# 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.

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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.
Thanks. You have given a method to find examples. For a general root of unity $\alpha$ often it happens that the quadratic number $\alpha+\bar \alpha$ has a higher degree expression as a polynomial in \alpha$. I am trying to prove this having a sizeable number of examples. – P Vanchinathan Jun 21 '13 at 0:48 If$\alpha$is a root of unity, then$\alpha+\overline\alpha$always has an expression as a polynomial in$\alpha$, since it's in the field you get by adjoining$\alpha$to the reals. So I'm not sure I understand your comment. – Gerry Myerson Jun 21 '13 at 0:59 I meant, by higher degree expression, a polynomial of degree more than degree$\deg \alpha/2$, violating my expectation in the original question. – P Vanchinathan Jun 21 '13 at 2:49 I like Gerry's answer, but I'd like to add that the scenario you described rarely happens. If you weaken your assumption that$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]\$.