Let R be a domain and let px qx epsilom Rx If p and q are ir

Let R be a domain, and let p(x), q(x) epsilom R[x]. If p and q are irreducible, prove that p | q if and only if there is ; q=up.

Solution

take any nonzero nonunit element x belong to R

with the induction N(x).

If N(x) is third smallest then s is irreducible.

so suppose x is nth smallest and suppose that every element of R that is of (n-1)st size or less is irreducible or

a product of irreducibles.

If p and q are irreducibles

suppose x= pq with p,q belongs to R with N(p or q)<N(x)=N(p or q)

such that q = a(pq) +r

then r= a(1-qb)

of 1-qb !=0 then

N(a)<= N(a(1-qb)) = N(r)

this is possible since N(r) < N(a)

so 1-qp = 0

here if u is a unit

so we cannot have N(a)=N(x)

We cannot have N(b)<N(x)

so now we can write p and q as product of irreducibles.

Thus express q as a product of irreducibles

therefore q= up