Field extensions of odd degree

Question

Problem: Let $L$ an extension of $K$ with $[L : K]$ odd. Show that $K(\alpha)=K\left(\alpha^{2}\right)$ for all $\alpha \in L \backslash K$

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Solution:Let $L$ an extension of $K$ with $[L : K]$ odd. Let $\alpha \in L \backslash K .$ The inclusions

$$ K \subset K\left(\alpha^{2}\right) \subset K(\alpha) \subset L $$ show that the degrees of each extension is odd by the formula of multiplicity of degrees. Let's look at $K\left(\alpha^{2}\right) \subset K(\alpha)$. The element $\alpha$ satisfies the quadratic equation $\alpha^{2}=\alpha^{2}$, thus $\left[K(\alpha) : K\left(\alpha^{2}\right)\right]$ because it can not be 2. Thus $K(\alpha)=K\left(\alpha^{2}\right)$.

I would like to understand the sentence in bold. If I understood correctly, the quadratic polynomial $t^2-\alpha ^2 \in K(\alpha ^2)[t]$ is satisfied only by $\alpha$. The thing I do not understand, shouldn't it implies that $[K(\alpha):K\left(\alpha^{2}\right)] \geq 2?$ He seems to say that $[K(\alpha):K\left(\alpha^{2}\right)] \leq 2$.

Answer 1

The extension $K(\alpha)/K(\alpha^2)$ has degree at most two, since it's obtained by adjoining $\alpha$, a root of a quadratic equation, to $K(\alpha^2)$. The quadratic is $X^2-\alpha^2=0$. But its degree is odd, so is one, that is $K(\alpha)=K(\alpha^2)$.

Answer 2

We have $[K(\alpha) :K(\alpha ^2)]=1$, since it is less than or equal to two ($\alpha$ satisfies $t^2-\alpha^2$), and it cannot be two since then $[L:K]=[L:K(\alpha)][K(\alpha),K(\alpha^2)][K(\alpha^2):K]=2\cdot [L:K(\alpha)][K(\alpha^2):K]$ is even.