Supplement: Proving the uniqueness of prime factorizations
This supplement completes the proof of the FTA (Thm 1.13)
Proof (Uniqueness of prime factorization: Direct) Fix n a natural. Let
k
k
k
l
l
l
p1 1 p2 2 ... pr r = n = q1 1 q2 2 ... qs s be factorizations of n with all of the notation as
per the statement of the FTA. To prove the uniqueness of the factorization, we
prove two things (in order) about these factorizations:
†
† †
†
(1) the lists of primes p1 < p2 < ... < pr and q1 < q2 < ... < qs associated with
† †
each of the factorizations agree; and
(2) k1 = l1 , k2 = l2 , ... , kr = lr (that is, the exponents attached to the primes are
†
†
†
† †
†
the same in each).
†
† † †
† †
To prove (1), just assume that the two lists of primes differ somewhere (and we’ll
try to reach a contradiction). We then have two cases to consider: either some pi
is not in the list q1 < q2 < ... < qs or some qi is not in the list p1 < p2 < ... < pr .
l
l
l
Case 1: pi is not in the list q1 < q2 < ... < qs . But pi |n, and so pi | q1 1 q2 2 ... qr r .
†
l1
l2
l
pi | †
q2 ... qr r by
However,
1.5.1 †
and 1.4.5, ( pi , q1 ) =
† † by exercises
†
†1. Thus
†
l
r
in this
†exercise 1.2.4. Continuing
† †
† way,†we eventually
† get
† p†i | qr †. With this
l
( pi , qr r ) = pi by the definition†of†the greatest common
However, we
† † divisor.
†
l
know ( pi , qr r ) = 1 by exercises 1.5.1 and 1.4.5. Thus
† †pi = 1, which is not
prime.
† †
†
k
k
k
Case 2: qi is not on the list p1 < p2 < ... < pr . But qi | p1 1 p2 2 ... pr r . As in the
† †
†
last case, this gives rise to another contradiction concerning qi .
Since both cases give rise to contradictions, it must be that every prime on the list
†
† †
†
† † †
†
p1 < p2 < ... < pr is also on the list q1 < q2 < ... < qs and, conversely, every prime
†
on the list q1 < q2 < ... < qs is on the list p1 < p2 < ... < pr . This tells us that the
†
†
two lists of primes are really the same. In particular this means r = s. That is, we
†
† †
†k
k
k
l
l
l
now know our factorizations are p1 1 p2 2 ... pr r = n = p1 1 p2 2 ... pr r .
† †
†
† †
†
† †
†
† †
1
†
We can now turn to part (2) of the proof. Just suppose that the exponents
attached to the individual primes don’t all agree. As above, let i denote the first
time they fail to agree. That is, let
k1 = l1 , k2 = l2 , ... and ki-1 = li-1 but ki ≠ li .
Case 1: ki < li . So 1 < li - ki . Using some common rules of exponential
k
k
k
k
k
k
l
l
notation we have p1 1 p2 2 ... pi-1 i -1 pi i pi+1 i +1 ... pr r = n = p1 1 p2 2 ...
† † † †
†
†
† †
l i -1
ki
l i -ki
l i +1
lr
... pr . Canceling common factors from both sides of
i+1
† pi-1† pi pi † p†
†
†
these expressions (don’t forget, k1 = l1 , k2 = l2 , ... and ki-1 = li-1 !) we have
† †
†
† †
†
† †
ki +1
kr
l i -ki
l i +1
lr
ki +1
k
pi+1 ... pr = pi
pi+1 ... pr . So pi | pi+1 ... pr r . As we have seen in part
† †
†
†
k
(1) of the proof, this will
us to†the statement
pi | pr r . The only way to
† lead
† †
†
†
avoid a contradiction is to have i = r (that is, all the exponents agree until the
†
†
†
†
† †
†
k
k
k
last one). However, if this is true then we have p1 1 p2 2 ... pr r = n =
† †
l1
l2
lr
k1
k2
kr
l r -kr
p1 p2 ... pr = p1 p2 ... pr pr
. Canceling under this condition would
l -kr
give 1 = pr r
. So pr |1 and pr < 1. But
r is prime,
† p†
† and thus pr > 1. This
final contradiction tells us that ki cannot be less than li .
† †
†
† †
† †
Case 2: li < ki . See the last case.
†
†
†
†
†
Since both cases give rise to contradictions, it must be that all of the exponents
†
†
agree. This uniqueness argument completes the proof of the FTA (Thm 1.13).
† †
2