Fix an algebraic integer $\alpha$ of degree $n$ 
such that the extension $K=\mathbf{Q}(\alpha)/\mathbf{Q}$ has intermediate fields.
(We can assume $K$ is Galois with non-simple Galois group.)
This $\alpha$, by its powers, gives rise to a co-ordinatisation for the affine space $K=\mathbf{Q}^n$. Now what is known about subset of algebraic numbers inside $K$ of
a fixed degree $m$ less than $n$? Is this a $\mathbf{Q}$-Zariski closed subset of $K$? 
Or should we consider subsets of elements of degree less than or equal to $m$?


If we fix the Galois group $G$ and the divisor $m$ of $[K:\mathbf Q]$ is the ideal for
degree m elements independent of $K$?



Lower Degree Elements in an Algebraic Number Field
Write an arbitrary element of $K$ as $x = x\_0 + x\_1 \alpha + \dots + x\_{n-1} \alpha^{n-1}$ with $x\_0, \dots, x\_{n-1} \in \mathbb{Q}$. Then you can write $1, x, \dots, x^m$ as 
$$
x^d = y\_{d,0} + y\_{d,1} \alpha + \dots + y\_{d,n-1} \alpha^{n-1},
$$
where the $y\_{d,k}$ are polynomial expressions in $x\_0, \dots, x\_{n-1}$ (and also in the coefficients of the minimal polynomial of $\alpha$). Then the condition that $x$ have degree $\leq m$ is exactly that the $(m+1) \times n$ matrix with coefficients $y\_{d,k}$ has rank strictly less than $m+1$ (then you can also find an element of the nullspace in $\mathbb{Q}$). This is an algebraic condition which can be cut out by the $(m+1) \times (m+1)$ minors. So indeed you get a Zariski closed subset for the elements of degree $\leq m$. I think the condition for degree equal to $m$ will be only locally closed in general. Also, at least from the above it appears that the ideal should depend on $K$. (I guess the case when $m$ is $1$ is an exception.)


