Is F2[x]/(x^8+x^4+x^3+x+1) a finite field?

(sorry i don't know how to make the html code work correctly)

Printable View

- Apr 20th 2009, 01:41 PMmlemilysIs F_2_[x] a finite field?
Is F2[x]/(x^8+x^4+x^3+x+1) a finite field?

(sorry i don't know how to make the html code work correctly)

- Apr 20th 2009, 02:20 PMmlemilys
or maybe someone could help explain the notation to me.

thanks! - Apr 20th 2009, 02:26 PMGammaIrreducible?
As long as that polynomial you are modding out by is irreducible you should be in business. If it is irreducible over $\displaystyle \mathbb{F}_2$, you have a field that is polynomials in $\displaystyle \mathbb{F}_2$ basically capped at degree 7 because you are associating your polynomial $\displaystyle p(x)=0$, so anytime you get a degree higher than 7 you can replace using $\displaystyle x^8=-x^4-x^3-x-1$. So you just have basically 2^7 elements in here as the binary choices of 1 or 0 for the coefficients on the different polynomials.

It is not immediate to me that the polynomial is irreducible though, it kind of looks like a cyclatomic polynomial, but maybe it is given to you or someone else will see a good reason why. - Apr 21st 2009, 06:20 PMmlemilys
i am not sure what cyclatomic means

i do know that because you can not factor it is irreducible

and that it is a finite field. i talked to my professor about it today and he said there is a more "algebraic" way to show that it is irreducible other than not factoring....

does anyone know why

i am having a hard time making sense of the chapter in my text

A Computational Introduction to Number Theory and Algebra

chapter 19 finite fields

thanks for any help anyone can offer! - Apr 21st 2009, 08:23 PMGammaIrreducibility
That is all it means to be irreducible is that it does not factor into smaller polynomials (constants don't count, has to be at least degree 1). So if you showed it can't factor in $\displaystyle \mathbb{F}_2$ you are done, it is irreducible, and since $\displaystyle \mathbb{F}_2[x]$ is a Euclidean Domain, and more importantly a PID, the ideal generated by your irreducible polynomial $\displaystyle (p(x))$ is prime and therefore maximal. Finally, this means $\displaystyle \frac{\mathbb{F}_2[x]}{(p(x))}$ is a field, my argument before explains why it is finite just by counting arguments and order considerations size $\displaystyle 2^7$.

I am not sure what other considerations he is talking about unless he means if you know a priori that $\displaystyle \frac{\mathbb{F}_2[x]}{(p(x))}$ is a field, then $\displaystyle (p(x))$ is maximal and therefore prime and therefore irreducible, but that is the point of the question as far as I can tell.

Hope that was some help.

PS cyclotomic are a particular variety of irreducible polynomial that come from multiplying roots of unity, they often take forms similar to that, but I am fairly certain that is actually not one.