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Question:

If I want to show irreducibility of $x^4-2x^2+9$ over $\mathbb Q$ can I do it like this:

I show irreducibility in $\mathbb Z$ because by Gauss the polynomial will be also irreducible in $\mathbb Q$. The gcd of the coefficients is $1$, hence its not possible to factor an integer out of the polynomial.

The polynomial also has no roots in $\mathbb Z$, hence its not possible to write it as the product of a polynomial with degree $1$ and a polynomial with degree $3$.

If I write down $x^4-2x^2+9=(a_1x^2+b_1x+c_1)(a_2x^2+b_2x+c_2)$ then I will get a contradiction because:

We have the following equations:

$(1)$ $a_1a_2=1\Rightarrow a_1=a_2=1$

$(2)$ $a_1b_2+a_2b_1=0\Rightarrow b_1=-b_2$

$(3)$ $a_1c_2+b_1b_2+a_2c_1=-2\Rightarrow c_2-b_1^2+c_1=-2 $

$(4)$ $c_1b_2+b_1c_2=0\Rightarrow c_1b_2-c_2b_2=0 $

$(5)$ $c_1c_2=9 \Rightarrow c_1=\frac{9}{c_2} $

$(4) \text{ and } (5)$ gives $b_2(9-c_2^2)=0$ if $b_2=0$, then $b_1=0$ and (3) and (5) could not be true at the same time, hence $c_1=c_2=\pm 3$

This gives in $(3): b_1^2=2+2c_1$ in any case of $c_1$ there in no solution in terms of $b_1$ a contradiction.

So $x^4-2x^2+9$ is irreducible over $\mathbb Q$

Can someone go trough it and tell me if this is correct?

glS
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Duke
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    The idea is correct. I didn't check your computations, but I spotted a mistake: $a_1a_2=1\color{red}{\Rightarrow} a_1=a_2=1$. In any case, on any field if a monic polynomial is reducible, then it can be written as the product of monic polynomials. – Git Gud Feb 14 '15 at 13:10
  • I did this because I want to show irreducibility over $\mathbb Z$. So if $ab$=1 the only possibility for $a$ and $b$ is that $a=b=1$. Okay I see now that $a=b=-1$ is also possible. But is this still wrong? – Duke Feb 14 '15 at 13:13
  • @duke If that's the first thing you've done, you can assume without loss of generality that they are both positive, since flipping them to negative would just flip the sign of all the other terms. – Alan Feb 14 '15 at 13:15
  • So my solution is correct? At least the idea and the first steps? – Duke Feb 14 '15 at 13:27
  • Easy way As described here, shifted Eisenstein applies since $2$ divides the discriminant of $f$ and $f$ is a prime power $\bmod 2,,$ viz. $(x+1)^4\ \ $ – Bill Dubuque Feb 28 '23 at 08:22

2 Answers2

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We just need to find a finite field $\mathbb{F}_p$ for which the quadratic polynomial $x^2-2x+9$ is irreducible, i.e. a prime $p$ for which $\Delta=-32$ is not a quadratic residue, or, in terms of the Legendre symbol: $$\left(\frac{-2}{p}\right)=-1, $$ that is equivalent to $p\in\{5,7\}\pmod{8}$. If we take a prime number $p$ in such a set, $$ p(x) = x^4-2x^2+9 $$ splits as: $$ p(x) = (x^2+ax\pm 3)\cdot(x^2-ax\pm 3)\tag{1} $$ over $\mathbb{F}_p$, where: $$a^2\mp 6 \equiv 2\pmod{p}.\tag{2}$$ Assuming that $p(x)$ splits over $\mathbb{Q}$, then it splits like in $(1)$, where: $$ a^2 \in \{-4,8\}\pmod{p}\tag{3}$$ for an infinite number of primes $p$, so that: $$ a^2\in \{-4,8\}\tag{4}. $$ However, that gives a contradiction, since no squared rational number can be equal to $-4$ or $8$.

Jack D'Aurizio
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Following your idea and making it as explicit as possible.

You can start by assuming that there is a decomposition $$x^4-2x^2+9=(x^2+b_1x+c_1)(x^2+b_2x+c_2).$$ Then

$(1)$ $b_1+b_2=0\Rightarrow b_1=-b_2$

$(2)$ $c_1+b_1b_2+c_2=-2\Rightarrow c_1-b_1^2+c_2=-2 $

$(3)$ $c_1b_2+b_1c_2=0\Rightarrow c_1b_2-c_2b_2=0 $

$(4)$ $c_1c_2=9$

From $(3)$ you get $$b_2(c_1-c_2)=0,$$ so there are two cases:

(i) $b_2=0\stackrel{(1)}\Rightarrow b_1=0\stackrel{(2)}\Rightarrow c_1+c_2=-2\Rightarrow c_2=-2-c_1\stackrel{(4)}\Rightarrow c_1(-2-c_1)=9\Rightarrow c_1^2+2c_1+9=0$ and this is an equation which doesn't have real solutions.

(ii) $c_1=c_2\stackrel{(4)}\Rightarrow c_1^2=9\Rightarrow c_1=\pm3\stackrel{(2)}\Rightarrow b_1^2=8$ or $b_1^2=-4$, a contradiction.

user26857
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