Irreducible polynomial of ζ_6, ζ_8, ζ_9 over the field Q(ζ_3).

In summary: Q}$. Thus, the irreducible polynomial of $\zeta_6$ over $\mathbb{Q}$ is also $x^2 - x + 1$. In summary, there are multiple methods that can be used to find the irreducible polynomial of $\zeta_6, \zeta_8, \zeta_9$ over the field $\mathbb{Q}(\zeta_3)$. One approach is to construct field extensions and use the degrees of the extensions, while another is to use the fact that $\zeta_n$ is a primitive $n$th root of unity and construct
  • #1
kalish1
99
0
Last edited:
Physics news on Phys.org
  • #2


Hello,

Thank you for your question. Finding the irreducible polynomial of $\zeta_6, \zeta_8, \zeta_9$ over the field $\mathbb{Q}(\zeta_3)$ can be approached in a few different ways. One possible method is indeed to construct field extensions and use the degrees of the extensions to determine the irreducible polynomial. However, there are also other methods that can be used.

One approach is to use the fact that $\zeta_n$ is a primitive $n$th root of unity, meaning that it is a complex number that, when raised to the $n$th power, gives the result of 1. Using this property, we can write $\zeta_6 = e^{2\pi i/6} = \cos(\pi/3) + i\sin(\pi/3) = 1/2 + i\sqrt{3}/2$. Similarly, $\zeta_8 = e^{2\pi i/8} = \cos(\pi/4) + i\sin(\pi/4) = \sqrt{2}/2 + i\sqrt{2}/2$, and $\zeta_9 = e^{2\pi i/9} = \cos(2\pi/9) + i\sin(2\pi/9)$.

Using these expressions, we can then construct a polynomial of degree 6 that has $\zeta_6$ as a root, a polynomial of degree 8 that has $\zeta_8$ as a root, and a polynomial of degree 9 that has $\zeta_9$ as a root. These polynomials will necessarily be irreducible, as they will have the minimal degree needed to have these roots.

Another approach is to use the fact that the minimal polynomial of $\zeta_n$ over $\mathbb{Q}$ is the cyclotomic polynomial $\Phi_n(x) = \prod_{d|n} (x^d - 1)^{\mu(n/d)}$, where $\mu$ is the Möbius function. This means that we can find the irreducible polynomial of $\zeta_n$ over $\mathbb{Q}$ by finding the factors of $\Phi_n(x)$ that are irreducible over $\mathbb{Q}$. For example, the cyclotomic polynomial for $n=6$ is $\Phi_
 

FAQ: Irreducible polynomial of ζ_6, ζ_8, ζ_9 over the field Q(ζ_3).

What is an irreducible polynomial?

An irreducible polynomial is a polynomial that cannot be factored into a product of two polynomials with lower degree. In other words, it cannot be broken down into simpler components.

What does ζ_6, ζ_8, and ζ_9 represent in this context?

In this context, ζ_6, ζ_8, and ζ_9 represent primitive complex roots of unity. These are complex numbers that, when raised to a certain power, equal 1. In this case, ζ_6, ζ_8, and ζ_9 are the primitive complex roots of unity for the fields Q(ζ_3), Q(ζ_3), and Q(ζ_3) respectively.

What is the significance of Q(ζ_3) in this equation?

Q(ζ_3) is the field extension of the rational numbers (Q) by the primitive complex root of unity ζ_3. This means that all elements of Q(ζ_3) can be expressed as a combination of rational numbers and powers of ζ_3.

How do you determine the irreducible polynomial of ζ_6, ζ_8, and ζ_9 over Q(ζ_3)?

To determine the irreducible polynomial of ζ_6, ζ_8, and ζ_9 over Q(ζ_3), we need to find the minimal polynomial for each primitive complex root of unity. This can be done by finding the smallest degree polynomial that has the root as a solution and cannot be factored further.

What is the importance of finding the irreducible polynomial in this equation?

Knowing the irreducible polynomial of ζ_6, ζ_8, and ζ_9 over Q(ζ_3) is important because it helps us understand the structure of the field extension Q(ζ_3). It also allows us to simplify calculations and solve equations involving these primitive complex roots of unity.

Similar threads

Replies
16
Views
3K
Replies
1
Views
1K
Replies
9
Views
1K
Replies
1
Views
1K
Replies
24
Views
4K
Replies
1
Views
1K
Back
Top