Eisenstein polynomial and field extension

[mathmari]

Hey!

Let $f = x^4−2x^2−1 \in \mathbb{Q}[x]$.

We have that $f(x+1)=(x+1)^4-2(x+1)^2-1=x^4+4x^3+6x^2+4x+1-2(x^2+2x+1)-1=x^4+4x^3+4x^2-2$

We have that $p=2$ divides all the coefficients $4,4,-2$ and $p^2=4$ does not divide the constant term $-2$.

So, the polynomial $f(x+1)$ is Eisenstein. This means that $f(x+1)$ is irreducible, or not? (Wondering)

Knowing that, can we conclude that $f(x)$ is irreducible as follows:

Suppose that $f(x)$ is reducible, then there are non-constant polynomials $g(x),h(x)\in \mathbb{Q}[x]$ such that $f(x)=g(x)h(x)$. Then $f(x+1)=g(x+1)h(x+1)$. Since the polynomials are non-constant that would mean that $f(x+1)$ is reducible, a contradiction.
So, $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Is this correct? (Wondering)

Let $\rho\in \mathbb{C}$ be a root of $f$. I want to find a basis and the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$.

We have that $\text{Irr}(\rho, \mathbb{Q})=f$ and $\text{Irr}(\sqrt{2},\mathbb{Q}[\rho])=x^2-2$, or not?

Therefore, the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$ is $\deg f \cdot \deg (x^2-2)=4\cdot 2=8$.

Is this correct? (Wondering)

How could we find a basis of the extension? (Wondering)

 

[caffeinemachine]

Hey!

Let $f = x^4−2x^2−1 \in \mathbb{Q}[x]$.

We have that $f(x+1)=(x+1)^4-2(x+1)^2-1=x^4+4x^3+6x^2+4x+1-2(x^2+2x+1)-1=x^4+4x^3+4x^2-2$

We have that $p=2$ divides all the coefficients $4,4,-2$ and $p^2=4$ does not divide the constant term $-2$.

So, the polynomial $f(x+1)$ is Eisenstein. This means that $f(x+1)$ is irreducible, or not? (Wondering)

Knowing that, can we conclude that $f(x)$ is irreducible as follows:

Suppose that $f(x)$ is reducible, then there are non-constant polynomials $g(x),h(x)\in \mathbb{Q}[x]$ such that $f(x)=g(x)h(x)$. Then $f(x+1)=g(x+1)h(x+1)$. Since the polynomials are non-constant that would mean that $f(x+1)$ is reducible, a contradiction.
So, $f(x)$ is irreducible in $\mathbb{Q}[x]$.

Is this correct? (Wondering)

This is correst. Basically what we are doing is this: Consider the ring map $R[x]\to R[x]$ which sends $x$ to $x+1$. This is a ring isomorphism. Thus $f(x)$ in $R[x]$ is irreducible iff and only if its image, which is $f(x+1)$, is irreducible.

Let $\rho\in \mathbb{C}$ be a root of $f$. I want to find a basis and the degree of the extension $\mathbb{Q}[\sqrt{2},\rho]$.

We have that $\text{Irr}(\rho, \mathbb{Q})=f$ and $\text{Irr}(\sqrt{2},\mathbb{Q}[\rho])=x^2-2$, or not?

It is not clear why $x^2-2$ is irreducible over $\mathbf Q(\rho)$. This must be true but it requires a proof. And I am unable to see it myself.

 

[mathmari]

It is not clear why $x^2-2$ is irreducible over $\mathbf Q(\rho)$. This must be true but it requires a proof. And I am unable to see it myself.

So that $x^2-2$ is irreducible over $\mathbf Q(\rho)$, the root $\pm \sqrt{2}$ does not have to be an element of $\mathbf Q(\rho)$, right?
So we have to know that the real part of $\rho$ is not equal to $\pm \sqrt{2}$, or not?

