Finite fields

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 13 June 2011

Finite fields

Let $m$ and $n$ be positive integers. Let $R$ be a ring.

• A finite field is a field $𝔽$ such that the set $𝔽$ is finite.

For $p\in {\mathbb{Z}}_{>0}$ prime let ${𝔽}_p$ be the field $\mathbb{Z}/p\mathbb{Z}$ and let ${\overline{𝔽}}_p$ be the algebraic closure of ${𝔽}_p$.

• The pth Frobenius map is the function

  $\begin{array}{cccc}F:& {\overline{𝔽}}_p& ⟶& {\overline{𝔽}}_p\\ & x& ⟼& x^p\end{array}$

The map $F$ is a field homomorphism since

${(x+y)}^p=x^p+y^p\text{and}{\left(xy\right)}^p=x^p{y}^p\text{in characteristic}p.$

HW: Show that ${𝔽}_4={𝔽}_2\left[t\right]/(t^2+t+1)$ since $t^4-t=t(t-1)(t^2+t+1)$ and give the complete multiplication table of ${𝔽}_4=\{a,b,c,d\}$.

Sketches of proofs

Proof. Let $𝔽$ be a finite field. Let

$p=\mathrm{char}\left(𝔽\right)$

be the characteristic of $𝔽$. If $\phi :\mathbb{Z}\to 𝔽$ is the ring homomorphism determined by $\phi \left(1\right)=1$ then $\mathrm{im}\phi$ is a finite integral domain. So $\mathrm{im}\phi$ is a field, and hence, $\ker \phi$ is a maximal ideal in $\mathbb{Z}$. So $p$ is prime.

The field $𝔽$ is an extension of ${𝔽}_p$. Let

$k=[𝔽:{𝔽}_p]$.

Then $\mathrm{Card}\left(𝔽\right)=p^k$ and $F^{\times }=𝔽-\left\{0\right\}$ is a group of order $p^k-1$. Thus, if $\alpha \in 𝔽$ and $\alpha \ne 0$ then ${\alpha }^{p^k-1}=1$. Thus, if $\alpha \in 𝔽$ then ${\alpha }^{p^k}=\alpha$. So

$𝔽\subseteq \{\alpha \in {\overline{𝔽}}_p|\alpha \text{is a root of}{x}^{p^k}-x\}$.

Since $𝔽$ has $p^k$ elements and the polynomial $x^{p^k}-x$ has at most $p^k$ roots,

$𝔽=\{\alpha \in {\overline{𝔽}}_p|\alpha \text{is a root of}{x}^{p^k}-x\}={\left({\overline{𝔽}}_p\right)}^{F^k}.\square$

Proof.

(a) Let $R$ be a finite integral domain. Let $x\in R$, $x\ne 0$. The map

$\begin{array}{cccc}{L}_x:& R& ⟶& R\\ & r& ⟼& xr\end{array}$

is a homomorphism of groups (not of rings) and $\ker L_x=\left\{0\right\}$ since $R$ is an integral domain. So $\mathrm{im}{L}_x=R$, since $R$ is finite. So there exists $r\in R$ such that $xr=1$. So $x$ is invertible.

(b) Let $𝔻$ be a finite division ring. Then the center of $𝔻$,

$Z\left(𝔻\right)$     is a field,

and, for every $x\in 𝔻$, the centralizer of $x$ in $𝔻$,

$Z_{𝔻}\left(x\right)=\{r\in 𝔻|rx=xr\left\}\text{is a vector space over}Z\right(𝔻)$.

Say $Z\left(𝔻\right)={𝔽}_q$. Then

$\mathrm{Card}\left(𝔻\right)=\mathrm{Card}\left(Z_{𝔻}\right(1\left)\right)=q^n,\text{for some}n\in Z_{>0}$.

Now ${𝔻}^{\times }=𝔻-\left\{0\right\}$ is a group and, if ${𝒞}_x$ is the conjugacy class of $x$ in ${𝔻}^{\times }$, then

$\begin{array}{rl}{𝔻}^{\times }& =q^n-1=\mathrm{Card}\left(Z\right({𝔻}^{\times }\left)\right)+\sum _{\left|{𝒞}_x\right|>1}\mathrm{Card}\left({𝒞}_x\right)\\ & =(q-1)+\sum _{\left|{𝒞}_x\right|>1}{\displaystyle \frac{\mathrm{Card}\left({𝔻}^{\times }\right)}{\mathrm{Card}\left(Z_{𝔻}\right(x\left)\right)-1}}\\ & =(q-1)+\sum _{\left|{𝒞}_x\right|>1}{\displaystyle \frac{q^n-1}{q^{d\left(x\right)}-1}},\end{array}$

where $d\left(x\right)$ is the dimension of $Z_{𝔻}\left(x\right)$ as a vector space over $Z\left(𝔻\right)$. The cyclotomic polynomial ${\Phi }_n\left(t\right)$ divides $t^n-1$ and divides $(t^n-1)/(t^{d\left(t\right)}-1)$, so

${\Phi }_n\left(q\right)$     (an integer)   divides    $q^n-1$   and   ${\displaystyle \frac{q^n-1}{q^{d\left(x\right)}-1}}$.

So ${\Phi }_n\left(q\right)$ divides $q-1$. So $n=1$ and $\left|{𝔻}^{\times }\right|=q-1$. So $𝔻=Z\left(𝔻\right)$. $\square$

Notes and References

These notes are taken from notes of Arun Ram from 1999. One nice reference is [Mac, Ch. IV §1] and another is [Bou, Ch. V §12].

References

[Bou] N. Bourbaki, Algebra II, Chapters 4–7 Translated from the 1981 French edition by P. M. Cohn and J. Howie, Reprint of the 1990 English edition, Springer-Verlag, Berlin, 2003. viii+461 pp. ISBN: 3-540-00706-7. MR1994218

[Mac] I.G. Macdonald, Symmetric functions and Hall polynomials, Second edition, Oxford University Press, 1995. ISBN: 0-19-853489-2 MR1354144

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