Saturday, December 3, 2016

Why are the elements of a galois/finite field represented as polynomials?


I'm new to finite fields - I have been watching various lectures and reading about them, but I'm missing a step. I can understand what a group, ring field and prime field is, no problem.


But when we have a prime extension field, suddenly the elements are no longer numbers, they are polynomials. I'm sure there is some great mathematical tricks which show that we can (or must?) use polynomials to be able to satisfy the rules of a field within a prime extension field, but I haven't been able to find a coherent explanation of this step. People I have asked in person don't seen to know either, it's just assumed that that is the way it is.


So I have two questions:


What is a clear explanation of "why polynomials?".


Has anyone tried using constructs other than polynomials to satisfy the same field rules?


Thanks in advance.


Answer



In any ring, finite or infinite, we have two operations: $+$ and $\cdot$. The idea of a ring extension is this: let $R$ be a ring and $x$ be some element that we want to add to $R$ (maybe $R\subset S$ and $x\in S$, or $x$ is some formal element).




We need $R[x]$ to be closed under addition and multiplication.



This means that any element that can be formed by manipulating elements of the set $R\cup\{x\}$ by $+$ and $\cdot$ must be an element of $R[x]$.



Polynomials are a general way of expressing such manipulations.



An arbitrary polynomial in $x$ is a completely general manipulation of $x$ using only the operations valid in a ring. Moreover, any element of a ring can be expressed as a polynomial in terms of the generators of the ring.


Let's see how this works: start with some element $a\in R$. We can add or multiply by any other element of $R$, but this just gives us some $a'\in R$. Or we can multiply by $x$ any number of times to get $a'x^n$ for some $n$. And given different elements of this form, we can add them together to get a polynomial.


In many rings, because the elements $x$ satisfy non-trivial relations (e.g. in $\mathbb C=\mathbb R[i]$, $i^2+1=0$), there are neater ways to express elements, and we can avoid polynomials, even though they lurk in the background. In finite fields, polynomials happen to be the easiest and most intuitive way to express what's going on.



No comments:

Post a Comment

analysis - Injection, making bijection

I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...