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Does an algebraic function always ramify at a multiple root?

  1. May 8, 2013 #1
    Hi,

    Consider the algebraic function [itex]w(z)[/itex] given by the expression

    [tex]f(z,w)=a_0(z)+a_1(z)w+a_2(z)w^2+\cdots+a_n(z)w^n=0[/tex]

    where [itex]f(z,w)[/itex]is irreducible over the rationals, and the coefficients, [itex]a_i(z)[/itex], polynomials with rational coefficients . Let [itex]z_s[/itex] be a point such that [itex]f(z_s,w)=0[/itex] has roots with multiplicty greater than one. Will [itex]w(z)[/itex] always ramify at [itex]z_s[/itex] or can [itex]f(z_s,w)[/itex] have multiple roots and [itex]w(z_s)[/itex] have only regular coverings there?

    Edit:

    Wasn't hard to find one:

    [tex]f(z,w)=(10 z^3/7 -
    z^5) + (-3 z^4/5) w + (2 z - 8 z^2/7 + z^3) w^2 + (-5/7) w^3
    [/tex]

    That one obviously has a multiple root at the origin yet [itex]w(z)[/itex] is reqular there.
     
    Last edited: May 8, 2013
  2. jcsd
  3. May 10, 2013 #2
    May I ask why does Bliss in "Algebraic Functions" call these points singular if there is the possibility that nothing singular, as the example in my edit above demonstrates, is there. Or else I computed the monodromy incorrectly.

    And just to confirm my observations: root multiplicity is a necessary condition for branch sheet ramification, but it is not a sufficient condition. If this is true, then it begs the question: what is then?
     
    Last edited: May 10, 2013
  4. May 11, 2013 #3
    I believe that's because the derivative is indeterminate there even though the function is regular at the origin: The singularity is removable I suppose. Recall:

    [tex]w'(z)=-\frac{f_z}{f_w}[/tex]

    and note that [itex]f(0,0)=f_z(0,0)=f_w(0,0)=0[/itex]. Which is in the form of the classic definition of a singularity for an algebraic function.

    I believe that's the case anyway.
     
    Last edited: May 11, 2013
  5. May 12, 2013 #4

    Bacle2

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    Maybe it would be a good idea to PM Mathwonk with this question; he is the main AGeometry guy here, at least that I know off. If you do get an answer, could you please post it here? I'm interested, but have a pretty- limited knowledge in this area.
     
  6. May 14, 2013 #5

    mathwonk

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    picture a u shaped curve lying sideways, such as the graph of x = y^2. then projection onto the x axis is ramified at 0. i.e. there is only one value of y corresponding to x=0, and the intersection number of the vertical line x=0 with the graph is 2. and when you go around the point x=0 in the complex x axis along a circle, there are always two values of y for each x, and those values interchange when you go around once. i.e. there is non trivial monodromy, and ramification.

    Now picture the graph of y^2 - x^2 = 0. again there are two values of y for each value of x except for x=0 when there is only one value of y. again projection onto th x axis is 2:1 over every non zero value of x, but only one-one over x=0. again the intersection number of the line x=0 and the graph is 2 at the point (0,0). this time however it is not because of tangency, but because there are two branches of the curve at that point which cross at (0,0). going around the point (0,0) in the complex x axis, causes one to also go once around one branch of the graph in the plane and come back to the same point. the difference here is that there are two "branches" of the graph at (0,0) which can be separated by removing the point (0,0). The point (0,0) is also singular on the curve, but this is not the key issue.

    Another example is given by a "cusp" y^2 = x^3. Here projection on the x axis is again 2:1 except at (0,0), and again the point is singular on the graph, but there is only one branch of the curve at that point.

    In general the question of whether there is ramification or not is answered by whether or not the curve has only fewer branches than the multiplicity of the point. for a double point, there is ramification if and only if there is only one branch.

    a point with more than one branch is singular, but a point can also be singular when there is only one branch. I will try to attach some notes on computing branches.
     

