0

Our text makes a statement that by using the Euclidean Algorithm, it can be shown that $\mathbb{F}_p[x]/\langle f(x)\rangle$, with $f(x)$ being an irreducible polynomial of degree $m$, is a finite field $\mathbb{F}_q$ where $q=p^m$.

I have found a number of answers on the here to the fact that $q=p^m$, such as this question: Problem related polynomial ring over finite field of intergers

However, I cannot seem to find any concrete proof as to why $\mathbb{F}_p[x]/\langle f(x)\rangle$ is a field. The linked question makes the statement that it is because of the "irreducibility of $f(x)$, however it provides no proof of this.

Is this a known fact that we did not review, or is this something more complicated? Our text only has a very quick review of finite field theory, so I can easily see this being something the author left out.

Community
  • 1
user3784030
  • 1,013
  • 2
    The fact that $f(x)$ is irreducible makes $(f(x))$ a maximal ideal in $\mathbb F_p[x]$ – Mathmo123 Mar 02 '16 at 08:41
  • 1
    This is a general fact about fields: if $F$ is any field, then all irreducible polynomials $f \in F[x]$ generate a maximal ideal of $F[x]$ (i.e. $F[x]/\langle f \rangle $ is a field). – Crostul Mar 02 '16 at 09:44

2 Answers2

2

Here is another simple proof using some basic algebra: let $f \in \Bbb{F}_p[x]$ be irreducible of degree $m$. Then $R=\Bbb{F}_p[x] / \langle f \rangle$ is an $\Bbb{F}_p$-vector space generated by $1 + \langle f \rangle , x + \langle f \rangle , \dots , x^{m-1} + \langle f \rangle$. In particular $R$ is finite dimensional, and it has cardinality $p^m$.

In order to show that $R$ is a field, it is enough to show that $R$ is a domain (since every finite domain is a field): this is equivalent on proving that $\langle f \rangle$ is a prime ideal of $\Bbb{F}_p[x]$.

Suppose $g,h \in \Bbb{F}_p[x]$ are polynomials satisfying $gh \in \langle f \rangle$. Then there exists some $q \in \Bbb{F}_p[x]$ such that $gh=fq$. Since $f$ is irreducible, $f$ appears in the factorization of $g$ or $h$: hence $g \in \langle f \rangle$ or $h \in \langle f \rangle$. This proves that $\langle f \rangle$ is a prime ideal of $\Bbb{F}_p[x]$, and you are done.

Crostul
  • 37,500
1

Take an element $[g] = g + \langle f \rangle$ of $\mathbb{F}_p[x]/\langle f(x)\rangle$.

$[g] = [0]$ means that $f \mid g$. If $[g] \ne [0]$, thus, we have $f \nmid g$. Since $f$ is irreducible, that is, up to units, its divisors are only $1$ and $f$, this means that the gcd of $g$ and $f$ will be $1$.

And now, use the (extended) Euclidean algorithm to find $u, v \in \mathbb{F}_p[x]/\langle f(x)\rangle$ such that $$ g u + f v = 1. $$ Passing to $\mathbb{F}_p[x]/\langle f(x)\rangle$, you have $$ [g] \cdot [u] = [1], $$ so $[u]$ is the inverse of $[g]$.

Andreas Caranti
  • 70,862