FInd non-zero elements are primitive in a field

[HaLAA]

Homework Statement

Construct $\mathbb{F}_{16}$ as a quotient of $\mathbb{Z}_2[X]$. How many non-zero elements are primitive in this field? Calculate $|GL2_(\mathbb{F}_16)|$.

Homework Equations

Primitive Theorem

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

 

[Fredrik]

Just a LaTeX tip: Use ## instead of $. See https://www.physicsforums.com/help/latexhelp/ for more.

 

[micromass]

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?

 

[micromass]

Homework Statement

My guess is $\mathbb{Z}_2[X]/p(x)$ where $p(x)$ is not in $\mathbb{Z}_2[X]$.

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

 

[HaLAA]

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?

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

 

[micromass]

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

$p(x)$ 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?

x^3-x-1

That is not correct.

 

[HaLAA]

$p(x)$ 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.

x^2-2,

$p(x)$ 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.

$p(x)$ 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.

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

 

[micromass]

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.

 

[HaLAA]

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.

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)

 

[micromass]

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)

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.

 

[SammyS]

Sorry, but I was having difficulty reading the OP, so here it is with readable Latex:

Homework Statement

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}})}|\$ .

Homework Equations

Primitive Theorem

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\$

... and there it is.