Field extension-degree of the minimal polynom

[mathmari]

Hey!

Let $K\leq E$ a field extension and $a \in E$ an algebraic element over $K$. We suppose that $Irr(a,K)$ has an odd degree. Show that $K(a)=K(a^2)$. Does this also hold when the degree is even??

I have done the following:

We have that $K\leq K(a^2)\leq K(a)$.

$f(x)=x^2-a^2 \in K(a^2)[x]$ and $f(a)=0$

$Irr(a,K(a^2)) \mid x^2-a^2$, so $degIrr(a,K(a^2))=1 \text{ or } 2$

Since $[K(a):K(a^2)][K(a^2):K(a)]=[K(a):K]=\text{ odd }$ it should be $[K(a):K(a^2)]=degIrr(a,K(a^2))=\text{ odd }$.

Therefore, $degIrr(a,K(a^2))=[K(a):K(a^2)]=1$.

Is this correct?? Do we conclude from that, that $K(a)=K(a^2)$??

What can I say about the case where the degree is even??

 

[mathbalarka]

Is this correct?? Do we conclude from that, that $K(a)=K(a^2)$??

It looks good to me! Yes, it is true that if $[E : F] = 1$ for extension $E/F$, then $E = F$. This follows immediately from the notion of degree of extensions : $[E : F]$ can be thought of as the cardinality of a basis by which $E$ is spanned over $F$. If $[E : F] = 1$, the basis is a singleton set. But then $E \supseteq F$ so $E$ is naturally spanned over $F$ by $\{1\}$ (plus some more stuff). But since the basis is singleton, $E$ is spanned over $F$ by $\{1\}$ only, forcing $E = F$.

What can I say about the case where the degree is even??

This doesn't hold if $\text{min}_K(a)$ is even. For example, take $K = \Bbb Q$, and $a = \sqrt{2}$. Then $K(a^2) = \Bbb Q(\sqrt{2}^2) = \Bbb Q$ which is a strict subset of $\Bbb Q(\sqrt{2}) = K(a)$. If $a$ has an even degree, you can only say that $[K(a^2) : K(a)] = 2$.