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Let $\mathcal{A}$ be a Real division algebra of finite dimension over $\mathbb{R}$. Show that for each element $a \in \mathcal{A}, \exists \mu \in \mathbb{R}$ such that $a^2 + \mu a \in \mathbb{R}$.
What I tried to do is the following:

Let $a \in \mathcal{A} \setminus \mathbb{R},$ then $\mathbb{R}(a)=\{x+ay | x,y \in \mathbb{R}\}$ is a dimensional space over the field of real numbers with basis $\{1,a\},$ since $\dim_{\mathbb{R}}\mathcal{A}=n < \infty$. Since this extension is of finite degree, should it not be enough to show that $\mathbb{R}(a)$ is an algebraic extension of $\mathbb{R}$ and hence $a$ is algebraic over the reals? Because then we could find a $f$ polynomial with real coefficients such that $f(a)=0$. From that we can see that $a$ is a root of a second order polynomial (and we're done), because if $a$ were the root of a $1$st order polynomial, $a$ would be real, a contradiction.


Something feels off though, since I can't see where I used the condition that $\mathcal{A}$ is a division algebra, but I'm not really sure why what I've written above is not right. Using the division algebra condition of $\mathcal{A}$ for showing that $\mathbb{R}(a)$ is a commutative division algebra seems to me like a good idea, but I don't really see a reason on doing all that, supposing that my quoted text is OK.

separable ninja
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Your start is fine: $\Bbb R(a)$ is a two dimensional field extension of $\Bbb R$. In order to conclude it's a field, you need to at least know that $\Bbb R(a)$ has no zero divisors (see this for example), and so knowing that $\mathcal A$ is a division algebra is one way of doing that.

Then $\{1, a, a^2\}$ is a linearly dependent set over $\Bbb R$, so there exists $\alpha,\beta,\gamma\in \Bbb R$ such that $a^2=\beta a+\gamma 1$. Can you get it from here?

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rschwieb
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