# Construct a field

• Nov 24th 2007, 02:04 PM
Construct a field
Construct a field of order 25.

I don't really understand this question, am I suppose to list the polynomials of degree 5?
• Nov 24th 2007, 02:28 PM
ThePerfectHacker
Quote:

Construct a field of order 25

I don't really understand this question, am I suppose to list the polynomials of degree 5?

The question is saying to explicity create a field, we know a field of 25 elements exists but we want to explicitly state it.

I have to ways to to construct such a field I will do it with polynomials since that is what you mentioned and I assume that is what you want.

First we need to know a theorem: Let $\displaystyle p(x)$ be a (nonconstant) polynomial over a field $\displaystyle F$. Then $\displaystyle p(x)$ is irreducible over $\displaystyle F$ if and only if $\displaystyle F[x]/\left< p(x) \right>$ is a field.

Consider $\displaystyle F = \mathbb{Z}_5$, and the polynomial $\displaystyle p(x) = x^2 + 2$ is irreducible since it is of degree two and it has no zero (just check for zeros x=0,1,2,3,4).

Now since it is irreducible by the theorem $\displaystyle F[x]/\left< x^2+2 \right>$ is a field. How many elements are in this field? Note any element in this field can be written as $\displaystyle ax+b$ since of uniquneness $\displaystyle a$ and $\displaystyle b$ can be any elements of the field $\displaystyle F$ since there are five elements in total we can write $\displaystyle 5\cdot 5 = 25$.
• Nov 25th 2007, 06:04 AM
So if I want to construct a field of order 27. I pick $\displaystyle x^3 + 1 \in Z_{9}$?
No, $\displaystyle x^3+1$ is reducible, you can write $\displaystyle (x+1)(x^2+2x+1)$ or note that $\displaystyle x=2$ is a zero of this polynomial.