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### Proving a Polynomial is Irreducible using Eisenstein's Criterion

```Date: 10/18/2004 at 02:02:59
From: Gunnar
Subject: Irreducible polynomials over Q

Let p be a prime number.  Show that the polynomial x^p + px + (p-1)
is irreducible over Q if and only if p >= 3.

I had in mind two approaches for the question.  Either prove directly
(i.e let x^p + px + (p-1) = (...)(...) and prove for contradiction),
or using Eisenstein's criterion.  But I find it hard to even start
using either method, or maybe there's a better way of proving.  Can
you help?

```

```
Date: 10/18/2004 at 14:01:32
From: Doctor Pete
Subject: Re: Irreducible polynomials over Q

Hi Gunnar,

This question seems to be solved best by Eisenstein's criterion, which
states that the polynomial

f(x) = a[n] x^n + a[n-1] x^(n-1) + ... + a[1] x + a[0]

where a[n] is nonzero and a[k] are integers for all 0 <= k <= n, is
irreducible if there exists a prime p such that p divides a[0], a[1],
..., a[n-1] but not a[n], and furthermore, p^2 does not divide a[0].

However, Eisenstein's criterion doesn't directly apply to the
polynomial

f(x) = x^p + px + (p-1),

since as it is written, no such prime satisfies the conditions.
However, we can observe that f(x) is irreducible if and only if f(x-c)
is irreducible for any given integer c.  So let us consider the polynomial

f(x+1) = (x+1)^p + p(x+1) + (p-1).

Expanding the first term by the Binomial Theorem, we obtain

(x+1)^p = x^p + C[p,1] x^(p-1) + C[p,2] x^(p-2) + ... + C[p,p-1] x +
C[p,p],

where C[p,k] = p!/(k!(p-k)!) is the binomial coefficient counting the
number of ways to select k objects from a group of p objects.  So
C[p,p] = 1, and C[p,p-1] = p, and therefore

f(x+1) = x^p + C[p,1] x^(p-1) + ... + C[p,p-2] x^2 + px + 1 + px + p
+ p - 1
= x^p + C[p,1] x^(p-1) + ... + C[p,p-2] x^2 + 2px + 2p.

But since p is a prime, C[p,k] is divisible by p for all 1 <= k <= p-1
because none of the factors of k! or (p-k)! divide p.  Hence f(x+1)
satisfies the Eisenstein criterion for the prime p, except when p = 2,
where it fails because 2p = p^2 when p = 2.  Therefore, f(x+1), and
by our previous comment, f(x), are irreducible for p >= 3.

All that remains is to show that f(x) is reducible for p = 2, which is
trivial:

f(x) = x^2 + 2x + 1 = (x+1)^2.

What makes this question interesting, then, is the idea that a
polynomial may not satisfy the Eisenstein criterion as it is written,
but that a translation may allow one to show irreducibility.  Clearly,
Eisenstein's criterion is a sufficient, but not necessary, condition
for irreducibility.  It is especially important in the above example
that deg(f) = p, because this is precisely what leads to f(x+1) having
coefficients that are divisible by p.

- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 10/26/2004 at 22:59:40
From: Gunnar
Subject: Thank you (Irreducible polynomials over Q)

```
Associated Topics:
College Number Theory

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