The problem is to prove that the quintic $$x^5+10x^4+15x^3+15x^2-10x+1$$ is irreducible in the rationals.

I don't have much knowledge in group theory, and certainly not in Galois theory, and I'm pretty sure this problem can be solved without those tools.

I know about Eisenstein's criterion, but it cannot be applied to this particular quintic because $5$ does not divide the constant term. If we somehow manipulate the polynomial so that $5$ divides the constant term, we still have to make sure that $25$ doesn't.

So is there any other easy ("elementary") way to solve this?

  • $\begingroup$ Then equate the coefficients? I don't think solving a system of $5$ equations is exactly easy. $\endgroup$ – Edward Jiang Dec 15 '14 at 5:01
  • $\begingroup$ @Will $15$ and $1$. Let me think about this. $\endgroup$ – Edward Jiang Dec 15 '14 at 5:03
  • $\begingroup$ Actually, 10 and 1. $\endgroup$ – Will Jagy Dec 15 '14 at 5:09
  • $\begingroup$ wrote slightly more as an answer. $\endgroup$ – Will Jagy Dec 15 '14 at 5:59


First, prove that $f(x)$ is irreducible over a field $F$ $\iff$ $f(x+c)$ is also irreducible over $F$ for any $c \in F$.

Given this result, note that $f(x-1) = x^5 + 5x^4 - 15x^3 + 20x^2 - 30x + 20$.

  • $\begingroup$ Sorry, you lost me at "field". $\endgroup$ – Edward Jiang Dec 15 '14 at 5:00
  • 2
    $\begingroup$ That's ok. As far as this problem is concerned, wherever you see "field" or "F" you can replace it with "the rationals". $\endgroup$ – Kaj Hansen Dec 15 '14 at 5:01
  • $\begingroup$ Let me give it some thought $\endgroup$ – Edward Jiang Dec 15 '14 at 5:02
  • $\begingroup$ Sure thing! This "shift trick" is a somewhat important result as you go forward in abstract algebra. It is useful, for example, in proving that the $p^{\text{th}}$ cyclotomic polynomial has degree $p-1$. At any rate, you can also use Will Jagy's method in the comments, though I hate undetermined coefficients myself. :) $\endgroup$ – Kaj Hansen Dec 15 '14 at 5:07
  • 3
    $\begingroup$ @Lucian, Eisenstein's criterion is a blessing from the gods. $\endgroup$ – Kaj Hansen Dec 15 '14 at 5:14

Major simplification: result of Gauss that one may factor a polynomial with integer coefficients and get the same answer as with rational numbers, http://en.wikipedia.org/wiki/Gauss%27s_lemma_%28polynomial%29

This is Theorem 3.10.1 on page 160 of Topics in Algebra by Herstein.

So, the choices to finish are $$ (x^2 + a x + 1)(x^3 + b x^2 + c x + 1) = x^5 + 10 x^4 + 15 x^3 + 15 x^2 - 10 x + 1 $$ and $$ (x^2 + a x - 1)(x^3 + b x^2 + c x - 1) = x^5 + 10 x^4 + 15 x^3 + 15 x^2 - 10 x + 1 $$

It is really the same thing to point out the rational roots theorem, the only possible roots (and linear factors, therefore) are $\pm 1.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.