Unique factorization domain that is not a Principal ideal domain

In this article, we provide an example of a unique factorization domain – UFD that is not a principal ideal domain – PID. However, it is known that a PID is a UFD.

We take a field \(F\), for example \(\mathbb Q\), \(\mathbb R\), \(\mathbb F_p\) (where \(p\) is a prime) or whatever more exotic.

The polynomial ring \(F[X]\) is a UFD. This follows from the fact that \(F[X]\) is a Euclidean domain. It is also known that for a UFD \(R\), \(R[X]\) is also a UFD. Therefore the polynomial ring \(F[X_1,X_2]\) in two variables is a UFD as \(F[X_1,X_2] = F[X_1][X_2]\). However the ideal \(I=(X_1,X_2)\) is not principal. Let’s prove it by contradiction.

Suppose that \((X_1,X_2) = (P)\) with \(P \in F[X_1,X_2]\). Then there exist two polynomials \(Q_1,Q_2 \in F[X_1,X_2]\) such that \(X_1=PQ_1\) and \(X_2=PQ_2\). As a polynomial in variable \(X_2\), the polynomial \(X_1\) is having degree \(0\). Therefore, the degree of \(P\) as a polynomial in variable \(X_2\) is also equal to \(0\). By symmetry, we get that the degree of \(P\) as a polynomial in variable \(X_1\) is equal to \(0\) too. Which implies that \(P\) is an element of the field \(F\) and consequently that \((X_1,X_2) = F[X_1,X_2]\).

But the equality \((X_1,X_2) = F[X_1,X_2]\) is absurd. Indeed, the degree of a polynomial \(X_1 T_1 + X_2 T_2\) cannot be equal to \(0\) for any \(T_1,T_2 \in F[X_1,X_2]\). And therefore \(1 \notin F[X_1,X_2]\).

Leave a Reply