[proofplan] We prove directly that every nonzero residue class in $k[x]/(f)$ has a multiplicative inverse. Given a nonzero class $g+(f)$, the equality criterion in the quotient says $g \notin (f)$, so $f$ does not divide $g$. We then use the [division algorithm](/theorems/725) in $k[x]$ to prove a Bezout identity $a f + b g = 1$, and reducing this identity modulo $(f)$ gives an inverse for $g+(f)$. [/proofplan]

[step:Translate nonzero residue classes into nondivisibility by $f$] Let $A := k[x]$, and let $(f) \trianglelefteq A$ be the principal ideal generated by $f$. Since $f$ is irreducible, $f$ is nonzero and is not a unit in $A$. Let $g+(f) \in A/(f)$ be a nonzero element. By [citetheorem:8254], applied in the ring $A$ with ideal $(f)$, we have \begin{align*} g+(f) = 0+(f) \quad \iff \quad g-0 \in (f). \end{align*} Since $g+(f)$ is nonzero, it follows that $g \notin (f)$. Equivalently, $f$ does not divide $g$ in $A$. [/step]

[step:Construct a Bezout identity for $f$ and $g$]Define the ideal \begin{align*} J := (f,g) = \{a f + b g : a,b \in A\} \subset A. \end{align*} Since $f \in J$ and $f \neq 0$, the ideal $J$ contains a nonzero polynomial. Choose $h \in J$ nonzero of minimal degree among all nonzero elements of $J$. We claim that $h$ divides every element of $J$. Let $u \in J$. By the [polynomial division algorithm](/theorems/1706) over the field $k$, there exist polynomials $q,r \in A$ such that \begin{align*} u = qh + r \end{align*} and either $r=0$ or $\deg r < \deg h$. Since $u \in J$ and $h \in J$, the difference \begin{align*} r = u - qh \end{align*} also belongs to $J$. The minimality of $\deg h$ among nonzero elements of $J$ forces $r=0$. Hence $u=qh$, so $h$ divides $u$. Applying this to $u=f$ and $u=g$, we see that $h$ divides both $f$ and $g$. Since $f$ is irreducible and $h$ divides $f$, either $h$ is a unit in $A$ or $h$ is an associate of $f$. If $h$ were an associate of $f$, then the divisibility $h \mid g$ would imply $f \mid g$, contradicting $g \notin (f)$. Therefore $h$ is a unit. Because $h \in J$ and $h$ is a unit, $1 \in J$. Thus there exist $a,b \in A$ such that \begin{align*} a f + b g = 1. \end{align*}[/step]

[guided]The goal of this step is to manufacture an inverse for the class $g+(f)$. In a quotient by $(f)$, an equation of the form \begin{align*} b g \equiv 1 \pmod{(f)} \end{align*} is exactly what we need. This is why we try to prove a Bezout identity involving $f$ and $g$. Define \begin{align*} J := (f,g) = \{a f + b g : a,b \in k[x]\}. \end{align*} This is the ideal generated by $f$ and $g$. Since $f$ is irreducible, it is nonzero, so $J$ contains the nonzero polynomial $f$. Among all nonzero polynomials in $J$, choose one of least degree and call it $h$. We now show that this least-degree element behaves like a greatest common divisor. Let $u \in J$. Because $k$ is a field, the polynomial division algorithm applies in $k[x]$: there exist $q,r \in k[x]$ such that \begin{align*} u = qh + r \end{align*} and either $r=0$ or $\deg r < \deg h$. Since $u \in J$ and $h \in J$, and since $J$ is an ideal, the polynomial \begin{align*} r = u - qh \end{align*} also lies in $J$. If $r$ were nonzero, it would be a nonzero element of $J$ with degree smaller than $\deg h$, contradicting the choice of $h$. Hence $r=0$, so $u=qh$. Thus $h$ divides every element of $J$. In particular, $h$ divides $f$ and $h$ divides $g$. Since $f$ is irreducible, every divisor of $f$ is either a unit or an associate of $f$. If $h$ were an associate of $f$, then $h \mid g$ would imply $f \mid g$, meaning $g \in (f)$. But the previous step proved $g \notin (f)$. Therefore $h$ must be a unit. Because $h \in J$ and $h$ is invertible in $k[x]$, multiplying by $h^{-1}$ gives $1 \in J$. By the definition of $J$, this means there exist polynomials $a,b \in k[x]$ such that \begin{align*} a f + b g = 1. \end{align*} This is the Bezout identity we need.[/guided]

[step:Reduce the Bezout identity modulo $(f)$ to obtain an inverse] From the Bezout identity, choose $a,b \in A$ such that \begin{align*} a f + b g = 1. \end{align*} Passing to the [quotient ring](/page/Quotient%20Ring) $A/(f)$ gives \begin{align*} (a f + b g) + (f) = 1 + (f). \end{align*} Since $a f \in (f)$, [citetheorem:8254] gives \begin{align*} a f + (f) = 0 + (f). \end{align*} Therefore \begin{align*} (b+(f))(g+(f)) = b g + (f) = 1+(f). \end{align*} Thus $b+(f)$ is a multiplicative inverse of the arbitrary nonzero element $g+(f)$. Hence every nonzero element of $k[x]/(f)$ is invertible, so $k[x]/(f)$ is a field. [/step]