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Let $\mathfrak o$ be a Dedekind domain, $K$ its field of fractions, $L$ a finite separable field extension of $K$, $\mathfrak O$ the integral closure of $\mathfrak o$ in $L$. Let $\mathfrak p$ be a prime ideal of $\mathfrak o$. $A= \mathfrak O/{\mathfrak p \mathfrak O}$ is a finite dimensional commutative $\mathfrak o/\mathfrak p$-algebra.

Is it true that $A$ has only finitely many maximal ideals? What is the most general condition on $A$ for this to be true?

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$\newcommand{\mf}[1]{\mathfrak{#1}}$ Yes $A$ has only finitely many maximal ideals. Since $\mf{D}$ is Dedekind, $\mf{pD}$ admits a unique factorization into prime maximal ideals $\mf{p}_1\cdots\mf{p}_n$. Since the maximal ideals of $A$ are just the maximal ideals of $\mf{D}$ containing $\mf{pD}$, and in Dedekind domains a maximal ideal contains an ideal $I$ iff the maximal ideal divides $I$, the $\mf{p}_i$'s are precisely the maximal ideals of $A$.

This also follows from the finite dimensionality of $A$. Having infinitely many maximal ideals would mean that you can use the chinese remainder theorem to decompose $A$ into an infinite product of field extensions of $\mf{o}/\mf{p}$.

(Also your use of mathfrak is completely unnecessary.)

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In fact, your question is trivial: a finite dimensional algebra over a field is artinian, and artinian rings have only finitely many maximal ideals.

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