irreducibles as far as I can see, nor (do I think!) is it a unit.

(I've just found a bit on wikipedia saying 'any field is trivially a UFD' which now makes sense to me but obviously my example is not a field).

None of the fractional powers of 2 have inverses in this ring since negative powers are excluded. So two factorizations will be in-equivalent. How about two factorizations of the square root of 2?