Irreducible not implies universal side divisor

Statement

An irreducible element in an integral domain need not be a universal side divisor. In fact, an irreducible element need not be a universal side divisor even in a Euclidean domain.

Related facts

Converse

• Universal side divisor implies irreducible: In an integral domain, any universal side divisor is irreducible.
• Universal side divisor not implies prime: In an integral domain, a universal side divisor need not be prime.

Caveats

There are some Euclidean domains where every irreducible element is a universal side divisor. For instance:

• Suppose $k$ is an algebraically closed field. Then, the polynomial ring $k[x]$ has the property that every irreducible polynomial is a universal side divisor. This is because in a polynomial ring, the universal side divisors are precisely the nonconstant linear polynomials, and the field being algebraically closed is precisely equivalent to saying that these are the only irreducible polynomials.
• In a discrete valuation ring, the uniformizing parameter, which is the unique irreducible (up to associates) is a universal side divisor. (Note that since both the property of being irreducible and the property of being a universal side divisor are preserved up to associates in an integral domain, this makes sense).

Proof

Example of the ring of integers

Further information: ring of rational integers

In the ring of rational integers $\mathbb {Z}$, the only universal side divisors are the elements $\pm 2,\pm 3$. However, there are infinitely many primes. In particular, $\pm 5$ are irreducibles that are not universal side divisors.

Example of the polynomial ring over a field that is not algebraically closed

Further information: polynomial ring over a field

Let $k$ be a field that is not algebraically closed. In the polynomial ring $k[x]$, the universal side divisors are precisely the nonconstant linear polynomials. However, since $k$ is not algebraically closed, there exists an irreducible polynomial of degree greater than $1$ in the polynomial ring $k[x]$. Thus, there is an irreducible element that is not a universal side divisor.

For instance, if $k=\mathbb {R}$, the polynomial $x^{2}+1\in \mathbb {R} [x]$ is irreducible but not a universal side divisor.