# Finite Fields: check my description/derivation

I am preparing for my exam in Advanced Algebra and Galois Theory, and I am trying to find an efficient way to communicate main properties of Finite Fields. If someone could check my approach and comment I would be most grateful!

## Finite Fields

Let $F$ be a field with $q>1$ elements. Since the field is finite it must have finite characteristic: $$\underbrace{1+1+...+1}_p=0$$ Now $p$ has to be a prime number for otherwise $p=a\cdot b=0$ would make $a$ and $b$ zero divisors which is a contradiction to have in a field.

Clearly $\newcommand{\Zp}{\mathbb{Z}_p}\Zp$ can be isomorhpically embedded in $F$ so we have a finite field extension $F\supset\Zp$ which then must have a basis $v_1,...,v_n\in F$ so that any $\alpha\in F$ can be expressed uniquely as $$\alpha=a_1\cdot v_1+...+a_n\cdot v_n$$ with coefficients $a_i\in\Zp$. This shows that $F$ has $q=p^n$ elements.

Now since the multiplicative group $F^*$ has $p^n-1$ elements and we know that an element of a group raised to the order of the entire group yields the neutral element we have that all elements of $F^*$ satisfies $$X^{p^n-1}-1=0$$ or just as well after multiplying by $X$ that $X^q-X=0$. This shows that $F$ can be embedded as $$F\subseteq \mbox{split}(X^q-X,\Zp)$$ in the splitting field of $X^q-X$ over $\Zp$.

Since $(X^q-X)'=qX^{q-1}-1=-1$ (using that $q=p^n=0$ in characteristic $p$) we see that $X^q-X$ has no multiple roots. Furtermore the roots form a field since we know that $(\alpha+\beta)^q=\alpha^q+\beta^q$ in characteristic $p$ so that $$(\alpha+\beta)^q-(\alpha+\beta)=(\alpha^q-\alpha)+(\beta^q-\beta)=0+0=0$$ for roots $\alpha,\beta$ of $X^q-X$. Furthermore, if $\alpha,\beta$ are roots we can either have one of them zero and then $\alpha\beta=0$ will be a root as well. Suppose they are both non-zero. Then they must be a root of $X^{q-1}-1$ so $q-1$'th roots of unity. These are closed under multiplication (it is the multiplicative group of $F$).

All this shows that $X^q-X$ is a separable polynomial over $\Zp$ and that the splitting field has $q$ elements. Thus we must have $$F\simeq \mbox{split}(X^q-X,\Zp)$$

I hope this is it!

• $F$ is isomorphic to $\mathbf{Z}_p^n$ as what? As a group and as a $\mathbf{Z}_p$-vector space, yes. As a ring, no! – fkraiem Jan 11 '14 at 19:09
• $F \cong \mathbb{F}_p[x]/(X^n)$ is absolutely wrong. Think about it. – Martin Brandenburg Jan 11 '14 at 19:28
• There is no need to talk about isomorphisms at this point, the point is that $F$ is a finite vector space over $\mathbf{Z}_p$, and so it has $p^k$ elements for some integer $k$. – fkraiem Jan 11 '14 at 19:28
• You just need an irreducible polynomial of degree $n$, not $X^n$. (In practice it's good if it's also primitive, meaning its root generates the multiplicative group of the extension.) – Alexander Gruber Jan 11 '14 at 19:36
• Splitting field of $X^q-X$ is the standard construction. I don't know of a way to prove that irreducible polynomials of degree $n$ exist other than via the construction of the field extension as a splitting field first. – Dustan Levenstein Jan 11 '14 at 19:43

The theorem is that for each prime $p$ and natural number $n \ge 1$, there exists a unique field, up to isomorphism, of order $q = p^n$. You've already shown that every finite field must have order equal to such a $q$.
To prove uniqueness, assume $F$ is a field of order $q$, and prove that $F$ consists of exactly the roots of $X^q-X$. Therefore it must be a splitting field over $\mathbb F_p$ of $X^q-X$, and splitting fields are unique up to isomorphism.
To prove existence, you need to show that the splitting field over $\mathbb F_p$ of $X^q-X$ consits of exactly $q$ elements. That it contains no fewer follows from $X^q-X$ having no multiple roots, i.e. $X^q-X$ is separable. To show it contains no more than $q$ elements, prove that the roots of $X^q-X$ themselves form a field; this consists primarily of showing that they are closed under addition and multiplication.