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I want to determine the number of elements in $\mathbb{Z}_p[x]/ \langle f \rangle$ where $f \in \mathbb{Z}_p[x]$ is an irreducible polynomial with $k$ degree bigger than 2. Is the number of elements always $p^k$? And are the equivalence classes $[f_1],[f_2], \dots, [f_{p^k}]$ always given be the polynomials in $\mathbb{Z}_p$ with degree smaller than $k$?

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It seems that you really want to look at $\mathbb Z_p[x]/(f)$, no? For the question proper, think about polynomial division. – Dylan Moreland Jul 3 '12 at 20:05
Use the Euclidean algorithm! It works in $\mathbb{Z}/p\mathbb{Z}[x]$. – Cocopuffs Jul 3 '12 at 20:07
The cardinal of $\mathbb{Z}_p[x]/(f)$ is $p^k$ even if $f$ is not irreducible. But then, the ring is not a field. – francis-jamet Jul 3 '12 at 20:12
Please consider chopping the [factoring] tag: I don't think it applies here. – rschwieb Jul 3 '12 at 20:17
@francis In fact it works over any (finite) coefficient ring when $\rm\:f\:$ is monic, since the division algorithm works universally for monic polynomials. – Bill Dubuque Jul 3 '12 at 20:24
up vote 2 down vote accepted

Yes, you're on the right track of reasoning.

Use the fact that $\mathbb{F}[x]$ has a division algorithm to justify that all possible remainders (=all polynomials of degree strictly less than $k$) form a complete set of distinct residues.

To show that every polynomial is equivalent to a remainder, remember that modding out by $(f)$ is basically saying that $g\equiv h\iff f|(g-h)$. To show that the remainders are distinct, start by writing out what it would mean if two are equivalent, and show it's impossible.

After that, it's easy to count how many remainders! You have $p$ choices for each of the $k$ coefficients of a remainder.

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Thanks! That is exactly the argument what I was looking for. – joachim Jul 3 '12 at 20:16

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