    Attached Files:

    Last edited: May 14, 2013
  7. May 15, 2013 #6

    mathwonk

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    Notice the word "singularity" may have several meanings. As a function of two variables, x=y^2 has no singularity anywhere. But as a function of x, at x=0, y^2 = 0 has a multiple root at x=0. This says that the projection of the non singular curve y^2 = x onto the x axis has a ramification point at (0,0). However the fact that the curve is non singular implies that then the projection onto the y axis will not be ramified at that same point. I.e. the line x=0 intersects the curve doubly at x=0, but the line y=0 intersects it simply there. So here ramification depends on choice of variable. I.e. ramification is a feature of the projection, and that depends on the direction of the projection, i.e. of which axis one projects onto. For a non singular curve at most one direction of projection, that along the tangent line, will be ramified.

    Now if we take a singular curve, like y^2 = x^2 - x^3, we have a curve that looks like a cross at (0,0), i.e. it is singular and both partials vanish. Thus projection onto both the x axis and the y axis will have the property that there are (locally near (0,0)) two preimages of all non zero values, whether of x or of y, but only one preimage of 0. Here every line through (0,0) will have double intersection with the curve. This does not however imply "ramification"!
    I.e. the short answer to the title question is no, since y^2 = 0 has a multiple root here, but there is no true ramification, in the sense of non trivial monodromy.

    I.e. on needs to understand the meaning of ramification in the singular case. Here it no longer can be recognized simply be the degree of a map dropping, or the disappearance of a preimage. The point is that true ramification can only be recognized after separating the branches of the curve at the point. In the example y^2 = x^2 - x^3, there are two branches and after separating them, we obtain two points in place of the one singular point (0,0). Then both projections, both onto the x and the y axes, do have two preimages also over 0.

    The number of branches can be determined by calculating the monodromy, since that process avoids the singular point, and takes place entirely at nearby non singular points. But calculating monodromy is not always easy. Ramification occurs at (0,0) for the projection onto the x axis, precisely when the multiplicity of intersection of the line x=0 with the curve, is greater than the number of branches of the curve at the point (0,0).

    The attached pdf files to the previous post give examples of curves for which the number of branches can be calculated by hand, the so called cyclic singularities. If the curve is smooth, then ramification (of projection onto the x axis), occurs precisely whenever the line of projection x=0, is tangent to the curve at the point (0,0). I.e. ramification of a projection for a non singular curve occurs when the line of projection is tangent to the curve.

    For a smooth projective curve of degree d, projection along lines through a general point of the plane, will thus be ramified at precisely d(d-1) points of the curve. By Riemann - Hurwitz formula, it follows such a curve has genus g = (1/2)(d-1)d-2). Each simple crossing singularity introduces an apparent but not genuine ramification point which counts doubly, hence lowers this computation by one. E.g. the curve y^2 = x^2 - x^3, when projectivized, has d=3, but genus zero rather than one.
     
    Last edited: May 15, 2013
  8. May 15, 2013 #7
    Thanks for that mathwonk.

    There's another way to easily compute the monodromy around the singular points for the expression:

    [tex]y^m=(x-a)^{\alpha}(x-b)^{\beta}\cdots (x-m)^{\omega}[/tex]

    When we expand the function around each singular point via Newton polygon, we make the substitution [tex]f(x+x_s,y)[/tex] which gives us the expression:

    [tex]f(x,y)=ax^k+bx^{k+1}+\cdots+hx^{n}-y^m[/tex]

    Forming the first Newton polygon, we obtain the characteristic equation [itex]a-x^m=0[/itex] which has distinct roots so that the first Newton polygon gives us immediately the ramification around this singular point in terms of the slope [itex]k/m[/itex]. Take for example, the equation:
    [itex]y^6=x^3(x-1)^4(x+1)^2(x-2)^5[/itex] in your handout above. When we expand around the point x=1, we obtain [itex]-4x^4+4x^5+\cdots+x^{14}-y^6[/itex] which has as the (negative) slope of its lower Newton leg, 4/6=2/3 or all branches around this singular point are in terms of [itex]z^{1/3}[/itex] so that at this singular point, the function ramifies into two 3-cycle branches as you indicated.
     
    Last edited by a moderator: May 16, 2013
  9. May 16, 2013 #8

    mathwonk

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    nice! Obviously the newton polygon is a useful tool. No wonder it is so famous.
     
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