MHB The polynomial is irreducible over Q(i)

[mathmari]

Hey!

I want to show that the polynomial $x^4-2\in \mathbb{Q}[x]$ remains irreducible in the ring $\mathbb{Q}(i)[x]$.

I have done the following:

The polynomial is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion with $p=2$.

Then if $a$ is a root of $x^4-2$ then the degree of the extension $\mathbb{Q}(a)$ over the field $\mathbb{Q}$ is $4$ : $[\mathbb{Q}(a):\mathbb{Q}]=4$.

We assume that the polynomial $x^4-2$ is reducible over $\mathbb{Q}(i)$.

Is everything correct so far? If yes, how could we continue to get a contradiction? Do we have to check all possible factorizations of $x^4-2$ ?

(Wondering)

 

[I like Serena]

Hey!

I want to show that the polynomial $x^4-2\in \mathbb{Q}[x]$ remains irreducible in the ring $\mathbb{Q}(i)[x]$.

I have done the following:

The polynomial is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion with $p=2$.

Then if $a$ is a root of $x^4-2$ then the degree of the extension $\mathbb{Q}(a)$ over te field $\mathbb{Q}$ is $4$ : $[\mathbb{Q}(a):\mathbb{Q}]=4$.

We assume that the polynomial $x^4-2$ is reducible over $\mathbb{Q}(i)$.

Is everything correct so far? If yes, how could we continue to get a contradiction? Do we have to check all possible factorizations of $x^4-2$ ?

(Wondering)

Hey mathmari!

Do you have a link to some course material that relates to this problem? (Wondering)

As a possible approach, consider that $\mathbb Q(i)$ is a subfield of $\mathbb C$. So any reduction over $\mathbb Q(i)$ must also be a reduction over $\mathbb C$, mustn't it?
So if we list the reductions over $\mathbb C$, we can check if there is a rational one. (Thinking)

 

[mathmari]

Do you have a link to some course material that relates to this problem? (Wondering)

No, I don't have. (Thinking)

As a possible approach, consider that $\mathbb Q(i)$ is a subfield of $\mathbb C$.

What other approaches are there? (Wondering)

So any reduction over $\mathbb Q(i)$ must also be a reduction over $\mathbb C$, mustn't it?
So if we list the reductions over $\mathbb C$, we can check if there is a rational one. (Thinking)

Do you mean that we have to check the factorization of the polynomial in $\mathbb{C}$ ? (Wondering)

 

[I like Serena]

No, I don't have. (Thinking)

What other approaches are there? (Wondering)

I don't know. I'm not really familiar with field extensions and splitting fields.

Do you mean that we have to check the factorization of the polynomial in $\mathbb{C}$ ? (Wondering)

(Nod)

 

[mathmari]

(Nod)

We have the following: $$x^4-2=(x^2-\sqrt{2})(x^2+\sqrt{2})=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$$ or not?

How could we continue from here? I got stuck right now. (Wondering)

 

[I like Serena]

We have the following: $$x^4-2=(x^2-\sqrt{2})(x^2+\sqrt{2})=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$$ or not?

How could we continue from here? I got stuck right now. (Wondering)

As you said, we assume that the polynomial $x^4-2$ is reducible over $\mathbb Q(i)$.

Then we must be able to factorize it into $(x-a)p(x)$ or $(x^2+bx+c)q(x)$ with coefficients in $\mathbb Q(i)$.
And since $\mathbb Q(i)$ is a subfield of $\mathbb C$, the factorization must also hold in $\mathbb C[x]$. Furthermore, polynomial factorization over a field is unique up to multiplication with constants.
Is any of the possibilities that you found for $a$ in $\mathbb Q(i)$? (Wondering)

 

[mathmari]

As you said, we assume that the polynomial $x^4-2$ is reducible over $\mathbb Q(i)$.

Then we must be able to factorize it into $(x-a)p(x)$ or $(x^2+bx+c)q(x)$ with coefficients in $\mathbb Q(i)$.
And since $\mathbb Q(i)$ is a subfield of $\mathbb C$, the factorization must also hold in $\mathbb C[x]$. Furthermore, polynomial factorization over a field is unique up to multiplication with constants.
Is any of the possibilities that you found for $a$ in $\mathbb Q(i)$? (Wondering)

I got stuck right now. I haven't really understood the part with $\mathbb C[x]$. Do we check the factorization of $x^4-2$ in $\mathbb C[x]$ and then we check if one of these is also in $\mathbb{Q}(i)[x]$ ? (Wondering)

 

[I like Serena]

I got stuck right now. I haven't really understood the part with $\mathbb C[x]$. Do we check the factorization of $x^4-2$ in $\mathbb C[x]$ and then we check if one of these is also in $\mathbb{Q}(i)[x]$ ? (Wondering)

Okay, let's take this step by step...

As you said, we assume that the polynomial $x^4-2$ is reducible over $\mathbb Q(i)$.

Then we must be able to factorize it into $(x-a)p(x)$ or $(x^2+bx+c)q(x)$ with coefficients in $\mathbb Q(i)$.

This is true, isn't it? (Wondering)

And since $\mathbb Q(i)$ is a subfield of $\mathbb C$, the factorization must also hold in $\mathbb C[x]$.

If the coefficients are in $\mathbb Q(i)$, then those coefficients are also in $\mathbb C$, aren't they?

Furthermore, polynomial factorization over a field is unique up to multiplication with constants.

This is also true, isn't it? (Wondering)

 

[mathmari]

Okay, let's take this step by step...This is true, isn't it? (Wondering)If the coefficients are in $\mathbb Q(i)$, then those coefficients are also in $\mathbb C$, aren't they?This is also true, isn't it? (Wondering)

I think I got that now.

