Take the tour ×  I came across the following fact:  Let $m$ and
$n$ be natural numbers such that $n\geqslant m$ and $m\equiv n
\pmod 2$. Then, there exists an irreducible polynomial
$f\in\mathbb{Q}\left[X\right]$ of degree $n$ such that $f$ has
exactly $m$ real roots.  I guess it is well known for
algebraists. Unfortunately, I cannot find the proof of this
anywhere and I do not know how to prove it. Could you help me?
share|improve this question isn't this a consequence (or rather
an equivalent statement) of the fundamental theorem of algebra?
–  scibuff Apr 10 '12 at 15:16 Probably you are right, but I
have no idea how to use it to prove this.. –  dawid Apr 10 '12
at 15:19 @scibuff: Irreducible? –  Did Apr 10 '12 at 15:33 Just
a comment, not an answer. Write $n=m+2k$. Then the assertion is
equivalent to the existence of an algebraic field extension
$\Bbb Q\subset F$ such that $F$ admits exactly $m$ different
embeddings into $\Bbb R$ and $2k$ different embeddings into
$\Bbb C$. The equivalence is realized by