FInd non-zero elements are primitive in a field

  • #1
85
0

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
 

Answers and Replies

  • #3
22,129
3,297
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
22,129
3,297

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.
 
  • #5
85
0
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
 
  • #6
22,129
3,297
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

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

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.
 
  • #7
85
0
##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

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

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

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

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

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

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
 
  • #8
22,129
3,297
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
85
0
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)
 
  • #10
22,129
3,297
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.
 
  • #11
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,379
1,039
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.
 
Last edited:

Related Threads on FInd non-zero elements are primitive in a field

Replies
7
Views
884
Replies
7
Views
2K
  • Last Post
Replies
16
Views
2K
  • Last Post
Replies
5
Views
5K
Replies
8
Views
1K
Replies
3
Views
11K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
0
Views
2K
Top