Let $\rho=\sqrt{2}+bi$ then $f(\rho)=0 \Rightarrow (\sqrt{2}+bi)^4−2(\sqrt{2}+bi)^2−1=0 \\ \Rightarrow -1-10 b^2+b^4+i (4 \sqrt{2} b-4 \sqrt{2} b^3)=0 \\ \Rightarrow -1-10 b^2+b^4=0 \land 4 \sqrt{2} b-4 \sqrt{2} b^3=0 \\ \Rightarrow (b=\pm \sqrt{5+\sqrt{26}} )\land (b=0 \lor b=\pm 1)$
a contradiction

Let $\rho=-\sqrt{2}+bi$ then $f(\rho)=0 \Rightarrow (-\sqrt{2}+bi)^4−2(-\sqrt{2}+bi)^2−1=0 \\ \Rightarrow -1-10 b^2+b^4+i (-4 \sqrt{2} b+4 \sqrt{2} b^3)=0 \\ \Rightarrow -1-10 b^2+b^4=0 \land -4 \sqrt{2} b+4 \sqrt{2} b^3=0 \\ \Rightarrow (b=\pm \sqrt{5+\sqrt{26}} )\land (b=0 \lor b=\pm 1)$
a contradiction

Therefore, $\sqrt{2}\notin \mathbf Q(\rho)$.

So, $x^2-2$ is irreducible over $\mathbf Q(\rho)$.

Is this correct? Could I improve something? (Wondering)

 

[caffeinemachine]

So that $x^2-2$ is irreducible over $\mathbf Q(\rho)$, the root $\pm \sqrt{2}$ does not have to be an element of $\mathbf Q(\rho)$, right?
So we have to know that the real part of $\rho$ is not equal to $\pm \sqrt{2}$, or not?

The first statement is correct because a degree 2 polynomial is irreducible over $\mathbb Q(\rho)$ if and only if it does not have a root in $\mathbf Q(\rho)$. I am unable to see the logic behind your second statement. Can you elaborate on that?

 

[mathmari]

I am unable to see the logic behind your second statement. Can you elaborate on that?

What I said is wrong. I thought that we would have that $\overline{\rho}\in \mathbb{Q}(\rho)$, which is not true, is it?

So, we have to show that $\rho$ cannot be equal to $\sqrt{2}$ so that $\sqrt{2}\notin \mathbb{Q}(\rho)$, right? (Wondering)

 

[caffeinemachine]

What I said is wrong. I thought that we would have that $\overline{\rho}\in \mathbb{Q}(\rho)$, which is not true, is it?

So, we have to show that $\rho$ cannot be equal to $\sqrt{2}$ so that $\sqrt{2}\notin \mathbb{Q}(\rho)$, right? (Wondering)

Are you saying that $\rho\neq \sqrt{2}$ implies $\sqrt{2}\notin \mathbf Q(\rho)$?

 

[mathmari]

Are you saying that $\rho\neq \sqrt{2}$ implies $\sqrt{2}\notin \mathbf Q(\rho)$?

Yes. Is this wrong? (Wondering)

 

[caffeinemachine]

Yes. Is this wrong? (Wondering)

Yes. This implication is incorrect. Try supplying a proof. You'll see why.

 

[mathmari]

Try supplying a proof. You'll see why.

Should I prove that it doesn't hold that $\sqrt{2}\in \mathbb{Q}(\rho)\Rightarrow \rho=\sqrt{2}$ ? (Wondering)

It is not clear why $x^2-2$ is irreducible over $\mathbf Q(\rho)$. This must be true but it requires a proof. And I am unable to see it myself.

So, is there an other way to find the degree of the extension? (Wondering)

 

[caffeinemachine]

Should I prove that it doesn't hold that $\sqrt{2}\in \mathbb{Q}(\rho)\Rightarrow \rho=\sqrt{2}$ ? (Wondering)

Yes. Try showing that if $F(\alpha, \beta)$ is an algebraic extension of a field $F$, then it doesn't in general hold that $\beta\in F(\alpha)$ implies $\beta=\alpha$.

So, is there an other way to find the degree of the extension? (Wondering)

One way I can see how to show that $\sqrt{2}\notin \mathbf Q(\rho)$ is by using the artifice of trace of an extension. Are you familiar with it?

 

[mathmari]

One way I can see how to show that $\sqrt{2}\notin \mathbf Q(\rho)$ is by using the artifice of trace of an extension. Are you familiar with it?

