The goal of this lecture is to discuss unique factorization in its various forms.
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 a
1 a
2
. . .
a r
. Then p divides (at least) one of the a i
.
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 = p
1 p
2
. . .
p r 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 k
1 k
2 where 1 < k
1
≤ k
2
< k.
By the inductive hypothesis, we know that both k
1 and k
2 ca be written as a product of primes, so let us write k
1
= p
1
. . .
p r and k
2
= q
1
. . .
q s
, where the p i and q i are primes.
Then k = k
1 k
2
= p
1
. . .
p r q
1
. . .
q s
. So we have expressed k as a product of prime numbers.
To show uniqueness, we assume that n = p
1
. . .
p r
= q
1
. . .
q s and show that these are the same factorization (up to a reordering of the primes). In particular, p
1
| n = q
1
. . .
q s and thus by the Key Property Of Primes II, p
1 divides one of the q i
. We are allowed to renumber the primes, so without loss of generality, we may assume p
1
| q
1
. But q
1 is prime and therefore q
1
= p see that p
2
= q
2
, etc.
1
. Now, we note that p
2
. . .
p r
= q
2
. . .
q s and we proceed similarly to
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.
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 − a
1
) . . .
( x − a n
) , where
a and the a i are complex numbers and the a i 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 − a i
) 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 x
2
+ ax + b or more generally x n
+ . . .
). Therefore, one can deduce that f ( x ) can be written as a f
1
( x ) . . .
f r
( x ) where a ∈
R and each f i 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 f
1
( x ) . . .
f r
( x ) where c ∈
Q and the f i 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 ) = a n x n
+ a n − 1 x n − 1
+ . . .
+ a
1 x + a
0 be a polynomial in
Z
[ x ] . Assume that there exists a prime p which does not divide a n but does divide all of the other a i
. Furthermore, assume that p
2 does not divide a
0
. Then f ( x ) is irreducible in
Q
[ x ] .
Example: f ( x ) = 3x
3
Example: f ( x ) = x
17
+ 15x
− 2x
10
2 − 25x
+ 20x
3
+
+
10, p
200x
=
−
5
2, p = 2
Example: f ( x ) = x
3
−
7
3 x
2
+
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 ˜ ( x ) = 15 f ( x ) = 15x
3
− 35x
2
+ 42 is. But we can apply Eisenstein to ˜ ( x ) using the prime p = 7.
Note: Eisenstein is not an if and only if statement. In particular, f ( x ) = x
2
+ 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 a
0
= 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 p i
∈
( x ) ∈
Z
[ x ] can be written as f ( x ) = ± p
1 p
2
. . .
p s f
1
( x ) f
2
( x ) . . .
f r
( x ) where
Z are prime numbers and each f i 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 f i in the above decomposition must be x − n. However, the constant term of f ( x ) is given by p
1
. . .
p s a
1
. . .
a r
, where a i is the constant term of the irreducible polynomial f r
. 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.
1. Show that if n is a square number then all of the exponents in the prime factorization of n are even.
2. Use Exercise 1 to show that
√
2 is irrational. More generally, show that if m is an 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 ] .
• x
4 − 2x
2
+ 4x − 6
• 13
5 x
5 − 4x
3
+
2
3 x
2 − 8x − 2
7
• x
3 − 2x
2
+ 5
5. Show that if n = a × b is composite then 2 n
− 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
( y + 1 ) p
− 1 = y p
+ p
1
= y ( y p − 1
+ p
1 y p − 1
+ p
2 y p − 2
+ p
2 y p − 2
+ . . .
y p − 3
+
+ . . .
p p − 1
+ y p p − 1
)
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So it will suffice to show that y p − 1
+ p
1 irreducible. Do this using Eisenstein’s criterion.
y p − 2
+ p
2 y p − 3
+ . . .
+ p p
− 1 is
7. The rational roots test says that, given a polynomial f ( x ) = a n rational roots of the polynomial can be written as r / s, where r | a
0 x n
+ . . .
a
0
∈
Z
[ x ] , all and s | a n
. 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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