Let R be any ufd. The content of a polynomial p(x) in R[x] is the greatest common divisor of its coefficients. A primitive polynomial has content 1.

The product of two primitive polynomials is primitive.
If q prime divides the content of a(x)*b(x),
let a_{i} and b_{j} be the first coefficients
in a and b respectively
that are not divisible by q.
The coefficient i+j in the product is not divisible by q.
This holds for all q, hence the product has content 1.

Factor out the contents of two polynomials to derive the corollary: content(a*b) = content(a)*content(b).

The concept of a primitive polynomial generalizes to rings that are not ufds. The polynomial p(x) is primitive if its coefficients generate the entire ring. The product f*g is primitive iff f and g are primitive. If f*g is not primitive, embed its coefficients in a maximal ideal H, then reduce R mod H, giving a field. The corresponding polynomials over R/H form an integral domain. The product f*g = 0, so either f or g is 0, and its coefficients lie in H. Conversely, if the coefficients of f lie in H, so must the coefficients of f*g.

All these results hold for polynomials in multiple indeterminants.

Let R be a ufd and let F be the fraction field of R. For example, if R is the integers, F is the rationals. If p(x) primitive is reducible in R[x] it is certainly reducible in F[x]. Conversely, if p(x) is reducible in F[x], write p(x) = a(x)*b(x). Multiply through by all denominators, being units, and the equation lies in R[x]. Recall that content(a*b) = content(a)*content(b). Divide through by these contents. Now the product of the primitive polynomials a and b must be the primitive polynomial p. Thus primitive polynomials are irreducible in R[x] iff they are irreducible in F[x]. (This fails when R is not a ufd; click here for an example.)

Now factor p(x) in R[x], starting with the content, which lies in R and factors uniquely. All other factors are irreducible in F[x], and must be unique.

Applying Gauss' lemma twice, R[x][y] is a ufd.
Keep adjoining indeterminants, as many as you like,
and by induction the result is always a ufd.
Thus **Z**[x,y,z] is a ufd.
This is not a pid, since the ideal generated by x and y is not principal.
Every pid is a ufd, but not the converse.
Similarly, every ufd is a factorization domain by definition,
but there are factorization domains that are not ufds.
This was illustrated by adjoining
sqrt(-5) to the integers.

When two polynomials are multiplied together, the product of their highest terms, largest degree and largest indeterminants, is nonzero, hence S is an integral domain.

Suppose a polynomial p factors two ways in S.
These factorizations use a finite number of indeterminants,
which contradicts the fact that
R[x_{1},x_{2},…x_{n}] is a ufd.
Therefore all of S is a ufd.

Bring in the indeterminants one at a time and build an ascending chain of ideals.
The first is generated by x_{1}, the second by x_{1} and x_{2},
the third by x_{1} x_{2} and x_{3}, and so on.
At the same time, we can build a complementary chain of descending ideals.
Let the first ideal be generated by all indeterminants except x_{1}.
The second is generated by all indeterminants except x_{1} and x_{2}, and so on.

So, if you're at a party, and the question comes up, "Is there a ufd that is neither noetherian nor artinian?", you can say yes.

Consider the polynomial 3x^{3} - 2x^{2} + 12x - 8.
A rational root has numerator 1 2 4 or 8, and denominator 1 or 3.
And of course the root could be negative.
In this case the only rational root is 2/3.
The other roots are ±2i.

Let p = u^{2}-3uv+7v-1.
When factoring p, we may view it as a polynomial in u,
with coefficients taken from the ufd **Z**[v].
The lead coefficient is 1, and the constant coefficient is 7v-1.
If q is a factor of p its lead coefficient is 1
and its constant coefficient divides 7v-1, which is prime in **Z**[v].
Thus p is the product of u±1 times u±(7v-1).
This produces the term ±7uv,
which cannot equal -3uv,
hence p is irreducible.

Yet p does have rational solutions, lots of them.
Set u to any rational number and solve for v.
If we're lucky, and 3u-7 divides u^{2}-1,
u and v can be integers.
Examples include ±1,0 and 5,3.

If we are given a 2 variable polynomial where each term has the same degree d,
divide through by v^{d}, and let r = u/v.
The result is a polynomial in r, which must obtain a rational root.
The numerator of R divides the constant term, which was the coefficient on v^{d}.
Similarly the denominator is drawn from the coefficient on u^{d}.
In other words, v divides the coefficient of u^{d} and u divides the coefficient of v^{d}.
Consider p = u^{2} - uv - 2v^{2}.
The only nonzero coprime solutions are 2,1 and 1,-1.

Try to factor the polynomial x^{2}+y^{2}-z^{2}.
The only possible notrivial factor is x±y±z,
and whenever you multiply two of these factors together,
mixed terms remain.
It is not possible to cancel them out,
the way we did with (x+y)×(x-y) = x^{2}-y^{2}.
Therefore x^{2}+y^{2}-z^{2} is irreducible.
Yet it has infinitely many integer solutions,
namely the pythagorean triples.