Let f(x) be an irreducible polynomial of degree n in Zp[x]. Show that the field E =

Zp[x]/<f(x)> has exactly p^n elements.

f(x) is irreducible, so it cannot be factored into a product of polynomials of lower degree.

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- Jan 27th 2011, 05:49 PMmathematicIrreducible Polynomial
Let f(x) be an irreducible polynomial of degree n in Zp[x]. Show that the field E =

Zp[x]/<f(x)> has exactly p^n elements.

f(x) is irreducible, so it cannot be factored into a product of polynomials of lower degree. - Jan 27th 2011, 06:03 PMLoblawsLawBlog
Think about what the elements of $\displaystyle \mathbb{Z}_p[x]/\langle f(x)\rangle$ look like and what it means for them to be different. Maybe that's too vague. I mean, for example, consider the coset $\displaystyle a=(x^n+1)+\langle f(x)\rangle$. If you can write $\displaystyle x^n+1=f(x)q(x)+r(x)$, then isn't $\displaystyle r(x)+\langle f(x)\rangle$ the same coset as $\displaystyle a$?

And how many different remainders can you get when you divide by $\displaystyle f(x)$? - Jan 27th 2011, 06:25 PMmathematic
Well, I started by letting f(x)=a_nx^n+.....+a1x+a0

I don't know if that helps me any - Jan 27th 2011, 06:42 PMLoblawsLawBlog
That helps to at least figure out what kind of remainders you can get when dividing by $\displaystyle f(x)$. That division is the key to the whole problem because when dividing by an nth degree polynomial, $\displaystyle f(x)$ in this case, the remainder must be of degree strictly less than n. So if you have some coset $\displaystyle g(x)+\langle f(x)\rangle$, you can divide $\displaystyle g(x)$ by $\displaystyle f(x)$ and represent the same coset by the remainder, which is guaranteed to be a polynomial of degree at most $\displaystyle n-1$.

The nice thing is that you don't have to actually do any division at all (although this is one way to find inverses if you want). You just have to show that each coset can be represented by a polynomial of degree less than $\displaystyle n$. Now since the coefficients come from a finite field, how many different polynomials can you form? That's your answer.

You do need to show that each of these different polynomials does actually represent a different coset, but you can do that by contradiction. Say that two different polynomials, each of degree at most $\displaystyle n-1$ represent the same coset. Write out what this means and use the fact that $\displaystyle f(x)$ is irreducible to get the contradiction.

edit: Actually the fact that f(x) is irreducible doesn't really come into play in that last paragraph. I must have been thinking of using that fact to prove the existence of inverses, but you already know it's a field.