# Show that "In a UFD, if $p$ is irreducible and $p\mid a$, then $p$ appears in every factorization of $a$" is false

This statement below is false, but I cannot find any counterexamples or explain why. When I tried to give some reasoning, I ended up showing the statement is true.

In a UFD, if $p$ is irreducible and $p\mid a$, then $p$ appears in every factorization of $a$.

(Here UFD is the abbreviation for Unique Factorization Domain.)

So, my explanation goes like this: since $a$ is in a UFD, there is a unique factorization of it, and the factorisation consists of irreducibles, therefore if $p\mid a$ then it is clear that $p$ appears in every factorization of $a$ since the factorization is unique.

Could anyone kindly show me where did I go wrong?

Thanks!

• The question cannot be answered without a precise definition of "p appears in a factorization". Suppose $\,p = q_1\cdots q_n$ for $q_i$ irreducible. Does it mean that, for some $i,$ we have $p = q_i,$ or, instead, that $(p) = (q_i),$ i.e. that they are associate, i.e. they are equal up to a unit multiple. In other words, to test "appears", which equivalence realtion is used, equality, or associate-to? The latter is almost always what is meant by such imprecise statements. Commented Jun 9, 2013 at 15:30
• Which definition is understood by the author can be deciphered from the major context clue: the statement is billed as false, and there is only one interpretation (actual equality, not associate-to, as the equivalence relation) on which the statement is false. It is of course valuable to know of both interpretations though - in fact the associate-to interpretation of "appears in a factorization" is much more natural in the context of algebraic number theory.
– anon
Commented Jun 10, 2013 at 2:46

Remember that the factorizations are unique only up to a unit multiple. Thus, even in $\mathbb Z$--or any UFD in which there are nonidentity units and irreducibles--we can find an example of this phenomenon. E.g. $2\mid 4$ but also $4=(-2)\cdot (-2)$.