0
$\begingroup$

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.

Edited reading Dustan Levenstein's answer

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!

$\endgroup$
  • 1
    $\begingroup$ $F$ 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! $\endgroup$ – fkraiem Jan 11 '14 at 19:09
  • 2
    $\begingroup$ $F \cong \mathbb{F}_p[x]/(X^n)$ is absolutely wrong. Think about it. $\endgroup$ – Martin Brandenburg Jan 11 '14 at 19:28
  • 1
    $\begingroup$ 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 $k$. $\endgroup$ – fkraiem Jan 11 '14 at 19:28
  • 1
    $\begingroup$ 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.) $\endgroup$ – Alexander Gruber Jan 11 '14 at 19:36
  • 1
    $\begingroup$ 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. $\endgroup$ – Dustan Levenstein Jan 11 '14 at 19:43
1
$\begingroup$

Per the comments:

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.