So, we are trying to factorize the polynomial into $(x-a)p(x)$ or $(x^2+bx+c)q(x)$ with coefficients in $\mathbb{C}$ and then these coefficients are also in $\mathbb{Q}(i)$ but then we get a contradiction and so it follows that $x^4-2$ is irreducible in $\mathbb{Q}(i)$.

Have I understood that correctly? (Wondering)

 

[I like Serena]

I think I got that now.

So, we are trying to factorize the polynomial into $(x-a)p(x)$ or $(x^2+bx+c)q(x)$ with coefficients in $\mathbb{C}$ and then these coefficients are also in $\mathbb{Q}(i)$ but then we get a contradiction and so it follows that $x^4-2$ is irreducible in $\mathbb{Q}(i)$.

Have I understood that correctly?

Yes. (Nod)

None of the $a$ that you found for $\mathbb C$ is in $\mathbb Q(i)$.
It could still be that there is a reduction over $\mathbb Q(i)$ of the form $(x^2+bx+c)q(x)$ though.
So we also need to check if e.g $(x-\sqrt[4]2)(x+\sqrt[4]2)= (x^2-\sqrt 2)$ has coefficients in $\mathbb Q(i)$. (Thinking)

 

[mathmari]

Yes. (Nod)

None of the $a$ that you found for $\mathbb C$ is in $\mathbb Q(i)$.
It could still be that there is a reduction over $\mathbb Q(i)$ of the form $(x^2+bx+c)q(x)$ though.
So we also need to check if e.g $(x-\sqrt[4]2)(x+\sqrt[4]2)= (x^2-\sqrt 2)$ has coefficients in $\mathbb Q(i)$. (Thinking)

So, we assume that $x^4-2$ is reducible in $\mathbb{Q}(i)[x]$, so also in $\mathbb{C}[x]$.
We consider first the factorizations over $\mathbb{C}$.
We can factorize the polynomial either as $(x^2-\sqrt{2})(x^2+\sqrt{2})$ or as $(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$, where the coefficients are in $\mathbb{C}$.
But none of the coefficients is in $\mathbb{Q}(i)$.
That means that there is no factorization in $\mathbb{Q}(i)[x]$, and so the polynomial is irreducible in $\mathbb{Q}(i)[x]$.

Is everything correct and complete? Could I improve something? (Wondering)

 

[I like Serena]

So, we assume that $x^4-2$ is reducible in $\mathbb{Q}(i)[x]$, so also in $\mathbb{C}[x]$.
We consider first the factorizations over $\mathbb{C}$.
We can factorize the polynomial either as $(x^2-\sqrt{2})(x^2+\sqrt{2})$ or as $(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$, where the coefficients are in $\mathbb{C}$.
But none of the coefficients is in $\mathbb{Q}(i)$.
That means that there is no factorization in $\mathbb{Q}(i)[x]$, and so the polynomial is irreducible in $\mathbb{Q}(i)[x]$.

Is everything correct and complete? Could I improve something? (Wondering)

It could also be that the expanded forms of $(x-\sqrt[4]{2})(x-i\sqrt[4]{2})$ or $(x-\sqrt[4]{2})(x+i\sqrt[4]{2})$ have coefficients in $\mathbb Q(i)$, which they don't.
We can skip the remaining combinations, since if there was a suitable one, then one of the ones we've checked must also be suitable. (Nerd)

 

[mathmari]

It could also be that the expanded forms of $(x-\sqrt[4]{2})(x-i\sqrt[4]{2})$ or $(x-\sqrt[4]{2})(x+i\sqrt[4]{2})$ have coefficients in $\mathbb Q(i)$, which they don't.
We can skip the remaining combinations, since if there was a suitable one, then one of the ones we've checked must also be suitable. (Nerd)

Do you mean the following?

We assume that $x^4-2$ is reducible in $\mathbb{Q}(i)[x]$ and so it is in $\mathbb{C}[x]$.
We consider first the factorizations over $\mathbb{C}$.
We can write the polynomial either as a product of linear factors: $$(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$$ or as a product with a quadratic polynomial: $$[x^2-\sqrt{2}]p(x) \ \text{ or } \ [x^2-(1+i)\sqrt[4]{2}x+i\sqrt{2}]p(x) \ \text{ or } \ [x^2-(1-i)\sqrt[4]{2}x-i\sqrt{2}]p(x)$$ where the coefficients are in $\mathbb{C}$.
But none of these coefficients is in $\mathbb{Q}(i)$, since $\sqrt{2}\notin \mathbb{Q}(i)$, $\sqrt[4]{2}\notin \mathbb{Q}(i)$.
This means that there is no factorization in $\mathbb{Q}(i)[x]$, and so the polynomial $x^4-2$ is irreducible in $\mathbb{Q}(i)[x]$.

Is everything correct? (Wondering)

 

[I like Serena]

Yep. (Nod)

 

[mathmari]

Yep. (Nod)

Great! At the initial post, do we need the part:

The polynomial is irreducible in $\mathbb{Q}[x]$ by Eisenstein's criterion with $p=2$.

Then if $a$ is a root of $x^4-2$ then the degree of the extension $\mathbb{Q}(a)$ over the field $\mathbb{Q}$ is $4$ : $[\mathbb{Q}(a):\mathbb{Q}]=4$.

? (Wondering)

 

[I like Serena]

Nope. (Shake)

 

[mathmari]

Nope. (Shake)

Ok! Thank you! (Mmm)