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

Question

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} $

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$(4) \text{ and } (5)$ 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?

Answer 1

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 $8$.

Answer 2

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.