Algebraic Extensions exercises

[Genady]

Show that if f(x) is an irreducible polynomial over K and K ⊂ L such that the degree of L over K and the degree of f(x) are relatively prime, then f(x) is irreducible over L.

Let x₀ be a root of f(x) and n be a degree of f(x). Since f(x) is irreducible over K, the degree of K(x₀) over K is n.

Let r be the degree of L over K. We know that n and r are relatively prime.

Let's assume that f(x) is reducible over L. Then x₀ is root of one of its factorizing polynomials. Call its degree, m, where m < n. Since this polynomial is irreducible over L, the degree of L(x₀) over L is m.

Since L includes K, L(x₀) includes K(x₀). Call the degree of L(x₀) over K(x₀), t. We have the degree of L(x₀) over K equals nt, because of the inclusions K ⊂ K(x₀) ⊂ L(x₀), and equals rm, because of the inclusions K ⊂ L ⊂ L(x₀).

So, nt = rm. As n and r are relatively prime, n | m. But this contradicts that m < n. Thus, f(x) is irreducible over L.

QED.

Any doubts? 🙂

[Genady]

Write down a multiplication table for the elements of the finite field extension F₂(α) where α²+α+1 = 0.

All elements of the field F₂(α) have a form a+bα, where a and b are 0 or 1. Thus, the elements are, 0, 1, α, and 1+α. The multiplication table is:

0 x anything = 0

1 x anything = the same anything

α x α = 1+α

α x (1+α) = 1

(1+α) x (1+α) = α

The End.

 

[Genady]

Show that a field L with q = pⁿ elements contains a field K with r = p^m elements if and only if m | n.

 

A field with pⁿ elements has degree n over the prime field F_p, [L : F_p] = n. 

A field with p^m elements has degree m over the prime field F_p, [K : F_p] = m. 

If L contains K, then

[L : F_p] = [L : K] [K : F_p], i.e., n = [L : K] m.

Thus, m | n.

 

L consists of all roots of polynomial x^q-x. K consists of all roots of polynomial x^r-x. It can be proved by induction than if m | n, then all roots of the later are also roots of the former*. Thus, L contains K.

 

QED

 

* For example, if n = 2m then x^q = (x^r)^r.

[Genady]

The following are my understandings related to the exercise in the previous post.

A field of pⁿ elements consists of roots of the polynomial x^q-x, where q=pⁿ. These are combined roots of all irreducible polynomials over field F_p which divide x^q-x. These are irreducible polynomials of degrees m such that m | n. Each of them contributes m roots to the field.

The number of elements in the field, pⁿ, is sum of products m·N_p(m), where N_p(m) is number of irreducible polynomials of degree m over field F_p. This sum is for all m which divide n.

Edited Thursday at 09:04 AM by Genady

[Genady]

The last formula in the previous post can be used to calculate the numbers of irreducible polynomials without listing them.

For example, for p = 2 we get:

degree 1 polynomials, N₂(1) = 2¹/1 = 2,

degree 2 polynomials, N₂(2) = (2² - 1xN₂(1))/2 = 1,

degree 3 polynomials, N₂(3) = (2³ - 1xN₂(1))/3 = 2,

degree 4 polynomials, N₂(4) = (2⁴ - 1xN₂(1) - 2xN₂(2))/4 = 3,

degree 5 polynomials, N₂(5) = (2⁵ - 1xN₂(1))/5 = 6.

 

These numbers can be compared with the finding in this earlier exercise:

 

Edited Thursday at 01:36 PM by Genady

[Genady]

Let S ⊆ M ⊆ L be field extensions.

1. Let S ⊆ M and M ⊆ L be normal extensions. Is S ⊆ L normal?

Not necessarily. Here is an example of it being "abnormal." 
Let S = Q, M = Q(2^1/2), L = Q(2^1/4). 
Polynomial x²-2 is irreducible over S and has both of its zeroes in M. So, S ⊆ M is normal.
Polynomial x²-2^1/2 is irreducible over M and has both of its zeroes in L. So, M ⊆ L is normal.
However, polynomial x⁴-2 is irreducible over S but has only two of its four zeroes in L. Thus, S ⊆ L is not normal.

2. Let S ⊆ L be normal. Is M ⊆ L normal?

Yes. Any polynomial irreducible over M is irreducible over S. If this polynomial has one zero in L it has all zeroes in L, because S ⊆ L is normal. Since it has all its zeros in L, M ⊆ L is normal, too.

3. Let S ⊆ L be normal. Is S ⊆ M normal?

Not necessarily. 
Let S = Q, M = Q(2^1/4), L = Q(2^1/4, i). 
Polynomial x⁴-2 is irreducible over S and has all its zeros in L. So, S ⊆ L is normal. However, it has only two of its four zeroes in M. Thus, S ⊆ M is not normal.

 

[Genady]

Is the extension K ⊆ L normal, where K = Q, L = Q(2^1/3)?

2^1/3 is zero of polynomial x³-2, which is irreducible in Q. Other two zeroes of this polynomial are complex and thus ∉ L. Hence, this extension is not normal.

Is such extension normal for K = Q(2^1/2), L = Q(2^1/4)?

2^1/4 is zero of polynomial x⁴-2, which is reducible in K, i.e., x⁴-2 = (x² - 2^1/2)(x² + 2^1/2). The first factor has both zeroes in L, while the second does not have any zeroes in L. Hence, this extension is normal.

[Genady]

Find all subgroups of the group of automorphisms G(L/K) and all corresponding subfields between the fields K and L, where K = Q and L is splitting field of the polynomial f(x) = (x²-2)(x²-5).

Each factor of f(x) above is irreducible in K. The first factor has zeroes ±2^1/2, the second has zeroes ±5^1/2. Thus L = Q(2^1/2, 5^1/2). The order of the group G(L/K) = [L : K] = 4, and it has four automorphisms:

s₀: 2^1/2 → 2^1/2, 5^1/2 → 5^1/2

s₁: 2^1/2 → 2^1/2, 5^1/2 → -5^1/2

s₂: 2^1/2 → -2^1/2, 5^1/2 → 5^1/2

s₃: 2^1/2 → -2^1/2, 5^1/2 → -5^1/2.

The subgroups are: H₀ = {s₀}, H₁ = {s₀, s₁}, H₂ = {s₀, s₂}, H₃ = {s₀, s₃}.

The subfield corresponding to H₀ fixes all elements in L and thus, it is Q(2^1/2, 5^1/2).   

The subfield corresponding to H₁ fixes the element 2^1/2 and thus, it is Q(2^1/2).   

The subfield corresponding to H₂ fixes the element 5^1/2 and thus, it is Q(5^1/2).

The subfield corresponding to H₃ fixes the element (2^1/2×5^1/2) and thus, it is Q(10^1/2).

[Genady]

Apparently, an extension K ⊆ L being normal simply means that every irreducible polynomial over K has either all or none of its zeroes in L.