# Thread: Eisenstein polynomial and field extension

Originally Posted by caffeinemachine
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?

Is this the only way?

2. Originally Posted by mathmari
Not really. Could you explain it further to me?

Is this the only way?
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.

Originally Posted by caffeinemachine
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?

4. Originally Posted by mathmari
Is this the only way to show the desired result?
Given any mathematical result, I would never have the audacity to say that "this is the only way to show this". :P

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?

6. Originally Posted by 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 $(x^2-1-\sqrt{2})$ ?
Yes of course. If $\rho$ is a root of $f(x)g(x)$ then $\rho$ is a root of $f(x)$ or $g(x)$.

Originally Posted by mathmari
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?
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.

Originally Posted by caffeinemachine
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?

How could we continue to get a contradiction?

8. Originally Posted by mathmari
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?

How could we continue to get a contradiction?
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!).

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 $[\mathbb{Q}[\rho]:\mathbb{Q}]=4$.

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?

10. Originally Posted by 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 $[\mathbb{Q}[\rho]:\mathbb{Q}]=4$.

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

Originally Posted by mathmari
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?
Again, the only general way I know how to prove a statement like this is to use trace.