Factorization
Let \(R\) be a ring.
An element \(x\in R\) is irreducible if \(x\) is not a unit and for all \(y, z\in R\), \(x = y z\) implies \(y\) or \(z\) is a unit.
An element \(x \in R\) is prime if \(x \ne 0\), \(x\) is not a unit and for all \(y,z\in R\), we have \(x  yz \implies xy \text { or } xz\). Note \(x\) is prime if and only if \(x R\) is a prime ideal.
In an integral domain, all primes are irreducbile. The converse is not always true. For example, take \(R =\mathbb{Z}[\sqrt{5}]\). Then by using the norm, it can be deduced that the units of \(R\) are \(\pm 1\). We have \(6 = 2\cdot 3 = (1+\sqrt{5})(1\sqrt{5})\), where all the factors are irreducible but not prime.
An integral domain \(R\) is a unique factorization domain (UFD) if every nonzero nonunit of \(R\) can be expressed as a product of irreducibles and furthermore the factorization is unique up to order and associates. In a UFD, all irreducibles are prime.
Example: \(\mathbb{Z}\) is a UFD. All fields are trivially UFD’s.
Gauss' Theorem: If \(R\) is a UFD then the polynomial ring \(R[x]\) is a UFD.
Proof: Consult an abstract algebra textbook.
Example

Let \(R = F[x_1,...,x_n]\) where \(F\) is a field. By Gauss' Theorem \(R\) is a UFD. If \(p \in R\) is any irreducible polynomial over \(F\) then it is also prime, and the principal ideal \(p R\) is a prime ideal of \(R\).

Let \(R = \mathbb{Z}\) or \(R = F[x]\) for some field \(F\). By using the Euclidean algorithm, it can be seen that every ideal of \(R\) is principal. (In \(\mathbb{Z}\), the nonzero prime ideals are generated by a prime, while in \(F[x]\), the nonzero prime ideals are generated by irreducible polynomials.) In these rings, it turns out that all nonzero prime ideals are also maximal, for reasons we shall see below.
A principal ideal domain (PID) is an integral domain in which all ideals are principal.
Proposition: Let \(R\) be a PID. Then every nonzero prime ideal is maximal.
Proof: Let \(I = x R\) be some nonzero prime ideal. Suppose \(I\) is strictly contained in some ideal \(y R\) of \(R\). Then \(x = y z\) for some \(z\in R\). Since \(I\) is prime, we must have \(y \in I\) or \(z \in I\). The former would imply \(y R\) is contained in \(I\), a contradiction. So we must have \(z \in I\). In other words, \(z = x t\) for some \(t \in R\), hence \(x = x y t \). implying that \(y t = 1\). Thus \(y\) is a unit so \(y R = R\).
Example: Going back to the example \(R = F[x_1,...,x_n]\) for some field \(F\), set \(M = \{p\in Rp(0) = 0\}\) (the set of polynomials with zero constant term). We have \(A/M \cong F\) thus \(M\) is maximal. Now if \(n \gt 1\), \(M\) is not principal, because any set of elements that generates \(M\) must contain at least \(n\) elements.