# Elements in a field

Prove that there exists a field with $16$ elements.

I know that there is a theorem that states that a finite field can only have $$p^k$$ Where $p$ is a prime and $k$ is any positive integer. But I'm not sure how I would start going about proving this statement.

• Find a degree $4$ polynomial over $\Bbb F_2$ which is irreducible. – anon May 14 '15 at 5:37
• $\mathbb{F}_2[\alpha]$, where $\alpha^4=\alpha+1$ – alex.jordan May 14 '15 at 5:45

Finite fields will always have cardinality equal to a power of a prime $p$, and in particular the finite field with $p^k$ elements will be an extension of $\mathbb{Z}_p$.

Now recall that if a polynomial $f$ is irreducible over a field $F$, then $F[x]/ \langle f(x) \rangle \cong F[\alpha]$, where $\alpha$ is a root of $f$. Furthermore, $F[\alpha]$ is a degree $\deg(f)$ extension of $F$.

Using this information, you should construct an extension of $\mathbb{Z}_2$ with $16 = 2^4$ elements. If you first convince yourself that a degree $n$ extension of $\mathbb{Z}_p$ will have $p^n$ elements, you should be well on your way to finishing up.

If a field $F$ is finite then its characteristic is a prime number say $p$ then its prime subfield(i.e intersection of all subfields ) is isomorphic to $\mathbb Z_p$.

So we can think $F$ as a finite dimensional vector space over the field $\mathbb Z_p$ say $n$

Then any element of $F$ can be expressed uniquely as $c_1\alpha_1+c_2\alpha _2+...+c_n\alpha_n$ where $c_i$ has $p$ choices each.

So $F$ has $p^n$ elements