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This question already has an answer here:

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

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marked as duplicate by Jyrki Lahtonen abstract-algebra Sep 7 '17 at 5:53

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    $\begingroup$ 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. $\endgroup$ – reuns Sep 7 '17 at 4:15
  • $\begingroup$ Finding the dupe was a bit more taxing than I had hoped. The original used the term cardinality and escaped my searches initially. $\endgroup$ – Jyrki Lahtonen Sep 7 '17 at 5:54
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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.

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