Problem 7 - A Computational Introduction to Number Theory and Algebra

Question

Let \(E\) be a finite extension of a finite field \(F .\) Let \(K\) be an intermediate field, \(F \subseteq K \subseteq E .\) Show that for all \(\alpha \in E\) (a) \(\mathbf{N}_{E / F}(\alpha)=\mathbf{N}_{K / F}\left(\mathbf{N}_{E / K}(\alpha)\right),\) and (b) \(\operatorname{Tr}_{E / F}(\alpha)=\operatorname{Tr}_{K / F}\left(\operatorname{Tr}_{E / K}(\alpha)\right)\)

Step-by-Step Solution

Answer

Question: Show that for any element α in field extension E, the Norm of α with respect to field extension E/F equals the composed norm of α with respect to both the intermediate field K and field extension E/K. Additionally, show that the Trace of α with respect to field extension E/F equals the composed trace of α with respect to the intermediate field K and field extension E/K. Answer: We showed that the Norm of α with respect to field extension E/F is equal to the composed norm of α with respect to both the intermediate field K and field extension E/K by composing the automorphisms of K fixing F and the automorphisms of E fixing K. Similarly, the Trace of α with respect to field extension E/F is equal to the composed trace of α with respect to the intermediate field K and field extension E/K by composing the automorphisms of K fixing F and the automorphisms of E fixing K.

Step 1: Recall the definitions of Norm and Trace

The norm of an element \(\alpha\) with respect to the field extension \(E/F\) is given by the product of the conjugates of \(\alpha\) over \(\sigma\), the automorphisms of \(E\) fixing \(F\); that is, $$\mathbf{N}_{E / F}(\alpha)=\prod_{\sigma} \sigma(\alpha)$$ Similarly, the trace of an element \(\alpha\) with respect to the field extension \(E/F\) is given by the sum of the conjugates of \(\alpha\) over \(\sigma\), the automorphisms of \(E\) fixing \(F\); that is, $$\operatorname{Tr}_{E / F}(\alpha)=\sum_{\sigma} \sigma(\alpha)$$

Step 2: Show equality of the Norms

We have \(F \subseteq K \subseteq E\). We want to show that for all \(\alpha \in E\), \(\mathbf{N}_{E / F}(\alpha)=\mathbf{N}_{K / F}\left(\mathbf{N}_{E / K}(\alpha)\right)\). Let's compute the right side of the equality: $$\mathbf{N}_{K / F}\left(\mathbf{N}_{E / K}(\alpha)\right) = \prod_{\tau} \tau\left(\prod_{\sigma} \sigma(\alpha)\right)$$ where \(\tau\) are automorphisms of \(K\) fixing \(F\) and \(\sigma\) are automorphisms of \(E\) fixing \(K\). Since \(\tau\) and \(\sigma\) are automorphisms, they can be composed, \(\tau \circ \sigma\) is an automorphism of \(E\) fixing \(F\). Therefore, the product of these composed automorphisms acting on \(\alpha\) is equal to the \(\mathbf{N}_{E/F}(\alpha)\): $$\mathbf{N}_{E / F}(\alpha) = \prod_{\tau \circ \sigma} (\tau \circ \sigma)(\alpha)$$ Hence, we have shown that \(\mathbf{N}_{E / F}(\alpha)=\mathbf{N}_{K / F}\left(\mathbf{N}_{E / K}(\alpha)\right)\).

Step 3: Show equality of the Traces

We have \(F \subseteq K \subseteq E\). We want to show that for all \(\alpha \in E\), \(\operatorname{Tr}_{E / F}(\alpha)=\operatorname{Tr}_{K / F}\left(\operatorname{Tr}_{E / K}(\alpha)\right)\). Let's compute the right side of the equality: $$\operatorname{Tr}_{K / F}\left(\operatorname{Tr}_{E / K}(\alpha)\right) = \sum_{\tau} \tau\left(\sum_{\sigma} \sigma(\alpha)\right)$$ where \(\tau\) are automorphisms of \(K\) fixing \(F\) and \(\sigma\) are automorphisms of \(E\) fixing \(K\). Since \(\tau\) and \(\sigma\) are automorphisms, they can be composed, \(\tau \circ \sigma\) is an automorphism of \(E\) fixing \(F\). Therefore, the sum of these composed automorphisms acting on \(\alpha\) is equal to \(\operatorname{Tr}_{E/F}(\alpha)\): $$\operatorname{Tr}_{E / F}(\alpha) = \sum_{\tau \circ \sigma} (\tau \circ \sigma)(\alpha)$$ Hence, we have shown that \(\operatorname{Tr}_{E / F}(\alpha)=\operatorname{Tr}_{K / F}\left(\operatorname{Tr}_{E / K}(\alpha)\right)\).

Key Concepts

Field Automorphisms

In abstract algebra, automorphisms are structural preservations of algebraic structures. Specifically, a **field automorphism** is a bijective function from a field to itself that respects both field operations: addition and multiplication.

These functions are essential in understanding field extensions. They provide insights into the symmetries of a field.

• Automorphisms of a field extension \(E/F\) are functions that fix elements in \(F\) while mapping the field \(E\) onto itself.
• Thus, if \(\sigma\) is an automorphism of \(E\) fixing \(F\), for any \(\alpha \in E\), \(\sigma(\alpha)\) remains within \(E\).

By using these automorphisms, we can explore deeper properties of elements within the field extension, such as calculating norms and traces.

Norm in Field Extensions

The **norm** in field extensions gives a scalar representation of the behavior of an element within a field. It's derived by multiplying all of its conjugates over its automorphisms.

To define it clearly:

• The norm of an element \(\alpha\) concerning an extension \(E/F\) is denoted as \(\mathbf{N}_{E/F}(\alpha)\).
• This is calculated by taking the product of all conjugates, where conjugates are derived from automorphisms of \(E\) that fix \(F\), i.e., \(\sigma(\alpha)\).

When there exist intermediate fields, like in the case of \(F\) within \(K\) in \(F \subseteq K \subseteq E\), the **transitive property of norms** tells us that: \[\mathbf{N}_{E/F}(\alpha) = \mathbf{N}_{K/F}(\mathbf{N}_{E/K}(\alpha))\] This equation indicates the interconnected nature of norms over successive field extensions.

Trace in Field Extensions

The **trace** offers another essential metric for understanding field extensions. Like the norm, the trace is associated with how an element projects through its field conjugates. But instead of a product, we sum them.

For a more detailed look:

• The trace of an element \(\alpha\) with respect to the extension \(E/F\) is expressed as \(\text{Tr}_{E/F}(\alpha)\).
• This involves calculating the sum of all \(\sigma(\alpha)\)'s, where \(\sigma\) explores automorphisms that lock elements of \(F\) but let \(E\) free.

Considering layered field expansions, the **trace behaves additively** like so:\[\text{Tr}_{E/F}(\alpha) = \text{Tr}_{K/F}(\text{Tr}_{E/K}(\alpha))\]This highlights how trace operations cascade through intermediate fields, maintaining coherence across extended field levels.