# How are the fields $\mathbb{F}_k$(where $k$ is an integer) be generated?

What are elements like in the fields $\mathbb{F}_k$? Does $\mathbb{F}_k$ contain only $k$ elements? When $k$ is a composite integer, what will be different from that $k$ is a prime? Please help me.

• They only exist, if $k$ is a prime number or a power of a prime number. In the latter case with $k=q=p^n$ we can "construct" the field as the splitting field of the polynomial $x^q-x$ over $\Bbb{F}_p$, but this is not very useful for doing arithmetic. A more useful way is to get $\Bbb{F}_q=\Bbb{F}_p[x]/\langle f(x)\rangle$, where $f(x)$ is a monic irreducible polynomial of degree $n$ from $\Bbb{F}_p[x]$. There is no known general formula for such a polynomial $f(x)$ though. – Jyrki Lahtonen Jun 25 '14 at 5:48
• It might help to know where you get your information from. How have you discovered the existence of finite fields, particularly of non-prime orders? If it's from an algebra textbook, there is presumably a lot more info on them there. – Ryan Reich Jun 25 '14 at 5:48
• Here are two examples: 1, 2. – David Jun 25 '14 at 5:50
• Here you see $\Bbb{F}_4$, $\Bbb{F}_8$ and $\Bbb{F}_{16}$ with their elements listed. – Jyrki Lahtonen Jun 25 '14 at 5:50
• Such as field $\mathbb{F}_4\approx \mathbb{F}_2[X]/(X^2+X+1)$, now I know the composite is only when $k=p^m$. @RyanReich – nix Jun 25 '14 at 6:02

${F_k}$={o, 1, $x$, $x^2$, ... ,$x^{k-2}$} where $k=p^{m}$, $p$ is a prime number and $m$ is some integer. $p$ is characterstic and $m$ is dimension of the field. $x$ is primitive element if it is the root of the primitive polynomial and $x^{k-1}=1$. Yes, $F_k$ has $k$ elements and $F_p$ is contained inside $F_k$. The notation I have used is power notation which is useful for multiplication operations and ofcourse there is polynomial notation for the field which is useful in addition operations.