Not really. Could you explain it further to me? (Wondering)

Is this the only way? (Wondering)

 

[caffeinemachine]

Not really. Could you explain it further to me? (Wondering)

Is this the only way? (Wondering)

Trace takes quite some time to develop. It can be found in any textbook on Galois theory. But since you are learning the basics of field theory as of now, that material will not be accessible to you yet.

 

[mathmari]

Trace takes quite some time to develop. It can be found in any textbook on Galois theory. But since you are learning the basics of field theory as of now, that material will not be accessible to you yet.

Is this the only way to show the desired result? (Wondering)

 

[caffeinemachine]

Is this the only way to show the desired result? (Wondering)

Given any mathematical result, I would never have the audacity to say that "this is the only way to show this". :P

 

[mathmari]

We can also write it as follows:
$[\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}[\sqrt{2}]][\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]$

$[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]=\deg \text{Irr}(x^2-2)$

We have that $x^4-2x^2-1=(x^4-2x^2+1)-2=(x^2-1)^2-2=(x^2-1+\sqrt{2})(x^2-1-\sqrt{2})$.

We have that $\rho$ is the root of $x^4-2x^2-1$. Does this mean that $\rho$ is a root of one of the factors $(x^2-1+\sqrt{2})$ or $(x^2-1-\sqrt{2})$ ?

If this is true then we have to show that this factor is irreducible over $\mathbb{Q}[\sqrt{2}]$, right?

Suppose that it is reducible then it must be the product of two polynomials of first degree, $(x-c_1), (x-c_2)$,
where $c_1, c_2\in \mathbb{Q}[\sqrt{2}]$.
Is this correct so far? (Wondering)

 

[caffeinemachine]

We can also write it as follows:
$[\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}[\sqrt{2}]][\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]$

$[\mathbb{Q}[\sqrt{2}]:\mathbb{Q}]=\deg \text{Irr}(x^2-2)$

We have that $x^4-2x^2-1=(x^4-2x^2+1)-2=(x^2-1)^2-2=(x^2-1+\sqrt{2})(x^2-1-\sqrt{2})$.

We have that $\rho$ is the root of $x^4-2x^2-1$. Does this mean that $\rho$ is a root of one of the factors $(x^2-1+\sqrt{2})$ or $(x^2-1-\sqrt{2})$ ?

Yes of course. If $\rho$ is a root of $f(x)g(x)$ then $\rho$ is a root of $f(x)$ or $g(x)$.

If this is true then we have to show that this factor is irreducible over $\mathbb{Q}[\sqrt{2}]$, right?

Suppose that it is reducible then it must be the product of two polynomials of first degree, $(x-c_1), (x-c_2)$,
where $c_1, c_2\in \mathbb{Q}[\sqrt{2}]$.
Is this correct so far? (Wondering)

This is correct so far. To show that $f(x):=x^2-1\pm\sqrt{2}$ is irreducible over $\mathbf Q(\sqrt{2})$, one just needs to show that $f(x)$ does not have a root in $\mathbf Q(\sqrt{2})$. I haven't tried doing this but this seems doable.

 

[mathmari]

To show that $f(x):=x^2-1\pm\sqrt{2}$ is irreducible over $\mathbf Q(\sqrt{2})$, one just needs to show that $f(x)$ does not have a root in $\mathbf Q(\sqrt{2})$. I haven't tried doing this but this seems doable.

Suppose that $x^2-1\pm\sqrt{2}$ is reducible over $\mathbf Q(\sqrt{2})$, then there are $c_1,c_2\in \mathbb{Q}(\sqrt{2})$, such that $x^2-1\pm\sqrt{2}=(x-c_1)(x-c_2)$.
Since $c_1, c_2\in \mathbb{Q}(\sqrt{2})$ we have that $c_1=a_1+b_1\sqrt{2}$ and $c_2=a_2+b_2\sqrt{2}$, for $a_1,a_2,b_1,b_2\in\mathbb{Q}$.
$$x^2-1\pm\sqrt{2}=(x-(a_1+b_1\sqrt{2}))(x-(a_2+b_2\sqrt{2}))\\ =x^2-(a_1+a_2+(b_1+b_2)\sqrt{2})x+(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})\\ =x^2-(a_1+a_2+(b_1+b_2)\sqrt{2})x+(a_1a_2+2b_1b_2+(a_1b_2+a_2b_1)\sqrt{2})$$

