= Solution
{c}
An <irreducible polynomial> has positive degree, so its <ideal> is proper: otherwise $1=fh$ would contradict degrees. Every nonzero class $[g]$ has a representative not divisible by $f$. Irreducibility gives $\gcd(g,f)=1$, and the <Euclidean algorithm> supplies <polynomials> $u,v$ with $ug+vf=1$. Passing to the <quotient> gives $[u][g]=1$. Thus the <quotient> is a nonzero <commutative ring> in which every nonzero element has an inverse, hence \b[a <field>]. Constants embed injectively because no nonzero constant lies in the positive-degree <ideal> $(f)$.
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