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Let $p(x) = x^4 + 4x^3 + 11x^2 + 10x + 9$. How to how that it is irreducible over $\mathbb{Q}$?

The first thing I tried was Eisenstein, but that obviously doesn't work with the polynomial in that form.

Then I looked at $p$ in $\mathbb{F}_2$ and $\mathbb{F}_5$ in order to show that it irreducible in one of them and then to use Gauß' Theorem. However, it turns out that $p$ is reducible over both fields.

I've seen people use nifty substitutions in connection with the Eisenstein criterion. I tried to represent $p$ in terms of $x+1$, but that doesn't seem to work, either.

Now I'm at my wit's end. What else could I try?

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    $\begingroup$ factor in $\mathbb F_3$ - you get $x$ times a irreducible cubic polynomial. Hence $p$ (over $\mathbb Q$) can't be the product of two quadratic polynomials. $\endgroup$
    – user8268
    Mar 19, 2016 at 19:11
  • $\begingroup$ Hmm, could you elaborate on that? $\endgroup$
    – Miriam
    Mar 19, 2016 at 19:26
  • $\begingroup$ To clarify for future readers: Suppose we could factor $p(x)$ as a product of two irreducible quadratics in $\Bbb Q[x]$. Then, by Gauss' theorem, we would have $p(x) = a(x)b(x)$ for monic $a(x), b(x) \in \Bbb Z[x]$. Then, reducing modulo $3$, we would have $\bar{p}(x) = \bar{a}(x) \bar{b}(x)$. But in $\Bbb F_3$, we see that $\bar{p}(x)$ cannot have a quadratic factor since it factorises as a linear polynomial times an irreducible cubic. $\endgroup$ Jun 22, 2021 at 9:05

2 Answers 2

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It is a standard result that it is irreducible over Q iff it is irreducible over Z. If it factors as linear x cubic, then the linear factor would have to be $x+a$ for some integer $a$, so $a$ would be a root and a factor of 9. So you have to check 1, 3, -3, 9, -9. None of them are roots. If it factors as quadratic x quadratic, then it is $(x^2+ax+b)(x^2+cx+d)$.

So you have $a+c=4,bd=9,b+d+ac=11$. If $b=1,-1,9$ or -9, then $ac=1$ or 21, both incompatible with $a+c=4$. So $b=d=3$ or $b=d=-3$, so $ac=5$ or 17. Neither is compatible with $a+c=4$. So it is irreducible.

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    $\begingroup$ Ah, that's it. Thanks you. $\endgroup$
    – Miriam
    Mar 19, 2016 at 19:28
  • $\begingroup$ A variation of this argument is to consider $p(x - 1) = x^4 + 5 x^2 - 4 x + 7$; since it has no cubic term, any factorization into quadratics over $\Bbb Z$ has the form $(x^2 + \alpha x + \beta) (x^2 - \alpha x + \gamma)$. The constant term is $\beta \gamma = 7$, so we can take $\beta = \pm 1, \gamma = \pm 7$ (respectively). Then the linear coefficient is $4 = \alpha (\gamma - \beta) = \pm 6 \alpha$, but $6 \not\mid 4$, so there is no factorization. $\endgroup$ May 22, 2022 at 17:01
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If in addition to knowing that $p$ is reducible both modulo $2$ and modulo $5$ we have the factorizations of $p$ modulo those primes into irreducible polynomials, namely, $$p(x) \equiv (x^2 + x + 1)^2 \pmod 2 \qquad \textrm{and} \qquad p(x) \equiv (x^3 + x + 1)(x + 4) \pmod 5,$$ we see immediately that if $p$ were reducible it would have both an irreducible quadratic factor and an irreducible cubic factor, which (because $\deg p = 4$) is impossible, hence $p$ is irreducible over $\Bbb Q$. (As Aryaman Maithani points out in the comments, $p(x)$ also factors modulo $3$ into the product of a cubic factor and a linear one, and factoring modulo $3$ is typically faster that factoring modulo $5$, leading to a probably faster solution using the same reasoning.)

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