Field extension exercise

Question

Let $E/F$ be a field extension and $a \in E$ ,$a$ algebraic over $F$. Prove that if the degree of the minimal polynomia of $a$ is an odd number then $F(a)=F(a^2)$.

The first step is to consider the extensions $F \leqslant F(a^2) \leqslant F(a)$.

We need need to show that $[F(a):F(a^2)]=1$

If $[F(a):F(a^2)]=m>1$ then $m<n$ and $m=2k_0+1$ where $n=deg(irr(a,F))$($m$ cannot be an even number)and $[m/2]>k_0>1$.

So we have that $[F(a^2):F]=n/m=s$ for some $s<n$ is an odd number.

From the above we see that $deg(irr(a^2,F))=n/m=s$.

From this point i don't know how to proceed to derive a contradiction.

Can someone help me with this?

Answer 1

The theorem to be used here is the following: If $F \supset F(a^2) \supset F(a)$ is a chain of containment of fields , then if all these quantities are finite, $[F(a) : F] = [F(a):F(a^2)] \times [F(a^2) : F]$.

Now, we know that $[F(a) : F(a^2)]$ can either be one or two, since the element $a$ satisfies the equation $x^2-(a^2) = 0$ in $F(a^2)$, so the degree of the minimal polynomial satisfied by $a$ is atmost two. However, if $[F(a) : F(a^2)]$ is $2$, then $[F(a):F]$ is an even number (from the formula), which is a contradiction, since $[F(a):F]$ is equivalent to the degree of the minimal polynomial of $a$ in $F$, which is given to be odd. Hence, it follows that $[F(a) : F(a^2)]=1$, that is, they are equal.

Answer 2

Since $\alpha^2 \in F(\alpha)$, clearly $F(\alpha^2) \subset F(\alpha)$.

We prove $\alpha \in F(\alpha^2)$. Consider the polynomial $p(x)=x^2-\alpha^2$, so that $p(\alpha)=0$. Note that $\alpha \in F(\alpha^2)$ if and only if $p(x)$ is reducible in $F(\alpha^2)$. Suppose $p(x)$ is irreducible in $F(\alpha^2)$, so that $[F(\alpha):F(\alpha^2)]=2$. Thus $$ [F(\alpha):F] = [F(\alpha):F(\alpha^2)][F(\alpha^2):F]=2[F(\alpha^2):F],$$ so $[F(\alpha):F]$ is even, a contradiction. Therefore, $p(x)$ is reducible in $F(\alpha^2)$ and $\alpha \in F(\alpha^2)$.

Answer 3

An alternate solution which is more constructive: suppose the minimal polynomial of $a$ over $F$ is $p(x)$. Then you can split $p$ into even and odd degree terms to write $p(x) = f(x^2) + x g(x^2)$ for $f, g \in F[t]$. Substituting $x := a$ gives $f(a^2) + a g(a^2) = p(a) = 0$. Now $g(x^2)$ has the same leading (nonzero) coefficient as $p(x)$ does, and it has smaller degree. Therefore, $g(a^2) \ne 0$. It follows that $a = -\frac{f(a^2)}{g(a^2)} \in F(a^2)$.