Dimension of the field extension $F[x]/(f)=\deg(f)$ for an irreducible monic polynomial $f\in F[x]$

Question

Let $F$ be a field and let $f$ be an irreducible monic polynomial in $F$. Prove that the dimension of $F[x]/(f) = \deg(f)$.

Here is how I am proceeding. Let $$f(x) = x^n + a_{n-1}x^{n-1} + a_{n-2}x^{n -2} +\ldots+ a_1 x+a_0.$$

For any $g\in F[x]$, I can reduce its highest power using $x^n = -a_{n-1}x^{n-1} - a_{n-2}x^{n-2} - \ldots- a_0$, so obviously the degree of the polynomials in the field $F[x]/(f)$ is at most $\deg(f) - 1$, and the dimension will be of $\deg(f)$.

Am I going? Any other methods to prove this?

Answer 1

1. Denote $E=F[x]/(f)$. $F$ can be embedded in $E$ via the homomorphism $\pi:x\mapsto x+(f)$. Every extension field is a vector space over the smaller field, so $E$ is a vector space over $F$.
2. We want to prove that if $d$ is the degree of $f$, then the dimension of $E$ over $F$ is $d$ (usually this is denoted $[E : F]=d$). We will do so by finding a basis for $E$ over $F$. Consider the following set: $$ B=\left\{\overline{x^i}\mid i=0,\ldots,d-1\right\}=\{1+(f), x+(f), x^2+(f),\ldots,x^{d-1}+(f)\}\subseteq F[x]/(f),$$ where $\overline{x}$ denotes $\pi(x)$. We will now show that it is indeed a basis of $E$ over $F$.
3. Linear span: let $\overline{g}\in F[x]/(f)$ for some $g\in F[x]$ and we want to show that $\overline{g} \in \operatorname{span}(B)$. If $\deg(g)<\deg(f)$, say $g=\sum_{i=0}^{k} a_ix^i$ for some $k<d$, then since $\pi$ is a ring homomorphism, we have $$\begin{align}\overline{g}(x) &=\overline{a_0\cdot 1}+\overline{a_1\cdot x}+\ldots+\overline{a_k\cdot x^k} \\ &=a_0\left(1+(f)\right)+a_1(x+(f))+\ldots+a_k(x^k+(f))\in\operatorname{span}(B).\end{align}$$ So for $g\in F[x]$ of degree $<d$, we proved our claim. If $g$ is of degree $>d$, we can divide it by $f$, and the residue will be in the same coset of $g$ in $F[x]/(f)$.
4. Linear independence: assume that $\sum_{i=0}^{d-1} c_i (x^i+(f))=0$ for some $c_i\in F$. Denote $q(x)=\sum_{i=0}^{d-1} c_i x^i$, and we want to show that $q\equiv0$. By assumption (again using the fact that $\pi$ is ring homomorphism), $$\left(\sum_{i=0}^{d-1} c_i x^i\right)+(f)=0,$$ so we deduce that $$\sum_{i=0}^{d-1} c_i x^i \in (f),$$ and therefore there exists some $h\in F[x]$ so that $q(x)=f(x)\cdot h(x)$. Consider the degree of $f\cdot h$: $$d-1\ge \deg(q)=\deg(f\cdot h)=\deg(f)+\deg(h)=d+\deg(h)$$ and therefore $h\equiv0$, and so does $q$. This finishes the proof. ${\blacksquare}$