Finite local length exactly when no common local branch

Translating the coordinates by $p$ replaces ${\mathfrak{m}}_p$ by $(x,y)$ and induces a $k$-algebra automorphism of $R$ carrying $f,g$ to nonzero elements of $(x,y)$; lengths and primary-ness are unchanged, so assume $p=0$ and $\mathfrak{m}=(x,y)$. Then $R$ is a Noetherian UFD of dimension $2$, $\mathrm{ht}(\mathfrak{m})=2$, and $O$ is a two-dimensional regular local ring with maximal ideal $\mathfrak{m}O$.

Every prime $\mathfrak{q}$ of $R$ with $(f,g)\subseteq \mathfrak{q}\subseteq \mathfrak{m}$ has height $1$ or $2$: height $0$ is impossible because $f\mathrm{\ne }0$, and height at most $2$ since $\mathfrak{q}\subseteq \mathfrak{m}$ and $\mathrm{ht}(\mathfrak{m})=2$. A height-two prime contained in $\mathfrak{m}$ equals $\mathfrak{m}$, and a height-one prime is generated by an irreducible element $h$, which lies in $\mathfrak{m}$ exactly when $h(p)=0$.

The ring $O$ is a UFD: factor a numerator in $R$ and discard the irreducible factors outside $\mathfrak{m}$, which become units. Each remaining factor $h$ stays prime and nonunit, since $O\mathrm{/}(h)O$ is a localisation of the domain $R\mathrm{/}(h)$ at denominators disjoint from $(h)$. Clearing denominators and using factorisation in $R$ proves uniqueness. Also, if $I\subseteq O$ has radical $\mathfrak{m}O=(x,y)O$, then $x^a,y^b\in I$ for some positive $a,b$, and every monomial of degree $a+b-1$ is divisible by $x^a$ or $y^b$; hence $(\mathfrak{m}O{)}^{a+b-1}\subseteq I$. For a quotient $Q=O\mathrm{/}I$ with this containment, the finite filtration by powers of $\mathfrak{m}O$ has finite-dimensional $k$-vector space factors, each killed by $\mathfrak{m}O$. Refine each factor by a finite vector-space flag to obtain simple factors $k=O\mathrm{/}\mathfrak{m}O$. Therefore ${\mathrm{\ell }}_O(Q)={\dim }_{k}Q$. Finally, for any quotient map $O\twoheadrightarrow V$ and a $V$-module $M$, the $O$-submodules and $V$-submodules coincide, so ${\mathrm{\ell }}_O(M)={\mathrm{\ell }}_V(M)$.

By [F4], the primes of $O$ containing $(f,g)O$ are exactly the primes $\mathfrak{q}O$ with $(f,g)\subseteq \mathfrak{q}\subseteq \mathfrak{m}$, and $\mathfrak{q}O=\mathfrak{m}O$ exactly for $\mathfrak{q}=\mathfrak{m}$. Hence $(f,g)O$ is $\mathfrak{m}O$-primary, equivalently $\sqrt{(f,g)O}=\mathfrak{m}O$ by [F5], if and only if $\mathfrak{m}$ is the only prime of $R$ with $(f,g)\subseteq \mathfrak{q}\subseteq \mathfrak{m}$: here the radical is the intersection of the primes containing the ideal The radical of an ideal is the intersection of the prime ideals containing it.

By step 1.2, the condition of step 2.1 fails exactly when there is a height-one prime $(h)\subseteq \mathfrak{m}$ containing $(f,g)$, that is, exactly when $f$ and $g$ have a common irreducible factor $h$ with $h(p)=0$. This proves the equivalence of condition (3) with condition (1).

Assume the conditions hold, so $(f,g)O$ is $\mathfrak{m}O$-primary and $\sqrt{(f,g)O}=\mathfrak{m}O$ with $f,g\in \mathfrak{m}O$ and $\dim O=2$. Then $(f,g)$ is a system of parameters and $(f,g)O$ is a parameter ideal of the regular local ring $O$ [F5]. The quotient $O\mathrm{/}(f,g)O$ is Noetherian [F3], it is local with maximal ideal $\mathfrak{m}O\mathrm{/}(f,g)O$, and by [F4] its only prime is that maximal ideal; hence every prime of it is maximal, so it is Artinian and therefore of finite length as an $O$-module [F6]. In particular (1) implies (2), and the described quotient is zero-dimensional of finite length.

Conversely assume there is a common irreducible factor $h$ with $h(p)=0$, so that $(f,g)\subseteq (h)\subseteq \mathfrak{m}$ and $(h)$ has height one. Then $(h)O$ is a prime of $O$ containing $(f,g)O$ and different from $\mathfrak{m}O$ because $\dim O=2>\mathrm{ht}((h))=1$ [F1, F3]. Its image in $O\mathrm{/}(f,g)O$ is prime and not maximal [F4], so $O\mathrm{/}(f,g)O$ is not Artinian; by [F6] it cannot have finite length, so its length is infinite. Hence (2) implies (1), the length is infinite whenever the conditions fail, and the equivalence of (1), (2) and (3) is established.