Ramification and Puiseux series in Several Variables

Question

In Algebraically Closed Fields Analogous to Fields of Puiseux Series, Rayner makes the following observation: given the equation $Z^2 = (X+Y)Y$, we can solve this equation for $Z$ using a generalization of Puiseux series, over say, $\mathbb{C}$. The solution depends on what we mean by this. If we take a graded lexicographic ordering on the coefficients where $X < Y$, we obtain two solutions:

$$Z = \pm Y + \frac{1}{2}X - \frac{1}{8}X^2 Y^{-1} + \dotsb. $$

On the other hand, if we pick a different order, where $Y < X$, then we get two different solutions, in a new field of series. This time they are "genuine" Puiseux series in that the exponents are now rational numbers:

$$Z = \pm Y^{1/2} X^{1/2} + \frac{1}{2} Y^{3/2} X^{-1/2} - \frac{1}{8}Y^{5/2}X^{-3/2} + \dotsb.$$

Rayner shows how to construct a field that is analogous to Puiseux series in several variables, but the field's construction depends on a choice of (ordered!) basis for $k[x_1, \dotsc, x_n]$, of which there is one for each element of $\DeclareMathOperator\SL{SL}\SL(n, \mathbb{Z})$. This example shows that for one choice, we have a "smooth" parametrization, but for another, the parametrization is "ramified."

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I feel like my algebraic geometry is not strong enough to understand what's going on here beneath the surface. I guess I should think using the functor of points, varying the base. Fixing an initial basis, say $\beta = (X, Y)$, we can associate to each $M \in \SL(2, \mathbb{Z})$ the basis $\beta ' = M \beta$, and then we can form a field of Laurent series with respect to this basis, as well as their algebraic closure the field of Puiseux series, and Rayner tells me these come with valuations given by taking the smallest (suitably chosen) exponent of the series. We can pass to the ring of elements with non-negative valuation, and then I guess the presence of rational exponents comes from ramification over these valuation rings. These are what I should vary.

Since the family of rings is parametrized by an algebraic group, it feels reasonable to conjecture that the collection of points for which I can get a Laurent expansion rather than a genuine Puiseux expansion might itself be a variety, maybe some sort of smooth locus. On the other hand, I can't shake the intuition that most bases should show a singularity. I'd like to know how to find these points, if they exist. I have a question in $4$ variables in my own research that I would like to expand in a generalized Puiseux series in this same way, but this paper seems to show that there is some geometric meaning to the question of finding the 'correct' expansion to make calculations as nice as possible, and being a quadric (affine) surface, this seems like the simplest possible example to start understanding.

So, given an algebraically closed field $k$ of characteristic $0$, a variety $V$ over $k$ with an isolated singularity at the origin, as well as a particular transcendence basis $x_1, \dotsc, x_n$, which $\SL(n, \mathbb{Z})$ matrices correspond to bases admitting multi-Laurent expansions, and which give Puiseux expansions?

Answer 1

I'm not sure if this is entirely what you're asking, but let me add some thoughts.

Probably the most natural place for ramification phenomena to live is in valuation theory. From that point of view, it is fairly intuitive what is going on in the example $Z^2=(X+Y)Y$: the Laurent series (i.e. $\mathbf Z \times \mathbf Z$-indexed) version of Rayner's construction is some sort of "completion" $\widehat R$ of a higher-rank valuation ring, so every element of $1+\mathfrak m_{\widehat R}$ should be a square if you believe in Hensel's lemma/Newton's method (this is Theorem 1 in Rayner). Thus, the only obstruction for an element $r \in \widehat R$ to be a square is that $v(r) \in \mathbf Z \times \mathbf Z$ be divisible by $2$.

In the first lexicographic order $X < Y$, the element $X/Y$ is in $\widehat R$, which (somewhat counterintuitively) means $v(Y) < v(X)$. So the lowest nonzero term of $(X+Y)Y$ is $Y^2$, so $v\big((X+Y)Y\big) = (0,2)$, and therefore $(X+Y)Y$ is a square. In the second, the lowest nonzero term is $XY$, so $v\big((X+Y)Y\big) = (1,1)$ and we need to extract a root of $XY$ to solve $Z^2 = (X+Y)Y$.

In general, if you have some monic polynomial $g = \sum_{i=0}^n a_iz^i$ with $a_i \in \mathbf C[x_1,\ldots,x_n]$, define its Newton polygon as the lower convex hull of the pairs $(i,v(a_i))$ in $\mathbf N \times \mathbf Z^n$ with respect to the given lexicographic order: start with $(0,v(a_0))$, and given a point $(i,v(a_i))$ choose the next segment to the right with smallest slope, i.e. $j > i$ minimising $\tfrac{v(a_j)-v(a_i)}{j-i}$. Repeat until you reach $(n,v(a_n))$.

Then a necessary condition for all solutions to be in $\widehat R$ is that all Newton slopes are in $\mathbf Z^n$ (instead of $\mathbf Q^n$), and the converse should hold as well since the algebraic closure of $\operatorname{Frac} \widehat R$ is the "Puiseux series" field (Theorem 2 of Rayner).

If you want a geometric place where all these valuations are living, one natural candidate is the Zariski–Riemann space $\mathcal{ZR}(\mathbf C(X,Y))$. But this contains many other types of points, so is maybe not the most suited for this question.

For the example $Z^2=(X+Y)Y$, this does show us something: the valuation of $(X+Y)Y$ is either $(1,1)$ or $(0,2)$, depending on whether $v(X) < v(Y)$ or $v(Y) < v(X)$ (they cannot be the same when $\mathbf Z \times \mathbf Z$ is totally ordered). In particular, there will be many more valuations where $(X+Y)Y$ is a square!

I believe that the locus $\{v \in \mathcal{ZR}(\mathbf C(X,Y))\ |\ v(X) \leq v(Y)\}$ is open, but there's a caveat: the Zariski–Riemann space also allows valuations with a kernel. For this reason, multiplicative notation is often used instead of additive notation, i.e. $v(0) = 0$ and $v(1) = 1$ instead of $v(0) = \infty$ and $v(1) = 0$. Again, this is some indication that the Zariski–Riemann space is a little bit too general for this question.