Proposition 9.3.98D696ED2F2AC

Proof. Let $A$ be the representative on a parametrization polydisc. Two local parametrizations of $(A,0)$ have covering loci which are open and dense in $A$ near $0$; their intersection is therefore nonempty, and at its points $A$ is a manifold whose dimension equals both base dimensions. Hence $d$ is determined by the germ. A regular point of $A$ has a neighborhood in $A$ which is a manifold; this neighborhood meets the dense covering locus, which has dimension $d$, so the regular point has dimension $d$ as well. Now let $y\in A$, let $B_1,\ldots ,B_ p$ be representatives of the branches of $(A,y)$, and fix $i$. Since $B_ i$ is not contained in $\bigcup _{k\neq i}B_ k$, corollary 9.3.6 gives points of the covering locus of a parametrization of $B_ i$ outside the other branches. Near such a point, $A$ coincides with $B_ i$ and is a manifold of dimension $\dim B_ i$; such points form a nonempty open subset of $A$, which meets the covering locus of $A$. Hence $\dim B_ i=d$.

Let $(B,0)\subset (A,0)$ be germs with both irreducible, and let $J_ A\subset J_ B$ be their prime ideals. For linear coordinates $z$ and $0\leq k\leq n$, let $\mathcal O_ k$ denote the ring of germs depending only on $z_1,\ldots ,z_ k$. For a nonzero prime ideal $J$, Step 1 of the proof of theorem 9.3.4 chooses the coordinates downward from $k=n$. As long as $J\cap \mathcal O_ k\neq 0$, it makes a linear change of $z_1,\ldots ,z_ k$ in which a chosen nonzero element of $J\cap \mathcal O_ k$ is distinguished in $z_ k$; it stops at the largest $d$ with $J\cap \mathcal O_ d=0$, which is the base dimension. The rest of that proof applies to every such choice. Since finitely many nonzero germs become distinguished in the last coordinate after one suitable linear change of coordinates, we may make these changes for $J_ B$ and, at the stages $k$ with $J_ A\cap \mathcal O_ k\neq 0$, simultaneously for $J_ A$. This gives parametrizations of both germs in common coordinates (if $J_ A=0$, the germ $(A,0)$ is the germ of $\mathbb C^ n$, with $d=n$). Since $J_ A\subset J_ B$, the ideal $J_ B\cap \mathcal O_ k$ is nonzero whenever $J_ A\cap \mathcal O_ k$ is, so $\dim B\leq \dim A$.

Suppose that the dimensions are equal, so that both germs are finite over the same base, and let $\delta _ A$, $\delta _ B$ be the two discriminants. By lemma 9.2.9, the set $A'$ of points of the connected covering locus of $A$ lying over $\{ \delta _ B\neq 0\} $ is a connected manifold of dimension $d$, dense in $A$. The points of $B$ over $\{ \delta _ A\delta _ B\neq 0\} $ form a nonempty submanifold of $A'$ of the same dimension, hence an open subset of $A'$, since the inclusion is an injective immersion between manifolds of the same dimension and therefore open by the inverse function theorem. Thus $B\cap A'$ is an analytic subset of the connected manifold $A'$ with nonempty interior; since proper analytic subsets of connected manifolds are nowhere dense (lemma 9.2.8), $B\supset A'$. Since $B$ is closed and $A'$ is dense in $A$, we get $(B,0)=(A,0)$. Hence a proper subgerm has strictly smaller dimension. For a general proper subgerm $B$ of $(A,0)$, each branch of $B$ is a proper subgerm of $(A,0)$, and the same conclusion follows. $\square$