Unique Factorization Domains (UFDs)

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Section 10.3 Unique Factorization Domains (UFDs)

“Always remember that you are absolutely unique. Just like everyone else.”
―Margaret Mead

Definition 10.26. UFD.

A ring \(R\) is called a unique factorization domain, or UFD for short, if \(R\) is an integral domain and every element \(r \in R\) that is non-zero and not a unit can be written as a finite product

\begin{equation*} r = p_1 \cdots p_n \end{equation*}

of (not necessarily distinct) irreducible elements \(p_1, \dots, p_n\) of \(R\) in a way that is unique up to ordering and associates. That is, if

\begin{equation*} r = q_1 \cdots q_m \end{equation*}

also holds with each \(q_i\) irreducible, then \(m = n\) and there is a permutation \(\sigma\) such that, for all \(i\text{,}\) we have \(p_i\) and \(q_{\sigma(i)}\) are associates.

Example 10.27. Examples of UFDs.

\(\Z\) is a UFD by the Fundamental Theorem of Arithmetic.
\(F[x]\) where \(F\) is a field is a UFD. This is the case because \(F[x]\) is a Euclidean domain and Euclidean domains are UFD’s (we’ll prove this shortly).
We will eventually prove that if \(R\) is a UFD then so is \(R[x]\text{.}\) It follows that \(F[x_1, \dots, x_n]\) is a UFD for all \(n\text{.}\) Note that if \(n > 1\text{,}\) this ring is not a PID and hence not a Euclidean domain.

Exercise 10.28. \(\Z[x]\) is a UFD.

\(\Z[x]\) is not a PID hence also not a Euclidean domain. For example, this can be seen because the ideal \(\igen{2,x}\) is not a principal ideal. It is a UFD because \(\Z\) is a UFD (based on the result that if \(R\) is a UFD then so is \(R[x]\) which we will prove shortly).

Theorem 10.29. PIDs are UFDs.

If \(R\) is a PID then \(R\) is a UFD.

Theorem 10.30. Polynomial Rings over UFDs are UFDs.

If \(R\) is a UFD then so is \(R[x]\text{.}\)