Question is :
Prove that the polynomial $p(x)=x^4-4x^2+8x+2$ is irreducible over the quadratic field $F=\mathbb{Q}(\sqrt{-2})$. [Hint : first use the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$ . Conclude that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$ and hence show that $p(x)$ has no linear factor over $F$. Show similarly that $p(x)$ is not the product of quadratics with coefficients in $F$.]
What I have done so far is :
Suppose $\alpha \in\mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)=x^4-4x^2+8x+2$ we would then have :
$p(\alpha)=\alpha^4-4\alpha^2+8\alpha+2=0\Rightarrow 2=\alpha(-\alpha^3+4\alpha-8)$
i.e., $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$.
so, I have used the method of proposition $11$ for the U.F.D $\mathbb{Z}[\sqrt{-2}]$(cf.Exercise $8$, Section $8.1$) to show that if $\alpha \in \mathbb{Z}[\sqrt{-2}]$ is a root of $p(x)$ then $\alpha$ is a divisor of $2$ in $\mathbb{Z}[\sqrt{-2}]$
I do not understand why does he mentioned that $\mathbb{Z}[\sqrt{-2}]$ is U.F.D and all... I do not use that at all... It is unnecessarily confusing me or i am unnecessarily getting confused.. hint is actually misleading me :(
Now, I have to prove that $\alpha$ must be $\pm 1,\pm \sqrt{-2}$ or $\pm 2$
i.e., suppose I have $2=ab$ in $\mathbb{Z}[\sqrt{-2}]$ then, $N(2)=N(ab)\Rightarrow 4=N(a)N(b)$
i.e.,$N(a)=1\text{ or }2\text{ or } 4$ i.e., $p^2+2q^2=1\text{ or }2\text{ or } 4$ for $a=p+\sqrt{-2}q$
$p^2+2q^2=1\Rightarrow p=\pm 1 \Rightarrow a=\pm 1$
$p^2+2q^2=2\Rightarrow q=\pm 1\Rightarrow a=\pm\sqrt{-2}$
$p^2+2q^2=4\Rightarrow p=\pm 2\Rightarrow a=\pm 2$
Once I prove that those are the only divisors then I would consider :
- $p(1)=(1)^4-4(1)^2+8(1)+2\neq 0$
$p(-1)=(-1)^4-4(-1)^2+8(-1)+2\neq 0$
$p(\sqrt{-2})=(\sqrt{-2})^4-4(\sqrt{-2})^2+8(\sqrt{-2})+2\neq 0$
$p(-\sqrt{-2})=(-\sqrt{-2})^4-4(-\sqrt{-2})^2+8(-\sqrt{-2})+2\neq 0$
$p(2)=(2)^4-4(2)^2+8(2)+2\neq 0$
$p(-2)=(-2)^4-4(-2)^2+8(-2)+2\neq 0$
So, no divisor of $2$ is a root..
Thus $p(x)$ do not have a root in $\mathbb{Z}[\sqrt{-2}]$
suppose I have something like :
$$x^4-4x^2+8x+2=(x^2+ax+b)(x^2+cx+d)=x^4+(a+c)x^3+(b+d+ac)x^2+(ad+bc)x+bd$$
But then, $bd=2\Rightarrow \text { b,d are a divisors of 2 in $\mathbb{Z}\sqrt{-2}$}$
But then we have seen that only divisors of $2$ in $\mathbb{Z}\sqrt{-2}$ are $\pm 1,\pm \sqrt{-2}$ or $\pm 2$
So, only possibilities are
$x^4-4x^2+8x+2=(x^2+ax\pm 1)(x^2+cx\mp 2)$
$x^4-4x^2+8x+2=(x^2+ax\pm \sqrt{-2})(x^2+cx\mp \sqrt{-2})$
But these are not possible...
This only tells that $p(x)$ is irreducible in $\mathbb{Z}[\sqrt{-2}]$ but then how do i show this is irreducible in $\mathbb{Q}(\sqrt{-2})$ I was expecting gauss lemma to help but it only works for integers and rationals...
So, please help me to clear this..
Thank you...
P.S : Proposition $11$ is Rational root theorem and Exercise $8$ is that ring of integers of $\mathbb{Q}(\sqrt{-2})$ is an Euclidean domain.
Gauss lemma does work for $\mathbb{Z}[\sqrt{-2}]$