1. ## Finite Field

An irreducible polynomial P(x) of degree m over GF(q) can be used to construct an extension field of GF(q). What is actually meant by constructing an extension field?

and why is it important to ensure whether a monic polynomial of degree m over GF(2^n) is primitive???

lastly, there are 3 bases of finite field: polynomial, normal and dual base. in what condition each of the bases is favored over the others?

thank you very much for your help

2. Originally Posted by classic_phohe
An irreducible polynomial P(x) of degree m over GF(q) can be used to construct an extension field of GF(q). What is actually meant by constructing an extension field?
If $p(x)$ is irreducible and $\mathbb{F}_q = \text{GF}(q)$ then $\mathbb{F}_q[x]/(p(x))$ is a field. Furthermore, $\mathbb{F}_q$ can be identified as a subfield of this field by an embedding $a \mapsto a + (p(x))$. And so $\mathbb{F}_q[x]/(p(x))$ can be regarded as an extension field of $\mathbb{F}_q$. In fact it is an extension of degree $m$, do you see why?

And I am not sure about you other question, I do not know what you mean by 'primitive'?

3. Originally Posted by ThePerfectHacker
If $p(x)$ is irreducible and $\mathbb{F}_q = \text{GF}(q)$ then $\mathbb{F}_q[x]/(p(x))$ is a field. Furthermore, $\mathbb{F}_q$ can be identified as a subfield of this field by an embedding $a \mapsto a + (p(x))$. And so $\mathbb{F}_q[x]/(p(x))$ can be regarded as an extension field of $\mathbb{F}_q$. In fact it is an extension of degree $m$, do you see why?

And I am not sure about you other question, I do not know what you mean by 'primitive'?
I dun quite get it yet.

Im not quite sure what is primitive but what I read is that a monic polynomial of degree m with maximum order is said to be primitive

4. Originally Posted by classic_phohe
I dun quite get it yet.
What does not make sense? Have you ever talked about talking a ring modulo an ideal in class? This is precisely what is being down. The ideal here is $(p(x))$ (or in notation $\left< p(x)\right>$ is you perfer) and the ring here is $\mathbb{F}_q[x]$. I am asking if you form this factor ring you get a field, and this field is an extension field of $\mathbb{F}_q$. By this method you can construct extension fields if you are given an irreducible polynomial.

5. Originally Posted by ThePerfectHacker
What does not make sense? Have you ever talked about talking a ring modulo an ideal in class? This is precisely what is being down. The ideal here is $(p(x))$ (or in notation $\left< p(x)\right>$ is you perfer) and the ring here is $\mathbb{F}_q[x]$. I am asking if you form this factor ring you get a field, and this field is an extension field of $\mathbb{F}_q$. By this method you can construct extension fields if you are given an irreducible polynomial.
erm..nope..I didnt take this subject in class, Im reading this on my own interest. So I really found a lot of doubt in it. what about composite field? what is the purpose of having composite field?