Download 1 Unique Factorization of Integers

Unique Factorization
Darren Glass
November 22, 2004
The goal of this lecture is to discuss unique factorization in its various forms.
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Unique Factorization of Integers
We begin by proving a property of prime numbers which will be very important to what
we need. This is a special case of something that Professor Friedman did in the first
lecture, but it is worth going over again.
Key Property Of Primes I: Let p be a prime number and suppose that p divides ab
where a and b are integers. Then either p|a or p|b (or both).
Proof. If p|a then we are done. So we assume that p does not divide a and we will prove
that p|b.
We know that gcd(p, a) must divide p so it is either p or 1. But it cannot be p if p
does not divide a, and thus p and a must be relatively prime.
This tells us, from last time, that there exist integers x, y such that px + ay = 1. This in
turn implies that pxb + ayb = b. Now, p|pxb and p|ayb (as p|ab), and thus p must divide
b.
Note that this is not true for composite numbers: An example to illustrate this is that
6|150 = (10 × 15) but 6 does not divide either 10 or 15. This example illustrates what is
happening morally, that if a composite number divides ab then part of it can divide a and
the other part can divide b.
The previous theorem can be generalized in the following way. I will leave the proof
as an exercise.
Key Property Of Primes II: Let p be a prime number such that p divides the product
a1 a2 . . . ar . Then p divides (at least) one of the ai .
We are now ready to prove that integers factor in a unique way. More precisely, we
will prove the following:
Fundamental Theorem of Arithmetic: Every integer n ≥ 2 can be factored into a
product of primes n = p1 p2 . . . pr in a unique way, up to a reordering of the primes.
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This is a theorem that you have probably seen and used for years, but do you know
how to prove it?
Proof. First we prove existence. That is, we show that every number n can be written as a
product of primes. We will do this by induction after noting that it is true for a few small
values of n. In particular, we will assume that all integers less than k can be written as the
product of primes and show that k can be as well.
If k is prime, then it is its own prime factorization and we are done. So we can assume
that k is composite. This means that k can be written as k1 k2 where 1 < k1 ≤ k2 < k.
By the inductive hypothesis, we know that both k1 and k2 ca be written as a product of
primes, so let us write k1 = p1 . . . pr and k2 = q1 . . . qs , where the pi and qi are primes.
Then k = k1 k2 = p1 . . . pr q1 . . . qs . So we have expressed k as a product of prime numbers.
To show uniqueness, we assume that n = p1 . . . pr = q1 . . . qs and show that these are
the same factorization (up to a reordering of the primes). In particular, p1 |n = q1 . . . qs
and thus by the Key Property Of Primes II, p1 divides one of the qi . We are allowed to
renumber the primes, so without loss of generality, we may assume p1 |q1 . But q1 is prime
and therefore q1 = p1 . Now, we note that p2 . . . pr = q2 . . . qs and we proceed similarly to
see that p2 = q2 , etc.
So every number can be written in a unique way as the product of primes. But if I
give you a number, how do you write it as a product of primes? This turns out to be a
very difficult problem in general. In fact, just asking whether a given number is prime
is a difficult question which we will discuss in a later seminar. This is an example of a
theorem which is nonconstructive.
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Unique Factorization of Polynomials
A natural question to ask is whether an analogue to the fundamental theorem of arithmetic
works in other settings. Last week, Prof. Friedman talked about what divisibility means
with polynomials, and this is a natural setting for us to consider. We will consider the
case of polynomials in one variable with different coefficients.
C[x] - The fundamental theorem of algebra tells us that every polynomial of degree
n has exactly n roots when counted with multiplicity. Thus, over the complex numbers
any polynomial can be factored into linear factors as f (x) = a(x − a1 ) . . . (x − an ), where
a and the ai are complex numbers and the ai are the roots of f (x). Furthermore, this way
of expressing f (x) will be unique. In this setting, we therefore have an analogue to the
Fundamental Theorem of Arithmetic, where linear polynomials (x − ai ) take the roles of
the primes.
R[x] - Let f (x) ∈ R[x] be a polynomial of degree n with real coefficients. We know
that f (x) has n complex roots (again, counted with multiplicity). What can we say about
the roots? We know that if α is any complex (but not real) root then α, the complex
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conjugate of α is also a root and the polynomial (x − α)(x − α) is a quadratic polynomial
with real coefficients and leading coefficient one. We define a polynomial to be monic if
the coefficient of the term with highest degree is equal to one (in other words, it looks like
x2 + ax + b or more generally xn + . . .). Therefore, one can deduce that f (x) can be written
as a f1 (x) . . . fr (x) where a ∈ R and each fi is a monic linear or quadratic polynomial. This
again gives us an analogue to the Fundamental Theorem of Arithmetic.
