Prove particular quintic is irreducible

Question

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?

Answer 1

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

Answer 2

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 $$ (x^2 + a x - 1)(x^3 + b x^2 + c x - 1) = x^5 + 10 x^4 + 15 x^3 + 15 x^2 - 10 x + 1 $$

Let us see how these proposed dactorizations fail. Consider the first case above. Multiplying the factors on the right gives

$x^5+10x^4+15x^3+15x^2-10x+1=x^5+(a+b)x^4+(ab+c+1)x^3+(ac+b+1)x^2+(a+c)x+1$

Thus to match the $x^1$ terms we require $c=-10-a$, and to match the $x^4$ terms we need $b=10-a$. Then the $x^2$ terms give

$ac+b+1=15\implies a(-10-a)+(10-a)+1=15\implies 2a^2+11a+4=0,$

whose discriminant $89$ is not a square and thus a factorization $(x^2+ax+1)(x^3+bx^2+cx+1)$ over the integers cannot exist. The alternative factorization $(x^2+ax-1)(x^3+bx^2+cx-1)$ fails by a similar argument.

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