# Thread: Eisenstein polynomial and field extension

Suppose that $\sqrt{3}\in \mathbb{Q}[\rho]$ then we have the following:

$$\sqrt{3}=a+b\rho+c\rho^2+d\rho^3 \\ \Rightarrow 3=a^2+2 a b\rho+b^2\rho^2+2 a c\rho^2+2 b c\rho^3+c^2\rho^4+2 a d\rho^3+2 b d\rho^4+2 c d\rho^5+d^2\rho^6\\ \Rightarrow 3=a^2+2 a b\rho+(b^2+2 a c)\rho^2+(2 b c+2 a d)\rho^3+(c^2+2 b d)(2\rho^2+1)+2 c d\rho(2\rho^2+1)+d^2\rho^2(2\rho^2+1)\\ \Rightarrow 3=(a^2+c^2+2 b d)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3+2d^2\rho^4\\ \Rightarrow 3=(a^2+c^2+2 b d)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3+2d^2(2\rho^2+1) \\ \Rightarrow 3=(a^2+c^2+2 b d+2d^2)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+5d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3 \\ \Rightarrow (a^2+c^2+2 b d+2d^2-3)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+5d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3=0$$

Then $$a^2+c^2+2 b d+2d^2-3=2 a b+2 c d=b^2+2 a c+2c^2+4 b d+5d^2=2 b c+2 a d+4 c d=0$$

How could we solve this system?

From the second equation we have that $ab=-cd$.
If we substitute this in the last equation we get $b(c-a)+a(d-b)=0$.
How could we continue?

Or is this way not correct?

2. Originally Posted by mathmari
Suppose that $\sqrt{3}\in \mathbb{Q}[\rho]$ then we have the following:

$$\sqrt{3}=a+b\rho+c\rho^2+d\rho^3 \\ \Rightarrow 3=a^2+2 a b\rho+b^2\rho^2+2 a c\rho^2+2 b c\rho^3+c^2\rho^4+2 a d\rho^3+2 b d\rho^4+2 c d\rho^5+d^2\rho^6\\ \Rightarrow 3=a^2+2 a b\rho+(b^2+2 a c)\rho^2+(2 b c+2 a d)\rho^3+(c^2+2 b d)(2\rho^2+1)+2 c d\rho(2\rho^2+1)+d^2\rho^2(2\rho^2+1)\\ \Rightarrow 3=(a^2+c^2+2 b d)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3+2d^2\rho^4\\ \Rightarrow 3=(a^2+c^2+2 b d)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3+2d^2(2\rho^2+1) \\ \Rightarrow 3=(a^2+c^2+2 b d+2d^2)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+5d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3 \\ \Rightarrow (a^2+c^2+2 b d+2d^2-3)+(2 a b+2 c d)\rho+(b^2+2 a c+2c^2+4 b d+5d^2)\rho^2+(2 b c+2 a d+4 c d)\rho^3=0$$

Then $$a^2+c^2+2 b d+2d^2-3=2 a b+2 c d=b^2+2 a c+2c^2+4 b d+5d^2=2 b c+2 a d+4 c d=0$$

How could we solve this system?

From the second equation we have that $ab=-cd$.
If we substitute this in the last equation we get $b(c-a)+a(d-b)=0$.
How could we continue?

Or is this way not correct?
I would find it very difficult to solve this problem this way. I suggest you leave this problem for now and come back to it when you have learnt more theory.

If we want to check if $f$ can be splitted in $\mathbb{Q}[\sqrt{3}]$ we do the following:

We have the following possibilities:

- $x^4-2x^2-1=(x+a)(x^3+bx^2+cx+d)$
When $a$ is a root then $-a$ is also a root, ans so $(x-a)$ is also a factor. Therefore, $x^2-a^2$ is a factor of $f$. We have that $a\in \mathbb{Q}[\sqrt{3}]$, so $a=q_1+q_2\sqrt{3}$. Then $a^2=(q_1+q_2\sqrt{3})(q_1-q_2\sqrt{3})=q_1^2-3q_2^2\in \mathbb{Q}[x]$. That means that there is a factor of $f$ in $\mathbb{Q}[x]$, so it is not irreducible in $\mathbb{Q}[x]$, a contradiction.

- $x^4-2x^2-1=(x^2+ax+b)(x^2+cx+d)$
We have that $bd=-1$, where $b,d\in \mathbb{Q}[\sqrt{3}]$, so $b=q_1+q_2\sqrt{3}, d=\tilde{q_1}+\tilde{q_2}\sqrt{3}$. Then $$bd=-1 \Rightarrow q_1\tilde{q_1}+3q_2\tilde{q_2}+(q_1\tilde{q_2}+\tilde{q_1}q_2)\sqrt{3}=-1$$
Does the following have to stand?
$$q_1\tilde{q_1}+3q_2\tilde{q_2}=-1 \\ q_1\tilde{q_2}+\tilde{q_1}q_2=0$$

4. Originally Posted by mathmari
... so $a=q_1+q_2\sqrt{3}$. Then $a^2=(q_1+q_2\sqrt{3})(q_1-q_2\sqrt{3})=q_1^2-3q_2^2\in \mathbb{Q}[x]$.
Why is $a^2 = (q_1+q_2\sqrt{3})(q_1-q_2\sqrt{3})$. Shouldn't it simply be $(q_1+q_2\sqrt{3})^2$?

