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

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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

Other related facts


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.