The elements in a finite field.

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 $n$.

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).

• This makes better sense now. Other than polynomial arithmetic modulo an irreducible polynomial in F_p, are there any other examples of finite fields? In other words can you give me an example of a set with p^n elements and two operations that obey the field axioms? – Geoffrey Critzer Dec 27 '15 at 21:02
• @GeoffreyCritzer If you are asking for more examples up to isomorphism, there aren't any; all finite fields arise as such a quotient. If you just mean to ask whether there are other interesting ways to present finite fields, one place where finite fields of order $2^{2^n}$ arise is in the study of the game of nim. – Slade Dec 27 '15 at 21:21

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).

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.

• Please edit out the first sentence, which does not belong in an answer and has a tone which is certainly not acceptable in this site. – Mariano Suárez-Álvarez Dec 27 '15 at 20:39
• Mod $q$? I think that's wrong - $q$ is usually a prime power. – Thomas Andrews Dec 27 '15 at 20:43
• @MarianoSuárez-Alvarez I found particularly odd that advice not to write “add” in capitals was followed by “add” in capitals in the answer. I suggest to use italics. – egreg Dec 27 '15 at 21:13
• The addition in a finite field of order $q$ is not in general computed $\bmod{q}$. – Morgan Rodgers Dec 27 '15 at 21:57
• Two objections: $$(x+y)^4\equiv x^4+2x^2y^2+y^4\not\equiv x^4+y^4\pmod 4,$$ so that congruence does not hold modulo a prime power. It only holds modulo a prime. Thankfully that is sufficient here. And brings me to the second point - what Morgan Rodgers said. – Jyrki Lahtonen Dec 28 '15 at 8:51