2

Let $f$ be a irreducible polynomial in $\mathbb{F}_p[x]$ of degree $n$.

Why are there $p^n$ elements in the quotient field $\mathbb{F}_p [x]/ (f)$

I am having some difficulty convincing myself this is true.

Any help or insight is deeply appreciated

PIandpie
  • 2,177
  • 1
    You don't need $f$ to be irreducible for saying $\mathbb{F}p[x]/(f(x)) = { \sum{k=0}^{n-1} a_k x^k + (f(x)), a_k \in \mathbb{F}_p}$ is a ring with $p^n$ elements. If $f$ is irreducible then $\mathbb{F}_p[x]/(f(x))$ is a finite integral domain and hence a finite field. – reuns Sep 07 '17 at 04:15
  • Finding the dupe was a bit more taxing than I had hoped. The original used the term cardinality and escaped my searches initially. – Jyrki Lahtonen Sep 07 '17 at 05:54

1 Answers1

3

Since $f(x)$ has degree $n$, every element of $\mathbb{F}_p[x]/(f)$ can be written uniquely in the form $$ c_0+c_1x+\dots+c_{n-1}x^{n-1}+(f)$$ with $c_0,\dots,c_{n-1}\in\mathbb{F}_p$. Since there are $p$ choices for each $c_i$, it follows that $\mathbb{F}_p[x]/(f)$ has $p^n$ elements.

carmichael561
  • 54,793