# FInd non-zero elements are primitive in a field

Tags:
1. Jun 3, 2015

### HaLAA

1. The problem statement, all variables and given/known data
Construct $\mathbb{F}_{16}$ as a quotient of $\mathbb{Z}_2[X]$. How many non-zero elements are primitive in this field? Calculate $|GL2_(\mathbb{F}_16)|$.

2. Relevant equations

Primitive Theorem

3. The attempt at a solution

For the first question, I don't know how to construct $\mathbb{F}_{16}$ as a quotient of $\mathbb{Z}_2[X]$. My guess is $\mathbb{Z}_2[X]/p(x)$ where $p(x)$ is not in $\mathbb{Z}_2[X]$.

And the second part which should be 16^4-16^3

2. Jun 3, 2015

### Fredrik

Staff Emeritus
3. Jun 3, 2015

### micromass

Staff Emeritus
Right, you will have to construct $\mathbb{F}_{16}$ as $\mathbb{Z}_2[X]/(p(x))$ for some polynomial. Now there are two essential facts:

1) The quotient will have to be a field
2) The quotient will have to have 16 elements.

Which conditions on $p(x)$ will guarantee this?

4. Jun 3, 2015

### micromass

Staff Emeritus
If $p(x)$ is not in $\mathbb{Z}_2[X]$, then the quotient makes no sense. So that is not correct.

5. Jun 3, 2015

### HaLAA

P (x) would be a maximal ideal and irreducible, I think x^3-x-1

6. Jun 3, 2015

### micromass

Staff Emeritus
$p(x)$ is just a polynomial, and a polynomial being a maximal ideal makes no sense. You are correct that $p(x)$ will have to be irreducible. Indeed, if $p(x)$ is irreducible, then

$$\mathbb{Z}_2[X]/(p(x))$$

will be a field, which is what you want. But you also want the field to have $16$ elements, under what conditions on $p(x)$ will that happen?

That is not correct.

7. Jun 3, 2015

### HaLAA

x^2-2,
If I didn't make mistakes, p (x)=x^2 - 3

8. Jun 3, 2015

### micromass

Staff Emeritus
Can you stop guessing and actually think about it? In either case, if you put up a proposal, at least try to motivate it and say why it is true.

9. Jun 3, 2015

### HaLAA

P (x)=x^4+1, we have p (0) mod 2 =1 and p (1)=2 which show p (x) is irreducible in Z_2, hence F_16 isomorphic with Z_2 [x]/p (x)

10. Jun 3, 2015

### micromass

Staff Emeritus
First of all, $p(1) = 0$ since $2=0$ in $\mathbb{Z}_2$.
Second of all, saying that the polynomials $p(x)$ has no roots in $\mathbb{Z}_2$ is not at all the same as saying that it is irreducible.
Third of all, even if $p(x)$ were irreducible, that would only show that $\mathbb{Z}_2[X]/(p(X))$ is a field, you still need to find an argument for why it has $16$ elements.
Fourth of all, $\mathbb{Z}_2[X]/p(X)$ makes no sense. You cannot quotient out an element of $\mathbb{Z}_2[X]$, you can only quotient out an ideal.

11. Jun 3, 2015

### SammyS

Staff Emeritus
Sorry, but I was having difficulty reading the OP, so here it is with readable Latex:
Construct $\ \mathbb{F}_{16}\$ as a quotient of $\ \mathbb{Z}_2[X]\ .\$ How many non-zero elements are primitive in this field? Calculate $\ |GL2_{({{\mathbb{F}}_{16}})}|\$ .

2. Relevant equations
Primitive Theorem

3. The attempt at a solution
For the first question, I don't know how to construct $\ \mathbb{F}_{16}\$ as a quotient of $\ \mathbb{Z}_2[X]\$. My guess is $\ \mathbb{Z}_2[X]/p(x)\$ where $\ p(x)\$ is not in $\ \mathbb{Z}_2[X]\$.

... and there it is.

Last edited: Jun 3, 2015