# Lower Degree Elements in an Algebraic Number Field

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 $K$?

-
Please, try to use existing tags if suitable ones already exist, and in general please connect individual words that only together convey a specific meaning by a hyphen. Thanks in advance. –  quid Jun 19 '13 at 19:47
Thanks, quid. I'll keep this in mind. –  P Vanchinathan Jun 20 '13 at 0:17

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.)

-
+1. Just a remark: for any intermediate field $\mathbf{Q} \subset K' \subset K$ of degree $m$ we have that $K'$ is an $m$-dimensional sub-$\mathbf{Q}$-vector space of $K \cong \mathbf{Q}^n$. Hence, under the correspondence from the OP, the elements of $K'$ correspond to an $m$-dimensional linear affine subvariety $V_{K'}$ of $V = \mathbf{A}^n_{\mathbf{Q}}$ (affine $n$-space over $\mathbf{Q}$). We can conclude that each set $V_m$ of elements of degree precisely $m$ is a finite boolean combination of linear affine subspaces of $V$. In particular, $V_m$ will not be closed unless $m=1$. –  René Jun 19 '13 at 19:14
good point! that's a much cleaner description of the situation than in my answer :-) –  Abhinav Kumar Jun 19 '13 at 20:30
Thanks Abhinav Kumar for the answer and to Ren\'e Pannekoek for throwing more light on it. Originally I had a different, but related question (see recent questions) and I thought I need to get this one out first. –  P Vanchinathan Jun 20 '13 at 0:28