So, the following must hold:
$$a_1+a_2+(b_1+b_2)\sqrt{2}=0 \ \ \land \ \ a_1a_2+2b_1b_2+(a_1b_2+a_2b_1)\sqrt{2}=-1\pm \sqrt{2} \\ \Rightarrow \left ( a_1+a_2=0 \land b_1+b_2=0 \right ) \ \ \land \ \ \left (a_1a_2+2b_1b_2=-1 \land a_1b_2+a_2b_1=\pm 1\right ) \\ \Rightarrow a_1=-a_2 \land b_1=-b_2 \land \left (-a_2^2-2b_2^2=-1 \land -2a_2b_2=\pm 1\right ) \\ \Rightarrow a_1=-a_2 \land b_1=-b_2 \land \left (a_2^2+2b_2^2=1 \land a_2b_2=\mp \frac{1}{2}\right )$$

Is this correct so far? (Wondering)

How could we continue to get a contradiction? (Wondering)

 

[caffeinemachine]

Suppose that $x^2-1\pm\sqrt{2}$ is reducible over $\mathbf Q(\sqrt{2})$, then there are $c_1,c_2\in \mathbb{Q}(\sqrt{2})$, such that $x^2-1\pm\sqrt{2}=(x-c_1)(x-c_2)$.
Since $c_1, c_2\in \mathbb{Q}(\sqrt{2})$ we have that $c_1=a_1+b_1\sqrt{2}$ and $c_2=a_2+b_2\sqrt{2}$, for $a_1,a_2,b_1,b_2\in\mathbb{Q}$.
$$x^2-1\pm\sqrt{2}=(x-(a_1+b_1\sqrt{2}))(x-(a_2+b_2\sqrt{2}))\\ =x^2-(a_1+a_2+(b_1+b_2)\sqrt{2})x+(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})\\ =x^2-(a_1+a_2+(b_1+b_2)\sqrt{2})x+(a_1a_2+2b_1b_2+(a_1b_2+a_2b_1)\sqrt{2})$$

So, the following must hold:
$$a_1+a_2+(b_1+b_2)\sqrt{2}=0 \ \ \land \ \ a_1a_2+2b_1b_2+(a_1b_2+a_2b_1)\sqrt{2}=-1\pm \sqrt{2} \\ \Rightarrow \left ( a_1+a_2=0 \land b_1+b_2=0 \right ) \ \ \land \ \ \left (a_1a_2+2b_1b_2=-1 \land a_1b_2+a_2b_1=\pm 1\right ) \\ \Rightarrow a_1=-a_2 \land b_1=-b_2 \land \left (-a_2^2-2b_2^2=-1 \land -2a_2b_2=\pm 1\right ) \\ \Rightarrow a_1=-a_2 \land b_1=-b_2 \land \left (a_2^2+2b_2^2=1 \land a_2b_2=\mp \frac{1}{2}\right )$$

Is this correct so far? (Wondering)

How could we continue to get a contradiction? (Wondering)

This seems correct (thought I haven't checked the calculations carefully). What we want to show, in particular, is that $\sqrt{1+\sqrt{2}}$ is not in $\mathbf Q(\sqrt{2})$.

A simpler problem is to decide whether or not $\sqrt[4]{2}$ is in $\mathbf Q(\sqrt{2})$. The answer is no. This is because by Eisenstien $x^4-2$ is irreducible over $\mathbf Q$ and thus the minimal polynomial of $\sqrt[4]{2}$ over $\mathbf Q$ has degree $4$. If $\sqrt[4]{2}$ were in $\mathbf Q(\sqrt{2})$, then the degree of the minimal polynomial of $\sqrt[4]{2}$ over $\mathbf Q$ would be at most $2$.

I will get back to you on what to say about $\sqrt{1+\sqrt{2}}$. Give me some time (and remind me in a few days if you do not hear from me!).

 

[mathmari]

Could we maybe do it as follows?

