1. ## Another field contruction

Construct a field of order 125.

I don't know, this chapter is just not coming to me for some reasons, I can't really understand how to construct a field. So I pick a field, then pick a poly that is irreducible and has order of 125, how do I do that?

Construct a field of order 125.

I don't know, this chapter is just not coming to me for some reasons, I can't really understand how to construct a field. So I pick a field, then pick a poly that is irreducible and has order of 125, how do I do that?
Again remember what I said. Let $\displaystyle F$ be a finite field $\displaystyle \mathbb{Z}_p$. Let $\displaystyle p(x) = a_nx^n+...+a_1x+a_0$ be an irreducible polynomial over $\displaystyle F$. Then $\displaystyle F/\left< p(x) \right>$ is a field. Now any element in $\displaystyle F/\left< p(x) \right>$ can be written uniquely as $\displaystyle b_0+b_1x+...+b_{n-1}x^{n-1}$ now for $\displaystyle b_i$ (for $\displaystyle 0\leq i \leq n-1$) we have $\displaystyle p$ choices for the coefficients. Thus, in total there are $\displaystyle p^n$ such elements in this larger field.

1)Let $\displaystyle F=\mathbb{Z}_5$
2)Let $\displaystyle p(x)$ be an order $\displaystyle 3$ irreducible polynomial in $\displaystyle F$.
3)The factor ring $\displaystyle F/\left< p(x)\right>$ is a field with $\displaystyle 5^3=125$ elements.

So the thing remaining now is for you to find an irreducible degree polynomial.

3. $\displaystyle (x-2)^3$, it has zero in there. would that be it?

$\displaystyle (x-2)^3$, it has zero in there. would that be it?