Lemma 37.36.1. Let R = \mathop{\mathrm{colim}}\nolimits R_ i be colimit of a directed system of rings whose transition maps are faithfully flat. Then the number of minimal primes of R taken as an element of \{ 0, 1, 2, \ldots , \infty \} is the supremum of the numbers of minimal primes of the R_ i.
37.36 Étale neighbourhoods and branches
The number of (geometric) branches of a scheme at a point was defined in Properties, Section 28.15. In Varieties, Section 33.40 we related this to fibres of the normalization morphism. In this section we discuss a characterization of this number in terms of étale neighbourhoods.
Proof. If A \to B is a flat ring map, then \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A) maps minimal primes to minimal primes by going down (Algebra, Lemma 10.39.19). If A \to B is faithfully flat, then every minimal prime is the image of a minimal prime (by Algebra, Lemma 10.39.16 and 10.30.7). Hence the number of minimal primes of R_ i is \geq the number of minimal primes of R_{i'} if i \leq i'. By Algebra, Lemma 10.39.20 each of the maps R_ i \to R is faithfully flat and we also see that the number of minimal primes of R is \geq the number of minimal primes of R_ i. Finally, suppose that \mathfrak q_1, \ldots , \mathfrak q_ n are pairwise distinct minimal primes of R. Then we can find an i such that R_ i \cap \mathfrak q_1, \ldots , R_ i \cap \mathfrak q_ n are pairwise distinct (as sets and hence as prime ideals). This implies the lemma. \square
Lemma 37.36.2. Let X be a scheme and x \in X a point. Then
the number of branches of X at x is equal to the supremum of the number of irreducible components of U passing through u taken over elementary étale neighbourhoods (U, u) \to (X, x),
the number of geometric branches of X at x is equal to the supremum of the number of irreducible components of U passing through u taken over étale neighbourhoods (U, u) \to (X, x),
X is unibranch at x if and only if for every elementary étale neighbourhood (U, u) \to (X, x) there is exactly one irreducible component of U passing through u, and
X is geometrically unibranch at x if and only if for every étale neighbourhood (U, u) \to (X, x) there is exactly one irreducible component of U passing through u.
Proof. Parts (3) and (4) follow from parts (1) and (2) via Properties, Lemma 28.15.6.
Proof of (1). Let \mathop{\mathrm{Spec}}(A) be an affine open neighbourhood of x and let \mathfrak p \subset A be the prime ideal corresponding to x. We may replace X by \mathop{\mathrm{Spec}}(A) and it suffices to consider affine elementary étale neighbourhoods (U, u) in the supremum as they form a cofinal subsystem. Recall that the henselization A_\mathfrak p^ h is the colimit of the rings B_\mathfrak q over the category of pairs (B, \mathfrak q) where B is an étale A-algebra and \mathfrak q is a prime lying over \mathfrak p with \kappa (\mathfrak q) = \kappa (\mathfrak p), see Algebra, Lemma 10.155.7. These pairs (B, \mathfrak q) correspond exactly to the affine elementary étale neighbourhoods (U, u) by the correspondence between rings and affine schemes. Observe that irreducible components of \mathop{\mathrm{Spec}}(B) passing through \mathfrak q are exactly the minimal prime ideals of B_\mathfrak q. The number of minimal primes of A_\mathfrak p^ h is the number of branches of X at x by Properties, Definition 28.15.4. Observe that the transition maps B_\mathfrak q \to B'_{\mathfrak q'} in the system are all flat. Since a flat local ring map is faithfully flat (Algebra, Lemma 10.39.17) we see that the lemma follows from Lemma 37.36.1.
Proof of (2). The proof is the same as the proof of (1), except that we use Algebra, Lemma 10.155.11. There is a tiny difference: given a separable algebraic closure \kappa ^{sep} of \kappa (x) for every étale neighbourhood (U, u) we can choose a \kappa (x)-embedding \phi : \kappa (u) \to \kappa ^{sep} because \kappa (u)/\kappa (x) is finite separable (Morphisms, Lemma 29.36.15). Hence we can look at the supremum over all triples (U, u, \phi ) where (U, u) \to (X, x) is an affine étale neighbourhood and \phi : \kappa (u) \to \kappa ^{sep} is a \kappa (x)-embedding. These triples correspond exactly to the triples in Algebra, Lemma 10.155.11 and the rest of the proof is exactly the same. \square
We will need a relative variant of the lemma above.
Lemma 37.36.3. Let X \to S be a morphism of schemes and x \in X a point with image s. Then
the number of branches of the fibre X_ s at x is equal to the supremum of the number of irreducible components of the fibre U_ s passing through u taken over elementary étale neighbourhoods (U, u) \to (X, x),
the number of geometric branches of the fibre X_ s at x is equal to the supremum of the number of irreducible components of the fibre U_ s passing through u taken over étale neighbourhoods (U, u) \to (X, x),
the fibre X_ s is unibranch at x if and only if for every elementary étale neighbourhood (U, u) \to (X, x) there is exactly one irreducible component of the fibre U_ s passing through u, and
X is geometrically unibranch at x if and only if for every étale neighbourhood (U, u) \to (X, x) there is exactly one irreducible component of U_ s passing through u.
Lemma 37.36.4. Let X \to S be a smooth morphism of schemes. Let x \in X with image s \in S. Then
The number of geometric branches of X at x is equal to the number of geometric branches of S at s.
If \kappa (x)/\kappa (s) is a purely inseparable1 extension of fields, then number of branches of X at x is equal to the number of branches of S at s.
Proof. Follows immediately from More on Algebra, Lemma 15.106.8 and the definitions. \square
[1] In fact, it would suffice if \kappa (x) is geometrically irreducible over \kappa (s). If we ever need this we will add a detailed proof.
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