Another field contruction

**tttcomrader** · November 26th 2007, 01:00 PM

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?

**ThePerfectHacker** · November 26th 2007, 06:15 PM

Originally Posted by tttcomrader

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 $F$ be a finite field $\mathbb{Z}_p$ . Let $p(x) = a_nx^n+...+a_1x+a_0$ be an irreducible polynomial over $F$ . Then $F/\left< p(x) \right>$ is a field. Now any element in $F/\left< p(x) \right>$ can be written uniquely as $b_0+b_1x+...+b_{n-1}x^{n-1}$ now for $b_i$ (for $0\leq i \leq n-1$ ) we have $p$ choices for the coefficients. Thus, in total there are $p^n$ such elements in this larger field.

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

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

**tttcomrader** · November 26th 2007, 07:25 PM

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

**ThePerfectHacker** · November 26th 2007, 08:12 PM

Originally Posted by tttcomrader

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

I am not sure what you are asking. Do you follow what I did? You need to find and irreducible polynomial of degree 3 over Z_5.

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