# Are there any other constructions of a finite field with characteristic $p$ except $\Bbb Z_p$?

I mean, $\Bbb Z_p$ is an instance of $\Bbb F_p$, I wonder if there are other ways to construct a field with characteristic $p$? Thanks a lot!

• As the answers say, there are infinitely many finite fields, of different size, with characteristic $p$ for each prime $p$. But as far as "$\mathbb{Z}_p$ is an instance of $\mathbb{F}_p$," it is the only instance; any two fields of the same finite cardinality are isomorphic. – Nick Matteo Apr 12 '13 at 17:52

## 5 Answers

There are also extensions of $\mathbb F_p$. For example $\mathbb F_2[t]/(t^2 + t + 1)$ is a field with $4$ elements. It has characteristic $2$.

You might try perusing the wikipedia page which has more examples.

For each power $p^r$ of $p$, there is a unique finite field of characteristic $p$ and size $p^r$, it is denoted by $\mathbb F_{p^r}$. For the construction, you take an irreducible polynomial $f\in\mathbb F_p[x]$ of degree $r$. Then $\mathbb F_{p^r} \cong \mathbb F_p[x]/(f)$. The smallest non-trivial example is $\mathbb F_4 = \mathbb F_{2^2}$. A concrete construction for it is given in the answer of Jim.

• I just want to mention that the field $F_{p^r}$ is the splitting field of the polynomial $X^{p^r}-X\in F_p[X]$. – Dave Apr 12 '13 at 7:31
• @DaveHartman: Right, thank you! In some situations, your representation is preferable, since it does not depend on the choice of the irreducible polynomial $f$. – azimut Apr 12 '13 at 7:44

Just to supplement the other answers: As stated in the other answers, for every prime power $p^r$, $r>0$, there is a unique (up to isomorphism) field with $p^r$ elements. There are also infinite fields of characteristic $p$, for instance if $F$ is any field of characteristic $p$ (e.g., $\mathbb Z_p$), the field $F(t)$ (the field of fractions of the polynomial ring $F[t]$, with $t$ an indeterminate), is an infinite field of characteristic $p$.

• Another example of an infinite field of characteristic $p$ would be the algebraic closure of any finite field of characteristic $p$. – David Ward Apr 12 '13 at 7:50
• +1: $\Bbb F_p(t)$ is one of the simplest (if not the simplest) examples of a field which is not perfect. – A.P. Apr 12 '13 at 9:40

As said in the other answers, if $P$ is an irreducible polynomial on $\mathbb{F}_p$ of degree $d$, then $k=\mathbb{F}_p[X]/ (P)$ is a field of cardinality $p^d$ and is of characteristic $p$. For a nice presentation of $k$, you can say that $k=\mathbb{F}_p(A) \subset M_d(\mathbb{F}_p)$ where $A$ is the companion matrix of $P$.

For instance, the example given by Jim becomes: $k= \left\{ \left( \begin{array}{cc} b & -a \\ a & b-a \end{array} \right) \mid a,b \in \mathbb{F}_p \right\}$.

There are two nice constructions I know, the first one is already mentioned. For every prime $p$ you can 'extend' it and construct fields with $p^n$ elements. (this depends on the fact that there are irreducible polynomials of every degree in $\mathbb{F}_p$, exercise: give some examples to show why this might be true). Then you can also take the algebraic closure of these fields and get a genuinely new field since an algebraic closed field can't be finite (you can prove this in the exact same way Euclid proved there are a infinite number of prime numbers), let's call this field $\overline{\mathbb{F}_p}$.

The second construction I know is first taking polynomials with coefficients in $\mathbb{F}_p$ i.e. $\mathbb{F}_p[x]$ and take its quotient field (as you do to construct $\mathbb{Q}$ from $\mathbb{Z}$) then get quotients of polynomials, this form a new field which contains $\mathbb{F}_p$ as a subfield (identified by the 'constant polynomials'). Let's denote this new field $\mathbb{F}_p(x)$.

One might think these two fields are isomorphic, but this post argues that it isn't the case.