The elements in a finite field.

Question

I am reading some lecture notes from MIT Open Courseware. One of the theorems states that the elements in a finite field of order $q$ are the $q$ distinct roots of the polynomial $x^q - x$. I can see that the nonzero roots of this polynomial would form a cyclic group under multiplication. I do not understand how these elements would add to form a group.

I think that one interpretation of the elements in a finite field $F_{p^n}$ is to let them be polynomials with coefficients in $F_p$ and the operations are modular polynomial arithmetic with modulus any irreducible polynomial in $F_p[x]$ with degree $n$.

Answer 1

The statement you are referring to is a description of the finite field with $q$ elements, not a construction. If $q$ is a prime power, you can define a field $F$ with $q$ elements by various methods, and whichever construction you choose, you will have the identity $x^q-x=\prod_{a\in F}(x-a)$ in the polynomial ring $F[x]$, which means that all elements of $F$ are simple roots of the polynomial $x^p-x$ (which happens to have its coefficients in the prime subfield of $F$). This description is not enough however to tell those roots apart, or to tell how they are added and multiplied in$~F$. That is part of the construction of $F$ which you have to do beforehand.

From this description it is not even clear that different constructions of a field with $q$ elements will all result in isomorphic fields, although that happens to be true. But given two constructions there may not be any canonical way to choose an isomorphism between the two resulting fields (unless $q$ is a prime number).

Answer 2

First of all elements do not "add", they "sum". Second if $k$ is any field of characteristic $p>0$ then it is an extension of $F_p:=\mathbb{Z}/(p)$ and in $k$, $$(x+y)^q=x+y$$ for all $x,y$ and all $q$ a power of $p$. This is because, using the binomial theorem, the coefficients belong to the ground field $F_p$ and all these coefficients except the first and last are divisible by $p$ and thus zero. A fact incidentally which follows trivially by induction, $$(x+y)^{p^n}\equiv (x+y)^{p^{n-1}} \pmod p.$$

So if in $k$ we take $k_q\subseteq k$ the set of elements satisfying $x^q=x$ we verify directly that $k_q$ is a field, and of course is finite. We then just need a field $k$ which contains all roots of $x^q=x$, but this is easily obtained from the general theory.

Answer 3

The roots of $x^q-x$, where $q=p^n$ for a prime $p$ and an integer $n>1$ form a subfield of the splitting field for this polynomial over the field $F_p$ with $p$ elements.

Indeed, if $a^q-a=0$ and $b^q-b=0$, then $$ (a+b)^q-(a+b)=a^q+b^q-(a+b)=a^q-a+b^q-b=0 $$ This is because the field has characteristic $p$, so $$ \binom{q}{k}=\binom{p^n}{k} $$ is divisible by $p$, for $0<k<p^n$, and so is zero over a field of characteristic $p$.

Also $(-a)^q-(-a)=-a^p+a=0$ if $p$ is odd; if $p=2$, then $-a=a$, so the relation holds as well.

Next $(ab)^q-ab=a^qb^q-ab=ab-ab=0$ and $(a^{-1})^q-a^{-1}=(a^q)^{-1}-a^{-1}=a^{-1}-a^{-1}=0$. Since also $0^q-0=0$ and $1^q-1=0$, we have the assert.

Note that $x^q-x$ has distinct roots and that the splitting field exists (and is unique up to isomorphisms).