# Question on UFDs

1. Jun 8, 2010

### logarithmic

I'm currently reading a textbook on algebra, and one of the lemmas is the following:

Let D be a UFD and let F be the field of quotients of D. Let f(x) in D[x] where degree(f(x)) > 0.
If f(x) is irreducible in D[x], then f(x) is also irreducible in F[x].
Also, if f(x) is primitive in D[x] and irreducible in F[x], then f(x) is irreducible in D[x].

It then says that this lemma shows that the irreducibles in D[x] are precisely the irreducibles in D, together with the nonconstant primitive polynomials that are irreducible in F.

Can someone explain how the lemma implies this, in particular it makes no mention of the irreducibles of D.

2. Jun 8, 2010

### Office_Shredder

Staff Emeritus
If b is an irreducible element in D, how can it possibly factor in D[x]? Well, it will be of the form b=f(x)g(x). Since b is degree 0, each of f(x) and g(x) must be of degree zero, i.e. must be in D itself. But then you've factored b which is impossible.

3. Jun 9, 2010

### logarithmic

I see. I've got another question. As a corollary to the above it states (using the same notation): For a nonconstant f(x) in D[x], f(x) can be factored into two polynomials of lower degree r and s in D[x] iff f(x) can be factored into two polynomials of lower degree r and s in F[x].

Since this is an iff statement, negating both sides of the iff statement is also true, this gives f(x) is irreducible in D[x] iff it is irreducible in F[x]. Is that correct? Seems strange because it's a stronger statement than the 3rd line of the original lemma.