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Let $L$ be the subfield of $\mathbb{R}$, of all reals that are algebraic over $\mathbb{Q}$: $L = \{ x\in \mathbb{R} : x \text{ is algebraic over } \mathbb{Q} \}, \;\;\; \mathbb{Q} \subseteq L$.

Let $K$ be finite field extension of $L$ such that $K \subseteq \mathbb{C}$. By the primitive element theorem $K = L(a)$. Denote by $f(x)$ the minimal polynomial of $a$ over $L$.

Prove $f(x)$ has no real roots.

Any hints ?

Edit: I made a mistake in the original question. $K/L$ is finite extension.

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What if the extension $K/L$ is infinite in degree? – user38268 Jul 15 '12 at 9:48
Sorry. $K/L$ need to be a finite extension. I fixed the question. – colge Jul 15 '12 at 10:09
I don't think you can use the PET for the reason Benja mentioned. Perhaps you want to talk about an algebraic element $\,a\in K-L\,$? – DonAntonio Jul 15 '12 at 10:10
Please edit your question to add that new, important info about $\,K/L\,$ – DonAntonio Jul 15 '12 at 10:11
Anyway, I think the result is true for any algebraic element $a\in K-L$. – colge Jul 15 '12 at 10:12
up vote 3 down vote accepted

Assuming $\,K/L\,$ is a finite non-trivial extension ( $\,K\neq L\,$) then clearly $\,a\in\Bbb C-\Bbb R\,$ .

If the minimal polynomial $\,f(x)\,$ of $\,a\,$ over $\,L\,$ had a real root $\,w\,$ then $$\,L(w)/L\,,\,L/\Bbb Q\,\,\text{are both algebraic}\, \,\Longrightarrow L(w)/\Bbb Q\,\,\text{is algebraic}$$ and thus $\,w\in L\Longrightarrow\,(x-w)\mid f(x)\,$ in $\,L[x]\,\Longrightarrow a\,$ is a root of $$g(x)=\frac{f(x)}{x-w}\in L[x]\,\,\,and\,\,\,\deg\frac{f(x)}{x-w}<\deg f(x)$$ which of course cannot be.

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The only algebraic extensions of $L$ are $L$ itself (which is a real closed field) and $ L(i) = \overline{\mathbf{Q}}$ (which is an algebraically closed field).

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