Let $\mathbb{F}$ be a field and $\mathbb{F\not=R}$. I want to show that there does not exist a field extension $\mathbb{F}/\mathbb{R}$ of odd degree. Here $\mathbb{R}$ is the field of real numbers.
My attempt: pick $\alpha\in\mathbb{F}, \alpha\notin\mathbb{R}$. By multiplicative nature of degree extensions, [$\mathbb{F:R}$]=[$\mathbb{F:R(\alpha)}$][$\mathbb{R(\alpha):R}$]. I have been trying to show that [$\mathbb{R(\alpha):R}$] is even to no avail.
If I try to show this by contradiction, then one of [$\mathbb{F:R(\alpha)}$] or [$\mathbb{R(\alpha):R}$] should be odd. But I again I get stuck.
Best Answer
Suppose $F$ has odd degree over $\mathbb{R}$; let $\alpha\in F$. Then, as you note, $$[F:\mathbb{R}] = [F:\mathbb{R}(\alpha)][\mathbb{R}(\alpha):\mathbb{R}].$$ Since $[F:\mathbb{R}]$ is odd, then so is its divisor $[\mathbb{R}(\alpha):\mathbb{R}]$. Therefore, the minimal irreducible polynomial of $\alpha$ over $\mathbb{R}$ has odd degree.
If $p(x) = x^{2n+1}+\cdots + a_1x + a_0$ is the minimal irreducible polynomial for $\alpha$, then note that $\lim\limits_{x\to\infty}p(x) =\infty$ and $\lim\limits_{x\to-\infty}p(x)=-\infty$. By the intermediate value theorem, $p(x)$ has a root $c$. Therefore, by the factor theorem $x-c$ divides $p(x)$. Since $p(x)$ is irreducible and monic, $p(x)=c-x$, so $\alpha=c$. Therefore, $\alpha\in\mathbb{R}$.
Since $\alpha\in\mathbb{R}$ for all $\alpha\in F$, we conclude that $F\subseteq\mathbb{R}$. Hence, $F=\mathbb{R}$. So the only extension of $\mathbb{R}$ of odd degree is $\mathbb{R}$ itself, which has degree $1$.