Let $A$ be a local ring, which we can assume is reduced. Let $k$ be the residue field of $A$.
In the Stacks project (https://stacks.math.columbia.edu/tag/06DT), I have learned some notion of the number of "geometric branches" of $A$ as being the number of minimal primes of the strict henselianization of $A$. Equivalently (as shown in that Stacks project page), it is the number of maximal ideals $\mathfrak{m}'$ of $A'$ ($A'$ being the integral closure of $A$ in its total ring of fractions), each weighted by the separable degree of $A'/\mathfrak{m}'$ over $k$.
This definition makes sense to me intuitively, on the one hand because if $A$ is the stalk of the structure sheaf of some curve $X$ at a point $x \in X$, then the etale topology on $X$ (the structure sheaf on which $A^{sh}$ is the stalk of) should be fine enough to distinguish between the branches of $X$ passing through $x$, and because on the other hand taking the normalization of a curve at a point (similarly to a blowup) is supposed to separate the branches.
However, I am having difficulty working out all of this in the simplest possible geometric example. For simplicity, let $k$ be an algebraically closed field. As we know from calculus or undergraduate algebraic geometry, if we have a (affine since we only care about the local ring) curve $X$ in $\mathbf{A}_k^2$ cut out by a polynomial
$$f = \sum_{i \geq 1} f_i \in k[X, Y],$$
$f_i$ being homogeneous of degree $i$ (the absence of constant term meaning we are assuming the curve passes through the origin), then the tangent lines to $X$ at the origin are cut out by $f_{i_0}$, where $i_0 \geq 1$ is the smallest $i$ such that $f_i \neq 0$. So we should expect that $\mathcal{O}_{X, (0, 0)}$ has at most $i_0$ branches according to the above definition, with equality if and only if $f_{i_0}$ factors over $k$ into $i_0$ distinct linear factors. Is this true ?
I can make some sense of this in one special case: if $i_0 = 2$ and $f_{i_0}$ has no repeated factors, then WLOG we can assume $f_{i_0} = (X-Y)(X+Y)$, and I can show that $\mathcal{O}_{X, (0, 0)}^{sh}$ is isomorphic to the strict henselianization to the localization at $(X, Y)$ of $k[X, Y]/(XY)$, because if we send $X$ to $\alpha X + \beta Y$ and $Y$ to $\alpha X – \beta Y$ then $XY$ maps to $\alpha^2X^2 – \beta^2Y^2$, and since $\mathcal{O}_{X, (0, 0)}^{sh}$ is henselian we can choose $\alpha, \beta \in 1 + \mathfrak{m}$ to be such that this is exactly $f$, and it is easy to check (thanks to where $\alpha, \beta$ live) that this will provide an isomorphism of henselian local rings. This means the number of branches is the same as if we had no terms after $f_{i_0}$, i.e. if $A = (k[X, Y]/(XY))_{(X, Y)}$, which is explicit enough that we can compute the number of branches to be $2$ (for example by using the equivalent definition involving the normalization).
But in the case where there are repeated linear factors, or more than 2 linear factors, I am completely stuck (in particular from examples it no longer seems to be true that the henselianization is indifferent to the terms after $f_{i_0}$). Is my claim about the branches still true ? Am I just missing some algebraic manipulation with the henselianization, or is there a conceptual step that I have not figured out ?
Best Answer
Let us assume that $k$ is separably closed. In the case of local rings of finite type $k$-schemes such as here, it is usually easier to look at the completion with respect to the maximal ideal. Note that such rings are excellent, so the number of geometric branches of $\mathcal{O}_{X, (0,0)}$ will coincide with the number of minimal primes in the completion $\widehat{\mathcal{O}}_{X, (0,0)}$ by https://stacks.math.columbia.edu/tag/0C2E.
The completion in this case is $k[[x,y]]/(f)$. Since $k[[x,y]]$ is a UFD, we can factor $f$ into a product of powers of distinct irreducibles $f = p_1^{n_1} \cdot p_2^{n_2} \cdot \ldots \cdot p_l^{n_l}$. The number of minimal primes in this case will be the number $l$ of distinct irreducible terms in the decomposition above. Let $\mathfrak{m}$ denote the maximal ideal of $k[[x,y]]$. For each irreducible power series $p_j$, we can write $p_j = h_j + g_j$, where $h_j$ is a homogeneous polynomial of degree the same as the order of vanishing $d_j = \text{ord}(p_j)$, and $g_j \in \mathfrak{m}^{d_j+1}$. By reducing the equality $$f = p_1^{n_1} \cdot p_2^{n_2} \cdot \ldots \cdot p_l^{n_l}$$ modulo $\mathfrak{m}^{i_0+1}$, we see that $$f_{i_0} = h_1^{n_1} \cdot h_2^{n_2} \cdot \ldots \cdot h_l^{n_l}$$ In paticular, by counting degrees, we see that $$ i_0 = \sum_{i=1}^l \text{deg}(h_i) \cdot n_i$$ Since none of the $p_i$ are units, we have $\text{deg}(h_i) \geq 1$for all $i$, and so the equality above makes it clear that the number of branches $l$ is indeed at most the degree of the lowest homogeneous term $i_0$.
On the other hand, equality does $\underline{not}$ guarantee that the polynomial $f_{i_0}$ splits into $i_0$ distinct linear factors. For an example, take $f = x^2 + xy^2$. The factorization over the completion is $$ f = x(x+y^2)$$ I claim that the irreducibles $x$ and $x+y^2$ do not differ by a unit, and so the number of branches is $2 = i_0$. Note however that $f_{i_0} = x^2$ does not factor into distinct terms.
In order to see the claim, suppose for the sake of contradiction that we have $x + y^2 = xu$ for a unit $u \in k[[x,y]]$. By looking at the linear terms, we see that we can write $u = 1+h$ for some $h \in \mathfrak{m}$. Using $x +y^2 = x(1+h)$, we see that $xh = y^2$. This would contradict uniqueness of factorization for $y^2$, since $x$ is an irreducible element that does not differ from $y$ by a unit.