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Let $\mathfrak{q}$ be a prime ideal of $C$. Its quotient ring $D=C/\mathfrak{q}$ is a nonzero integral domain and a finite-dimensional $K$-vector space. For $0\ne a\in D$, multiplication by $a$ is an injective $K$-linear map $D\to D$, because $D$ has no zero divisors. An injective endomorphism of a finite-dimensional vector space is surjective. In particular $ab=1$ for some $b\in D$. Every nonzero element is therefore invertible, so $D$ is a field and $\mathfrak{q}$ is a maximal ideal.

To prove finiteness, select any $r$ distinct maximal ideals ${\mathfrak{m}}_1,\dots ,{\mathfrak{m}}_r$. Distinct maximal ideals are comaximal: ${\mathfrak{m}}_i+{\mathfrak{m}}_j$ strictly contains ${\mathfrak{m}}_i$, so must equal $C$. The Chinese remainder theorem gives a surjectionHere is the needed surjectivity explicitly. For each $j\ne i$, choose $a_{ij}\in {\mathfrak{m}}_j$ with $a_{ij}\equiv 1(\mathrm{mod}{\mathfrak{m}}_i)$, using comaximality. Then $e_i={\prod }_{j\ne i}{a}_{ij}$ is one modulo ${\mathfrak{m}}_i$ and zero modulo the other selected ideals. Given any residue tuple represented by $b_i$, the element ${\sum }_i{b}_i{e}_i$ maps to that tuple. Thus the displayed map really is surjective.

Each nonzero field quotient has $K$-dimension at least one, soThere cannot be more than ${\dim }_{K}C$ maximal ideals, because otherwise one could select a finite list exceeding this bound. ConsequentlyThis proves the description of the spectrum of a finite-dimensional commutative algebra. If $C=0$, its spectrum is empty and the same conclusion holds.