Odd extension: Show that if [K(a) : K] is odd, then K(a) = K(a^2)

[Hill]

Homework Statement

Show that if [K(a) : K] is odd, then K(a) = K(a^2)

Relevant Equations

[M:K]=[M:L][L:K]

My solution:
1. $K(a^2) \subseteq K(a)$.
2. $a$ is zero of the quadratic polynomial $X^2 - (a^2)$, i.e., $[K(a) : K(a^2)] \leq 2$.
3. It is not 2 because $[K(a) : K] = [K(a) : K(a^2)][K(a^2) : K]$ is odd.
4. Thus, it is 1, and hence $K(a) = K(a^2)$.

Is this solid enough?

 

[fresh_42]

Homework Statement: Show that if [K(a) : K] is odd, then K(a) = K(a^2)

1. $K(a^2) \subseteq K(a)$.
2. $a$ is zero of the quadratic polynomial $X^2 - (a^2)$, i.e., $[K(a) : K(a^2)] \leq 2$.
3. It is not 2 because $[K(a) : K] = [K(a) : K(a^2)][K(a^2) : K]$ is odd.
4. Thus, it is 1, and hence $K(a) = K(a^2)$.

Looks ok to me. I would only add that $X^2-(a^2)\in K(a^2)[X]$ since it is important where the coefficients are from.

 

[Hill]

Looks ok to me. I would only add that $X^2-(a^2)\in K(a^2)[X]$ since it is important where the coefficients are from.

Of course. Thank you.

Now I struggle with finding a counterexample to the statement: if $K(a) = K(a^2)$, then $[K(a) : K]$ is odd.

 

[fresh_42]

Of course. Thank you.

Now I struggle with finding a counterexample to the statement: if $K(a) = K(a^2)$, then $[K(a) : K]$ is odd.

How about $\mathbb{Z}_2[X] / \bigl\langle X^2+X+1 \bigr\rangle$? We have $a^2=a+1$ so $K(a)=K(a^2)$ and the degree is two.

 

[Hill]

the degree is two

Is the degree what we are looking for? We need a dimension of $K(a)$ over $K$ to be even, but I don't even know how to find this dimension for $K=\mathbb{Z}_2[X] / \bigl\langle X^2+X+1 \bigr\rangle$.

 

[fresh_42]

Is the degree what we are looking for? We need a dimension of $K(a)$ over $K$ to be even, but I don't even know how to find this dimension for $K=\mathbb{Z}_2[X] / \bigl\langle X^2+X+1 \bigr\rangle$.

The polynomial is irreducible and prime, so $K$ is a field. $K$ is a field with four elements, i.e. a two-dimensional vector space over $\mathbb{Z}_2$ and so of degree two. It cannot be one since $X^2+X+1$ is irreducible. That the degree is therefore two is your own argument from post #1. What else could it be?

https://de.wikiversity.org/wiki/Endlicher_Körper/4/Operationstafeln

 

[mathwonk]

well, [C:R} = 2, and I bet you can find z such that R(z) = R(z^2) = C. (C = R(i), R = reals, i^2 = -1.). or adjoin a new cube root w of 1 to Q. ohh. this is fresh's example, since x^3-1 = (x-1)(x^2+x+1). i.e. w^3 = 1 implies w = (w^2)^2, so Q(w) = Q(w^2).

 

[fresh_42]

You can also replace $\mathbb{Z}_2$ by $\mathbb{Q}$ and it still works since $a^2=-a-1.$ You can calculate the roots of the polynomial as $\dfrac{1}{2}\left(-1\pm i \sqrt{3}\right)$ and work it out explicitly. And if you want higher even degrees, consider $X^{2n}+X+1.$