We have that $\rho$ is a root of $x^4-2x^2-1$ and that \begin{align*}x^4-2x^2-1&=(x^4-2x^2+1)-2\\ & =(x^2-1)^2-2\\ &=(x^2-1+\sqrt{2})(x^2-1-\sqrt{2}).\end{align*}
So, $\rho$ is also a root of oe of the factors $(x^2-1+\sqrt{2})$ ή $(x^2-1-\sqrt{2})$.
We have that $\rho^2-1\pm \sqrt{2}=0 \Rightarrow \sqrt{2}=\pm(1-\rho^2)\in \mathbb{Q}[\rho]$.
Therefore, $\mathbb{Q}[\rho, \sqrt{2}]=\mathbb{Q}[\rho]$.
So, $ [\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}]=[\mathbb{Q}[\rho]:\mathbb{Q}]$.

We have that $\rho$ is algebraic over $\mathbb{Q}$. So, we we he that the is an unique monic irreducible polynomal in $\mathbb{Q}[x]$ $p\in \mathbb{Q}[x]$ such that $p(\rho)=0$. This irreducible polynomial is the minimal polynomial.
Since irreducible monic polynomials ith common roots are equal, we have that $f=p$, so $f$ is the minimal polynomial of $\rho$ over $\mathbb{Q}$.
From a theorem that I found in my notes, we have that basis of the extension is $1, \rho, \rho^2, \rho^3$ and $[\mathbb{Q}[\rho]:\mathbb{Q}]=4$.

(Wondering) If we had $\sqrt{3}$ instead of $\sqrt{2}$, wouldwe do the same? Would we have to show that $\sqrt{3}\notin \mathbb{Q}[\rho]$ ? But how? (Wondering)

 

[caffeinemachine]

Could we maybe do it as follows?

We have that $\rho$ is a root of $x^4-2x^2-1$ and that \begin{align*}x^4-2x^2-1&=(x^4-2x^2+1)-2\\ & =(x^2-1)^2-2\\ &=(x^2-1+\sqrt{2})(x^2-1-\sqrt{2}).\end{align*}
So, $\rho$ is also a root of oe of the factors $(x^2-1+\sqrt{2})$ ή $(x^2-1-\sqrt{2})$.
We have that $\rho^2-1\pm \sqrt{2}=0 \Rightarrow \sqrt{2}=\pm(1-\rho^2)\in \mathbb{Q}[\rho]$.
Therefore, $\mathbb{Q}[\rho, \sqrt{2}]=\mathbb{Q}[\rho]$.
So, $ [\mathbb{Q}[\sqrt{2},\rho]:\mathbb{Q}]=[\mathbb{Q}[\rho]:\mathbb{Q}]$.

We have that $\rho$ is algebraic over $\mathbb{Q}$. So, we we he that the is an unique monic irreducible polynomal in $\mathbb{Q}[x]$ $p\in \mathbb{Q}[x]$ such that $p(\rho)=0$. This irreducible polynomial is the minimal polynomial.
Since irreducible monic polynomials ith common roots are equal, we have that $f=p$, so $f$ is the minimal polynomial of $\rho$ over $\mathbb{Q}$.
From a theorem that I found in my notes, we have that basis of the extension is $1, \rho, \rho^2, \rho^3$ and $[\mathbb{Q}[\rho]:\mathbb{Q}]=4$.

(Wondering)

Nice. This shows that $\sqrt{2}$ is in $\mathbf Q(\rho)$. Good job.

If we had $\sqrt{3}$ instead of $\sqrt{2}$, wouldwe do the same? Would we have to show that $\sqrt{3}\notin \mathbb{Q}[\rho]$ ? But how? (Wondering)

Again, the only general way I know how to prove a statement like this is to use trace.

 

[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? (Wondering)

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? (Wondering)

Or is this way not correct? (Wondering)

 

[caffeinemachine]

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? (Wondering) 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? (Wondering)

Or is this way not correct? (Wondering)

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 learned more theory.

 

[mathmari]

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?
$$q_1\tilde{q_1}+3q_2\tilde{q_2}=-1 \\ q_1\tilde{q_2}+\tilde{q_1}q_2=0$$
(Wondering)

 

[caffeinemachine]

... 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 $(q_1+q_2\sqrt{3})^2$?

 

[mathmari]

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 $f$ ? (Wondering)
- $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? (Wondering)