Q[x]: the space of all polynomials with rational coefficients. In this setting, we also
have unique factorization in the following sense. Given a polynomial f (x) ∈ Q[x] then
f (x) = c f1 (x) . . . fr (x) where c ∈ Q and the fi are monic polynomials in Q[x] which are irreducible in Q[x] (ie cannot be written as the product of two other polynomials of positive
degree in Q[x]).
It can be easy to see if a polynomial is reducible, by factoring it. In particular, polynomials of degree 2 or 3 are reducible if and only if they have a rational root. But how do
you prove that a polynomial is irreducible? The following theorem sometimes gives one
way.
Eisenstein Irreducibility Criterion (1850): Let f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0
be a polynomial in Z[x]. Assume that there exists a prime p which does not divide an but
does divide all of the other ai . Furthermore, assume that p2 does not divide a0 . Then f (x)
is irreducible in Q[x].
Example: f (x) = 3x3 + 15x2 − 25x + 10, p = 5
Example: f (x) = x17 − 2x10 + 20x3 + 200x − 2, p = 2
Example: f (x) = x3 − 73 x2 + 14
5 . At first glance we cannot use Eisenstein, as f (x)
is not in Z[x]. But we note that a polynomial is irreducible over Q if and only if any
multiple of it is also irreducible over Q. Thus, f (x) will be irreducible if we can prove
that f˜(x) = 15 f (x) = 15x3 − 35x2 + 42 is. But we can apply Eisenstein to f˜(x) using the
prime p = 7.
Note: Eisenstein is not an if and only if statement. In particular, f (x) = x2 + 1 is
irreducible over Q (the roots are ±i and a quadratic polynomial with no rational roots
must be irreducible over Q) but clearly no prime will work as a0 = 1 is not divisible by
any prime.
Z[x], the space of all polynomials with integer coefficients. In this setting, the analogous theorem says that we can write any polynomial in a unique way as the product of
irreducible polynomials.
Note that we can no longer require the polynomials to be monic, as there are irreducible polynomials in Z[x] which are not monic (such as 2x + 1). However, we do not
want to allow polynomials where all of the coefficients are divisible by the same number,
such as 2x + 2, or we will not be able to get uniqueness. The following definition will be
useful.
Definition: A polynomial in Z[x] is primitive if there is no prime which divides all of
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its coefficients.
We can now state precisely what we mean by unique factorization in this setting. Any
polynomial f (x) ∈ Z[x] can be written as f (x) = ±p1 p2 . . . ps f1 (x) f2 (x) . . . fr (x) where
pi ∈ Z are prime numbers and each fi is an irreducible primitive polynomial in Z[x].
A special case of this result is the Rational Roots Test. In particular, we note that if
an integer n is a root of a polynomial f (x) then one of the fi in the above decomposition
must be x − n. However, the constant term of f (x) is given by p1 . . . ps a1 . . . ar , where
ai is the constant term of the irreducible polynomial fr . In particular, n will divide the
constant term of this polynomial, so the only possible integral roots of f (x) must divide
the constant term of f (x). In the exercises, you will be asked to consider the case of all
rational roots.
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Exercises
1. Show that if n is a square number then all of the exponents in the prime factorization
of n are even.
√
More generally, show that if m is an
2. Use Exercise 1 to show that 2 is irrational.
√
integer then either m is a perfect square or m is irrational.
3. Is it possible for a prime p to divide both n and n + 1 for any integer n? What about
n and n + 2? n and n + 3? Formulate and prove a general theorem about when a prime p
can divide both n and n + k for some integer n.
4. Show whether or not the following polynomials are irreducible in Q[x].
• x4 − 2x2 + 4x − 6
•
2 2
2
13 5
3
5 x − 4x + 3 x − 8x − 7
• x3 − 2x2 + 5
5. Show that if n = a × b is composite then 2n − 1 is also composite.
6. You have probably seen the fact that x p − 1 factors as (x − 1)(x p−1 + x p−2 + . . . x + 1).
We want to show that, in the case where p is prime, this second factor is irreducible over
Q by using Eisenstein’s criterion. We cannot apply it directly, as all of the coefficients are
1 and thus are not divisible by any prime p. However, let us make the change of variables
x → y + 1. Then we have that
p
(y + 1) − 1 = y +
y
+
y
+...+
y
p−1
p
p
p
p−1
p−2
p−3
= y(y
+
y
+
y
+...+
)
1
2
p−1
p
p
p
1
p−1
4
p
2
p−2
p
p
p
p−2
p−3
So it will suffice to show that
+
y
+
y
+ ... +
is
1
2
p−1
irreducible. Do this using Eisenstein’s criterion.
y p−1
7. The rational roots test says that, given a polynomial f (x) = an xn + . . . a0 ∈ Z[x], all
rational roots of the polynomial can be written as r/s, where r|a0 and s|an . Note that r
and s are allowed to be either positive or negative. (But not all rational numbers of this
form are necessarily roots!). Prove this theorem using the unique factorization of f (x).
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