# The polynomial is irreducible over Q(i)

• MHB
• mathmari
In summary: So we can conclude that the polynomial $x^4-2$ is reducible over $\mathbb Q(i)$. (Wondering)Okay, let's take this step by step...As you said, we assume that the polynomial $x^4-2$ is reducible over $\mathbb Q(i)$.We can 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 sub
mathmari
Gold Member
MHB
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)

Last edited by a moderator:
mathmari said:
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)

I like Serena said:
Do you have a link to some course material that relates to this problem? (Wondering)

No, I don't have. (Thinking)
I like Serena said:
As a possible approach, consider that $\mathbb Q(i)$ is a subfield of $\mathbb C$.

What other approaches are there? (Wondering)
I like Serena said:
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)

mathmari said:
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.

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

(Nod)

I like Serena said:
(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)

mathmari said:
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)

Klaas van Aarsen said:
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)

mathmari said:
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...

Klaas van Aarsen said:
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)

Klaas van Aarsen said:
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?

Klaas van Aarsen said:
Furthermore, polynomial factorization over a field is unique up to multiplication with constants.

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

Klaas van Aarsen said:
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)

mathmari said:
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)

Klaas van Aarsen said:
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)

mathmari said:
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)

Klaas van Aarsen said:
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)

Yep. (Nod)

Klaas van Aarsen said:
Yep. (Nod)

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

Nope. (Shake)

Klaas van Aarsen said:
Nope. (Shake)

Ok! Thank you! (Mmm)

## 1. What does it mean for a polynomial to be irreducible over Q(i)?

When a polynomial is irreducible over Q(i), it means that the polynomial cannot be factored into two or more polynomials with coefficients in Q(i), where Q(i) represents the field of complex numbers with rational coefficients.

## 2. What is the significance of a polynomial being irreducible over Q(i)?

A polynomial being irreducible over Q(i) indicates that it cannot be simplified any further and is considered a "building block" in the field of complex numbers. It also means that the polynomial has no roots in Q(i), making it a fundamental element in the study of algebraic structures.

## 3. How can one determine if a polynomial is irreducible over Q(i)?

There are various methods for determining if a polynomial is irreducible over Q(i), including using the rational root theorem, checking for linear factors, and using the Eisenstein criterion. A combination of these methods can be used to determine if a polynomial is irreducible.

## 4. Can a polynomial be irreducible over Q(i) but reducible over another field?

Yes, a polynomial can be irreducible over Q(i) but reducible over another field. This is because different fields have different properties and structures, and a polynomial may have different factors in each field.

## 5. Are there any real-life applications of studying irreducible polynomials over Q(i)?

Yes, there are many real-life applications of studying irreducible polynomials over Q(i), such as in cryptography, coding theory, and signal processing. These fields require a deep understanding of algebraic structures, and irreducible polynomials provide a crucial foundation for their applications.

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