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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!

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  • $\begingroup$ 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. $\endgroup$
    – Key Ideas
    Commented Jun 9, 2013 at 15:30
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    $\begingroup$ 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. $\endgroup$
    – anon
    Commented Jun 10, 2013 at 2:46

1 Answer 1

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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 $4=(-2)\cdot (-2)$.

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  • $\begingroup$ Thanks so much! I understand it. I have deleted my previous comment. $\endgroup$
    – user71346
    Commented Jun 9, 2013 at 12:31
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    $\begingroup$ This may not apply depending upon what is meant by "p appears in a factorization". That may well mean that p is associate to one of the factors in the factorization (which is the denotation most authors intend when writing similar imprecise natural languages statements). $\endgroup$
    – Key Ideas
    Commented Jun 9, 2013 at 15:26

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