Algebraic Number Theory: Norm and Trace (Part 4)

January 31, 2011

Algebraic Number Theory: Norm and Trace (Part 4)

Filed under: Algebra,Galois Theory,Number Theory — by Masoud Zargar @ 8:41 pm

This is a continuation of the the previous post, regarding separability, in the Algebraic Number Theory series. Now that we have worked with separable polynomials, separable extensions, and separable elements, we will define norms and traces, and prove some results connecting this post to the previous posts in the series. Finally, I will apply all that we have learned so far to describe a method useful in combinatorics.

Suppose $E/F$ is a finite separable extension with $K$ an algebraic closed field with $F\hookrightarrow K$ the inclusion map. Then by separability, we have $|E:F|$ extensions $\sigma_i:E\hookrightarrow K$ of the inclusion map. Now if $a_1,\hdots, a_n$ is a basis for $E$ as an $F$ -vector space, then we define $V^*(a_1,\hdots,a_n)=\det(\sigma_i(a_j))_{i,j}$ . Now if $b_1,\hdots,b_n$ is another basis for the aforementioned vector space, we may write $b_j=\sum_{k=1}^n\mu_{jk}a_k$ for all $j$ . So it follows that $\sigma_i(b_j)=\sum_{k=1}^n\mu_{jk}\sigma_i(a_k)$ . Therefore, $V^*(b_1,\hdots,b_n)=V^*(a_1,\hdots,a_n)\det(\mu_{ij})$ .

The Theorem of Primitive Elements states that if $E=F(\alpha,\beta)$ , where $\alpha$ is separable and $\beta$ is algebraic, then $E$ is a simple extension of $F$ . It follows that $E/F$ is a finite separable extension iff $E=F(\alpha)$ , where $\alpha$ is a separable element. Finding this $\alpha$ for the above extension, we see that we can find a separable element $\alpha$ such that $\{\alpha^{j-1}\}_{j\in [n]}$ is a basis for $E$ . It then follows that $V^*(1,\alpha,\hdots,\alpha^{n-1})=V(\sigma_1(\alpha),\hdots,\sigma_n(\alpha))\neq 0$ . So $V^*\neq 0$ for any chosen basis.

We now define the trace $\text{tr}_{E/F}$ and norm $\text{N}_{E/F}$ of an element $\alpha$ by

$\text{tr}_{E/F}(\alpha):=\sum_{i}\sigma_i(\alpha)$

and

$\text{N}_{E/F}(\alpha):=\prod_i\sigma_i(\alpha)$

By the fact that $\sigma_i$ are field homomorphisms, $\text{tr}_{E/F}:E\rightarrow E$ is $F$ -linear, and $\text{N}_{E/F}:E^*\rightarrow E^*$ is a homomorphism. We also have the following: $\text{tr}_{E/F}(E), \text{N}_{E/F}(E)\subseteq F$ . We can extend $E/F$ to a Galois extension $L/F$ . Then since $G:=\text{Gal}(L|F)$ has the fixed field $F$ , and that $\tau|_{E}$ is an extension of $F\hookrightarrow K$ for all $\tau\in G$ , we see that $\{\tau|_E\sigma_i\}$ is a permutation of $\{\sigma_i\}$ for all $\tau\in G$ . So $\text{tr}_{E/F}(\alpha)$ and $\text{N}_{E/F}(\alpha)$ are in the fixed field of $G$ , which is $F$ , as required.

Given a basis $a_1,\hdots,a_n$ of $E/F$ , we define its discriminant by $d(a_1,\hdots,a_n)=\det(\text{tr}_{E/F}(a_ia_j))_{i,j}$ , which is in turn

$\det((\sigma_i(a_j))_{i,j}(\sigma_i(a_j))^T_{i,j})=V^*(a_1,\hdots,a_n)^2\neq 0$ .

So $(\text{tr}_{E/F}(a_ia_j))_{i,j}$ is non-singular, and so $\text{tr}_{E/F}:E\rightarrow F$ is nonzero somewhere. As a result of $F$ -linearity, trace is surjective. Having done these, it is now easy to prove that $\text{tr}_{E/L}=\text{tr}_{F/L}\circ\text{tr}_{E/F}$ , and a similar formula holds for $\text{N}_{E/L}$ , where we have $L\subseteq F\subseteq E$ .

We will now apply the tools that have been developed so far to describe a technique in combinatorics.

One way to study the divisibility of expressions regarding subsets of a set whose sums of elements are divisible by an integer $n$ is the following . Suppose we are given $n$ integers $a_1,\hdots,a_m$ . We take a polynomial $f\in\mathbb{Z}[z]$ and $n|\sum_{i\in I}a_i$ iff $\zeta_n^{\sum_{i\in I}a_i}=1$ , where $\zeta_n:=e^{\frac{2\pi i}{n}}$ . Then we consider the expression $\omega:=\prod_{i=1}^mf(\zeta_n^{a_i})$ , and so $\displaystyle \text{tr}_{\mathbb{Q}(\zeta_n)/\mathbb{Q}}(\omega)=\sum_{k\in \Pi(n):=\{(a,n)=1, 1\leq a\leq n\}}\prod_{i=1}^mf(\zeta_n^{ka_i})$ . If for example, $f(z)=a+bz$ , then we have

$\displaystyle\sum_{k\in \Pi(n)}\prod_{i=1}^m(a+b\zeta_n^{ka_i})=\sum_{\emptyset\subseteq I\subseteq [m]}a^{m-|I|}b^{|I|}\sum_{k\in\Pi(n)}\left(\zeta_n^{\left(\sum_{j\in I}a_j\right)}\right)^k$

Note that as soon as we let $n=p$ , $p$ a prime, then we are able to analyze properties related to the number of subsets of $\{a_i:i\in [m]\}$ whose sum of elements are divisible by $p$ .