Ideal generated by two prime elements in a unique factorization domain may be proper and not prime

Statement

It is possible to have a unique factorization domain $R$ and primes $p,q \in R$ such that $(p,q)$ is a proper ideal of $R$ that is not prime.

Related facts

• Unique factorization is not prime-quotient-closed: If $R$ is a unique factorization domain, and $P$ is a prime ideal, the quotient $R/P$ need not be a unique factorization domain. Examples of this where the prime ideal $P$ is principal also give examples of ideals generated by two elements that are proper and not prime.

Opposite facts

• Unique factorization implies every prime ideal is generated by prime elements
• Unique factorization implies every nonzero prime ideal contains a prime element
• Unique factorization and Noetherian implies every prime ideal is generated by finitely many prime elements

Other related facts

• Prime ideal need not contain a prime element

Proof

Example of a bivariate polynomial ring

Let $k$ be any field of characteristic not equal to $2$. Let <mah>R = k[x,y]</math>. The polynomials $x$ and $x^2 + y^2 - 1$ are irreducible in $R$, and hence prime. However, the ideal $I = (x,x^2 + y^2 - 1)$ is not prime in $R$. To see this, note that $R/I =R/(x,y^2 - 1) = k[x,y]/(x,y^2 - 1) \cong k[y]/(y^2 - 1)$, which is not an integral domain, because $y^2 - 1$ is not irreducible.

Example of a polynomial ring over the integers

Let $\mathbb{Z}$ be the ring of rational integers and $R = \mathbb{Z}[x]$. The elements $2, x^2 + 5$ are irreducible in $R$. However, the ideal $I = (2,x^2 + 5)$ is not a prime ideal in $R$. To see this, note that $I$ being prime is equivalent to the polynomial $x^2 + 5$ being irreducible in $\mathbb{Z}/2\mathbb{Z}[x]$, which is not true since the polynomial reduces as $(x+1)^2$ modulo $2$.