## 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 $p\in {ℤ}_{>0}$ prime let ${𝔽}_{p}$ be the field $ℤ/pℤ$ and let ${\stackrel{‾}{𝔽}}_{p}$ be the algebraic closure of ${𝔽}_{p}$.
• The pth Frobenius map is the function  $\begin{array}{cccc}F:& {\stackrel{‾}{𝔽}}_{p}& ⟶& {\stackrel{‾}{𝔽}}_{p}\\ & x& ⟼& {x}^{p}\end{array}$
The map $F$ is a field homomorphism since
 ${\left(x+y\right)}^{p}={x}^{p}+{y}^{p}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{\left(xy\right)}^{p}={x}^{p}{y}^{p}\phantom{\rule{2em}{0ex}}\text{in characteristic}\phantom{\rule{0.5em}{0ex}}p.$

(a)   The function
 $\begin{array}{ccc}\left\{\text{finite fields}\right\}& ⟶& \left\{{p}^{k}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}p\in {ℤ}_{>0}\phantom{\rule{0.5em}{0ex}}\text{is prime,}\phantom{\rule{0.5em}{0ex}}k\in {ℤ}_{>0}\right\}\\ 𝔽& ⟼& \mathrm{Card}\left(𝔽\right)\end{array}$
is a bijection.
(b)   The finite field ${𝔽}_{{p}^{k}}$ with ${p}^{k}$ elements is given by
 ${𝔽}_{{p}^{k}}$ is the extension of ${𝔽}_{p}$ of degree $k$,      ${𝔽}_{{p}^{k}}=\left\{\alpha \in {\stackrel{‾}{𝔽}}_{p}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}{\alpha }^{{p}^{k}}-\alpha =0\right\}$,      ${𝔽}_{{p}^{k}}={\left({\stackrel{‾}{𝔽}}_{p}\right)}^{{F}^{k}}$.

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

(a)   Every finite integral domain is a field.
(a)   Every finite division ring is a field.

## Sketches of proofs

(a)   The function
 $\begin{array}{ccc}\left\{\text{finite fields}\right\}& ⟶& \left\{{p}^{k}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}p\in {ℤ}_{>0}\phantom{\rule{0.5em}{0ex}}\text{is prime,}\phantom{\rule{0.5em}{0ex}}k\in {ℤ}_{>0}\right\}\\ 𝔽& ⟼& \mathrm{Card}\left(𝔽\right)\end{array}$
is a bijection.
(b)   The finite field ${𝔽}_{{p}^{k}}$ with ${p}^{k}$ elements is given by
 ${𝔽}_{{p}^{k}}$ is the extension of ${𝔽}_{p}$ of degree $k$,      ${𝔽}_{{p}^{k}}=\left\{\alpha \in {\stackrel{‾}{𝔽}}_{p}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}{\alpha }^{{p}^{k}}-\alpha =0\right\}$,      ${𝔽}_{{p}^{k}}={\left({\stackrel{‾}{𝔽}}_{p}\right)}^{{F}^{k}}$.

Proof. Let $𝔽$ be a finite field. Let

 $p=\mathrm{char}\left(𝔽\right)$
be the characteristic of $𝔽$. If $\phi :ℤ\to 𝔽$ 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, $\mathrm{ker}\phi$ is a maximal ideal in $ℤ$. So $p$ is prime.

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

 $k=\left[𝔽:{𝔽}_{p}\right]$.
Then $\mathrm{Card}\left(𝔽\right)={p}^{k}$ and ${F}^{×}=𝔽-\left\{0\right\}$ 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
 $𝔽\subseteq \left\{\alpha \in {\stackrel{‾}{𝔽}}_{p}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}\alpha \phantom{\rule{0.5em}{0ex}}\text{is a root of}\phantom{\rule{0.5em}{0ex}}{x}^{{p}^{k}}-x\right\}$.
Since $𝔽$ has ${p}^{k}$ elements and the polynomial ${x}^{{p}^{k}}-x$ has at most ${p}^{k}$ roots,
 $𝔽=\left\{\alpha \in {\stackrel{‾}{𝔽}}_{p}\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}\alpha \phantom{\rule{0.5em}{0ex}}\text{is a root of}\phantom{\rule{0.5em}{0ex}}{x}^{{p}^{k}}-x\right\}={\left({\stackrel{‾}{𝔽}}_{p}\right)}^{{F}^{k}}.\phantom{\rule{2em}{0ex}}\square$

(a)   Every finite integral domain is a field.
(a)   Every finite division ring is a field.

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 $\mathrm{ker}{L}_{x}=\left\{0\right\}$ 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 $𝔻$,
 ${Z}_{𝔻}\left(x\right)=\left\{r\in 𝔻\phantom{\rule{0.5em}{0ex}}|\phantom{\rule{0.5em}{0ex}}rx=xr\right\}\phantom{\rule{2em}{0ex}}\text{is a vector space over}\phantom{\rule{0.5em}{0ex}}Z\left(𝔻\right)$.
Say $Z\left(𝔻\right)={𝔽}_{q}$. Then
 $\mathrm{Card}\left(𝔻\right)=\mathrm{Card}\left({Z}_{𝔻}\left(1\right)\right)={q}^{n},\phantom{\rule{2em}{0ex}}\text{for some}\phantom{\rule{0.5em}{0ex}}n\in {Z}_{>0}$.
Now ${𝔻}^{×}=𝔻-\left\{0\right\}$ is a group and, if ${𝒞}_{x}$ is the conjugacy class of $x$ in ${𝔻}^{×}$, then
 $\begin{array}{rl}{𝔻}^{×}& ={q}^{n}-1=\mathrm{Card}\left(Z\left({𝔻}^{×}\right)\right)+\sum _{|{𝒞}_{x}|>1}\mathrm{Card}\left({𝒞}_{x}\right)\\ & =\left(q-1\right)+\sum _{|{𝒞}_{x}|>1}\frac{\mathrm{Card}\left({𝔻}^{×}\right)}{\mathrm{Card}\left({Z}_{𝔻}\left(x\right)\right)-1}\\ & =\left(q-1\right)+\sum _{|{𝒞}_{x}|>1}\frac{{q}^{n}-1}{{q}^{d\left(x\right)}-1},\end{array}$
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 $\left({t}^{n}-1\right)/\left({t}^{d\left(t\right)}-1\right)$, so
 ${\Phi }_{n}\left(q\right)$     (an integer)   divides    ${q}^{n}-1$   and   $\frac{{q}^{n}-1}{{q}^{d\left(x\right)}-1}$.
So ${\Phi }_{n}\left(q\right)$ divides $q-1$. So $n=1$ and $|{𝔻}^{×}|=q-1$. 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