# A non-UFD where we have different lengths of irreducible factorizations?

A unique factorization domain (UFD) is defined to be an integral domain where each element other than units and zero has a factorization into irreducibles, and if we have two such factorizations then they are of the same length (the sum of the powers of the irreducible factors) and the irreducibles involved are associated to each other.

My knowledge about algebra is extremely limited, but the only non-UFD integral domain I have seen is $\mathbb{Z}[\sqrt{-5}]$, where one shows $9=3\cdot 3=(2+\sqrt{-5})(2-\sqrt{-5})$ gives two different factorizations of $9$ into irreducibles with non-associated factors.

However in this case the two factorizations are of the same length.

So just curious, can we find integral domains where some elements have factorizations into irreducibles with different lengths?

Thanks!

From Algebraic Number Theory and Fermat's Last Theorem, pg $84$:
$\hskip 1.1in$
In addition to the number of irreducibles varying, naturally the exponents of irreducibles can vary too (even if the number of irreducibles counted with multiplicity remains constant). Note the factorizations take place in the ring of integers ${\frak O}_K$ of the number field $K/\Bbb Q$ listed on the left.
In particular look at the factorizations of $18$, $27$ and $30$ in the appropriate number rings.
• (+1) Of course, the amazing start to the field of algebraic number theory is the realization that if you put $()$ around all of these expressions (take principal ideals) then any differences in length of factorization, or any differences at all, disappears. Commented Oct 16, 2013 at 4:26
One standard example is the ring $\mathbb Q[X^2,X^3]$. This is a Noetherian domain and hence atomic (that is, every non-unit has a factorization into irreducibles). The elements $X^2$ and $X^3$ are irreducible and so $X^6$ has two factorizations of different length: $X^2 X^2 X^2 = X^6 = X^3 X^3$.