Proving a Polynomial is Irreducible using Eisenstein's CriterionDate: 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) Thank you. That's very helpful. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/