# The degree of finite field extensions.

I am revising for my algebraic number theory exam and was hoping I could get on some help on the following two questions:

a) Let $\alpha$ be algebraic over a field $K$ of odd degree. Show that $K(\alpha) = K(\alpha^2)$

b) Let $L = K(\alpha, \beta)$, with deg$_K(\alpha) = m, \,$ deg$_K(\beta) = n, \,$ and gcd$(m,n) = 1$. Show that $[L : K] = mn.$

I really don't know where to start with either of these problems.

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First problem has been answered many times in this group. Note that $\alpha^2\in K(\alpha)$, and $[K(\alpha):K(\alpha^2)]\leq 2$, since $\alpha$ satisfies the polynomial $x^2-\alpha^2\in K(\alpha^2)[x]$. Now use Dedekind's Product Theorem: $$[K(\alpha):K] = [K(\alpha):K(\alpha^2)][K(\alpha^2):K].$$
Second problem is a similar application of Dedekind's Product Theorem: $$[L:K] = [K(\alpha,\beta):K(\beta)][K(\beta):K]$$ and $$[L:K] = [K(\alpha,\beta):K(\alpha)][K(\alpha):K].$$ In particular, $n$ and $m$ divide $[L:K]$, so $\gcd(m,n)=1$ implies $mn|[L:K]$. Now show that the degree is at most $mn$ and you'll be done.