If we want to check if $f$ can be splitted in $\mathbb{Q}[\sqrt{3}]$ we do the following:

We have the following possibilities:

- $x^4-2x^2-1=(x+(a+b\sqrt{3}))p_3(x)$, where $p_3(x)\in \mathbb{Q}Q[\sqrt{3}](x)$
When $(a+b\sqrt{3})\in \mathbb{Q}[\sqrt{3}]$ is a factor of $f$ is its conjugate $(a-b\sqrt{3})$ also a factor of $f$ ?

- $x^4-2x^2-1=(x^2+(a+b\sqrt{3})x+(c+d\sqrt{3}))(x^2+(\alpha +\beta \sqrt{3})x+(\gamma+\delta \sqrt{3}))\\ =x^4+x^3\left [\alpha +\beta \sqrt{3}+a+b\sqrt{3}\right ]+x^2 \left [\gamma+\delta \sqrt{3}+ c+d\sqrt{3}+(a+b\sqrt{3})(\alpha +\beta \sqrt{3})\right ]+x\left [(a+b\sqrt{3})(\gamma+\delta \sqrt{3})+(c+d\sqrt{3})(\alpha +\beta \sqrt{3})\right ]+(c+d\sqrt{3})(\gamma+\delta \sqrt{3})$
where $a,b,c,d,\alpha, \beta, \gamma, \delta\in \mathbb{Q}$.

We have the following:

• $\alpha +\beta \sqrt{3}+a+b\sqrt{3}=0 \Rightarrow (\alpha+a)+(\beta+b)\sqrt{3}=0 \Rightarrow \alpha+a=0 \text{ and } \beta+b=0 \Rightarrow \alpha=-a \text{ and } \beta=-b$

• $\gamma+\delta \sqrt{3}+ c+d\sqrt{3}+(a+b\sqrt{3})(\alpha +\beta \sqrt{3})=-2 \Rightarrow \gamma+\delta \sqrt{3}+ c+d\sqrt{3}+(a+b\sqrt{3})(-a -b \sqrt{3})=-2 \\ \Rightarrow \gamma+\delta \sqrt{3}+ c+d\sqrt{3}-a^2 -3b^2 =-2 \Rightarrow (\gamma+c-a^2-3b^2)+(\delta+d)\sqrt{3}=-2 \Rightarrow \gamma+c-a^2-3b^2=-2\text{ and } \delta+d=0 \Rightarrow \gamma+c-a^2-3b^2=-2\text{ and } \delta=-d$

• $(a+b\sqrt{3})(\gamma+\delta \sqrt{3})+(c+d\sqrt{3})(\alpha +\beta \sqrt{3})=0 \Rightarrow (a+b\sqrt{3})(\gamma-d \sqrt{3})+(c+d\sqrt{3})(-a -b \sqrt{3})=0 \\ \Rightarrow a\gamma-ad\sqrt{3}+\gamma b\sqrt{3}-3bd-ac-bc\sqrt{3}-ad\sqrt{3}-3bd=0 \Rightarrow (a\gamma-ac-6bd)+(-2ad+\gamma b-bc)\sqrt{3}=0 \\ \Rightarrow a\gamma-ac-6bd =0\text{ and } -2ad+\gamma b-bc=0$

• $(c+d\sqrt{3})(\gamma+\delta \sqrt{3})=-1 \Rightarrow c\gamma-cd \sqrt{3}+\gamma d\sqrt{3}+3d\delta=-1 \Rightarrow (c\gamma+3d\delta)+(-cd+d\gamma )\sqrt{3} =-1 \\ \Rightarrow c\gamma-3d^2=-1 \text{ and } -d(c-\gamma)=0$

Therefore, we have the following equations:
1. $\gamma+c-a^2-3b^2=-2$
2. $a\gamma-ac-6bd =0$
3. $-2ad+\gamma b-bc=0$
4. $c\gamma-3d^2=-1$
5. $-d(c-\gamma)=0$

From the last equation we get either $d=0$ or $c=\gamma$.

-If $c=\gamma$ we get the following equations:
1. $c+c-a^2-3b^2=-2 \Rightarrow 2c-a^2-3b^2=-2$
2. $ac-ac-6bd =0\Rightarrow bd=0$
3. $-2ad+c b-bc=0\Rightarrow ad=0$
4. $c^2-3d^2=-1$

From the equation 3. suppose we have $d=0$, then from the equation 4. we get $c^2=-1$, a contradiction.
Therefore it must be $d\neq 0$. And so it must be $b=a=0$ from the equations 2. and 3. .
From the equation 1. we get then $c=-1$. And then from the equation 4. we get $3d^2=-2$, a contradiction.

Therefore the case $c=\gamma$ cannot hold.

-If $d=0$ we get the following equations:
1. $\gamma+c-a^2-3b^2=-2$
2. $a\gamma-ac =0 \Rightarrow a(\gamma -c)=0$
3. $\gamma b-bc=0 \Rightarrow b(\gamma-c)=0$
4. $c\gamma=-1$

From above we have that it cannot hold that $d=0$ and $\gamma=c$. So from the equations 2. and 3. we get that $a=b=0$.
Therefore, from the equation 1. we get $\gamma+c=-2 \Rightarrow \gamma=-2-c$. And so from the equation 4. we get $c(-2-c)=-1 \Rightarrow -2c-c^2=-1 \Rightarrow c^2+2c-1=0$.
So, we have to find the roots of $x^2+2x-1=0$. $\Delta=4+4=8$. $x_{1,2}=\frac{-2\pm 2\sqrt{2}}{2}=-1\pm \sqrt{2}\notin \mathbb{Q}$.

Therefore the case $d=0$ cannot hold.

Thus, there are no rationals so that the above relation hold. Therefore $f(x)$ cannot be factorized into two polynomials of degree $2$.

Is evrything correct?