# How to construct a field with exactly 125 elements

How to construct a field with exactly n elements in general? Is there any method to do so? And In case no such field exists, how do you determine that?

• Yes, $\mathbb Z_5[x]$ mod irreducible poly of deg 3. – user2345215 Nov 30 '14 at 15:09
• How about $5$ elements? Or $5^2$? – Simon S Nov 30 '14 at 15:09
• @user2345215 Yeah, your answer seems right, but how did you find this? – L887 Nov 30 '14 at 15:12
• This is in every book on Galois Theory, or just google "finite fields"; the famous result is that every finite field has $p^k$ elements and for each prime $p$ and $k\in\mathbb{N}$, there is a unique such field. However, it is a long story to write it all with details as an answer. – Peter Franek Nov 30 '14 at 15:17
• For a concrete example of an irreducible cubic over $\Bbb{F}_5$ consider the following. The polynomial $x^3-x$ takes values $0,0,1,4,0$ modulo $5$ at $x=0,1,2,3,4$. Therefore $p(x)=x^3-x+2$ has no zeros in $\Bbb{F}_5$. Therefore it has no linear factors in $\Bbb{F}_5[x]$. Because it's a cubic it has to be irreducible. ... – Jyrki Lahtonen Nov 30 '14 at 15:33

Let $\mathbb{K}$ be a field anf $f$ an irreducible polynomial of $\mathbb{K}[x]$. If $\alpha$ is a rooth of $f$ the the field $\mathbb{K}(\alpha)$ is isomorphisc to $\mathbb{K}[x]/(f)$
It is now clear that the field $\mathbb{K}(\alpha)$ is a vectorial space over $\mathbb{K}$ of dimension equal to the degree of $f$ .
Now let suppose that our field $\mathbb{K}$ is finite. We obtain:
If $f \in \mathbb{F}_p$ is an irreducible polynomial of degree $d$ and $\alpha$ is one of its roots, then the field $\mathbb{F}_q(\alpha)$ is isomorphic to $\mathbb{F}_p[x]/(f)$ ans so it has $p^d$ elements.