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

Example of a bivariate polynomial ring
Let $$k$$ be any field of characteristic not equal to $$2$$. Let R = k[x,y]. 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 $$2$$.