Polynomials giving Lower Degree Elements in an Algebraic  Number Field My earlier related  question 
Lower Degree Elements in an Algebraic Number Field
has been given a clean answer for the first part. My present question is below:
Take a number field  $K=\mathbf{Q}(\alpha)$ of degree $n$  admitting intermediate fields, so $n$ is a composite number $n=mk$, and there are algebraic numbers of degree $m$ inside $K$. Is there a polynomial $f(x)\in \mathbf{Q}[x]$ of degree $k$ such that $f(\alpha)$ is an algebraic number of degree $m$?
I am looking for a way of reaching lower degree elements from the primitive elements.
A 'generic polynomial' evaluates to another primitive element, there must be some special property that will give rise to lower degree elements. `Complementary degree' does not seem to be enough. 
Example: A primitive $p^{a+b}$-th root of unity, a number of degree $p^{a+b-1}(p-1)$, when raised to the power $p^a$,  leads to a primitive $p^b$-th root of unity, a number of desired degree I am looking for.
 A: Let $\beta=2^{1/4}$, and let $\alpha=\beta^3+\beta^2+1$. Then ${\bf Q}(\alpha)={\bf Q}(\beta)$ has degree 4 over the rationals. Then $\alpha^2=2\beta^3+4\sqrt2+4\beta+3$, and there is no quadratic polynomial $f$ with rational coefficients such that $f(\alpha)$ has degree 2 over the rationals. 
For another example, let $\alpha$ be a primitive 5th root of unity. The quadratic subfield of ${\bf Q}(\alpha)$ is real, so if $f(\alpha)$ is in it then it equals its complex conjugate. For $f$ quadratic, this leads to a degree 3 equation for $\alpha$, contradiction. 
A: I like Gerry's answer, but I'd like to add that the scenario you described rarely happens.
If you weaken your assumption that $f(x) \in \mathbb{Q}[x]$ to the assumption that $f(x) \in \mathbb{Q}(f(\alpha))[x]$, you have already forced $f(x) - f(\alpha)$ to be the fundamental polynomial for $\alpha$ over one of the (finitely many) intermediate fields of degree $m$.  If each of those fundamental polynomials has a non-constant term with an irrational coefficient, you won't be able to find $f(x) \in \mathbb{Q}[x]$.
