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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?

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  • $\begingroup$ 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
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Hint:

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 $f(x-1) = x^5 + 5x^4 - 15x^3 + 20x^2 - 30x + 20$.

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  • $\begingroup$ Sorry, you lost me at "field". $\endgroup$ – Edward Jiang Dec 15 '14 at 5:00
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    $\begingroup$ 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
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    $\begingroup$ @Lucian, Eisenstein's criterion is a blessing from the gods. $\endgroup$ – Kaj Hansen Dec 15 '14 at 5:14
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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 $\pm 1.$

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