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

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

## marked as duplicate by Jyrki Lahtonen abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 7 '17 at 5:53

• 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 7 '17 at 4:15
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.