Irreducibility of Polynomials over the Integers
Claudiu Raicu
April 27, 2010
1 Let’s warm up!
1. Is X2 + 4X + 3 reducible? What about X2 + 3X + 4?
2. Show that X3 − 5X + 14 is irreducible. What about X3 − 51X + 14?
3. Show that X4 + 1 is irreducible. Is X4 + 4 reducible?
4. Show that X5 + 6X4 + 6X3 + 24X + 72 is irreducible.
2 Take a look at the roots!
1. Is the polynomial X18 − 18 irreducible? What about X18 − 36? And X18 − 72?
2. Let a, n be integer numbers, n ≥ 1, and let p be a prime number, p > |a| + 1. Show that
the polynomial
Xn
+ aX + p
is irreducible.
3. Let n ≥ 2 be an odd integer, and let p be a prime number. Assume that all the roots of
the polynomial
Xn
+ an−1Xn−1
+ · · · + a1X + p2
have the same absolute value. Show that the polynomial g(X) = f(X2) is irreducible.
Theorem 2.1. (Perron’s criterion) Let f = Xn + a1Xn−1 + · · · + an be a polynomial with
integer coefficients. If |a1| > 1 + |a2| + · · · + |an|, then f is irreducible.
Quick question: is the polynomial Xn + 5Xn−1 + 3 reducible?
3 What if our polynomials take lots of small values?
1. Show that if a1, · · · , an are distinct integers, then the polynomial
(X − a1) · · · (X − an) − 1
is irreducible.
1
2. Show that if a1, · · · , an are distinct integers, then the polynomial
(X − a1)2
· · · (X − an)2
+ 1
is irreducible.
3. Let g be a polynomial of degree k with integer coefficients, and let d1, · · · , dk be distinct
integers. Show that
|g(di)| ≥
k!
2k
for at least one value of i ∈ {1, · · · , k}. (You’ll need this result for the next problem)
4. (Polya) Let f be a polynomial of degree n with integer coefficients, and set m = �(n +
1)/2�. Suppose there exist n distinct integer numbers a1, · · · , an which are not roots of
f, for which
f(ai) <
m!
2m
.
Prove that f is irreducible.
4 Reducing mod p...starting to feel the heat?
Theorem 4.1. (Eisenstein’s criterion) Let p be a prime number, and f = anXn+· · ·+a1X+a0
a polynomial with integer coefficients. Assume that
• p � an
• p|an−1, p|an−2, · · · , p|a1, p|a0.
• p2 � a0.
Then f is irreducible.
1. Show that if p is a prime number and q is not divisible by p, then Xn − pq is irreducible.
(Can you do this without Eisentein’s criterion?)
2. Let p be a prime number. Show that
Xp−1
+ Xp−2
+ · · · + X + 1
is irreducible.
3. Show that the polynomial
(X2
+ X)2n
+ 1
is irreducible.
4. Let p be a prime number, and a an integer not divisible by p. Show that the polynomial
Xp
− X + a
is irreducible over the integers (by showing the stronger statement that it is irreducible
over Z/(p)[X].
5. Show that X4+1 is irreducible in Z[X], but reducible in Z/(p)[X] for every prime number
p.
6. Is the polynomial Xn + 5Xn−1 + 3 reducible? (have you seen this before?)
5 Enter the Heroes: Newton Polygons
Theorem 5.1. (Dumas) The Newton polygon of a product of polynomials is the union of the
Newton polygons of the factors.
1. Let’s try this again: show that X5 + 6X4 + 6X3 + 24X + 72 is irreducible. Maybe try
something easier before: is X5 + 2X3 + 2X + 4 irreducible?
2. Is the polynomial Xn + 5Xn−1 + 3 reducible? (I thought we’ve answered this already...)
3. Let n ≥ 2 be an odd integer, and let p be a prime number. Assume that all the roots of
the polynomial
Xn
+ an−1Xn−1
+ · · · + a1X + p2
have the same absolute value. Show that the polynomial g(X) = f(X2) is irreducible.
(am I repeating myself?)
4. Show that the polynomial
Xn
n!
+
Xn−1
(n − 1)!
+ · · · +
X
1!
+ 1
is irreducible. (For those of you who know about power series, the above polynomials
form, as n ranges over the natural numbers, the truncations of the power series expansion
of eX.)
6 The Masterpiece
We’ll try to put together (some of) the techniques we’ve learnt so far to prove the irreducibility
of an interesting collection of polynomials. Consider the polynomials
fn =
(X + 1)n − Xn − 1
X
,
for n ≥ 1. We will show that f2p is irreducible whenever p is a prime number. Note that the
problem of establishing the irreducibility of fn is NOT solved for arbitrary n, so if you wanna
get rich...spiritually...you should give it a try!
Here are the steps for the irreducibility of f2p:
(a) Explain why if α is a root of fn, then so is 1/α.
(b) From here on, we assume that n = 2p with p a prime number. Consider the Newton
polygon of fn with respect to the prime number p. Show that fn is either irreducible, or it can
be written as a product of two irreducible polynomials of degree (p − 1), fn = g · h.
(c) Without loss of generality, assume that g and h have positive leading coefficients. If we
let gcd(g) and gcd(h) be the greatest common divisors of the coefficients of the two polynomials
(also known as the contents of the polynomials), show that gcd(g) =gcd(h) = 1. Show that,
under our assumption, the constant terms in g, h are positive.
(d) Show that if α is a root of g, then 1/α is NOT a root of g.
(e) Prove that g is the reciprocal of h, i.e.
g(X) = Xn−1
h(1/X).
(f) Show that the situation in (e) cannot occur, and conclude that fn is irreducible (when
n = 2p).