On the Luroth semigroup of smooth Plane curves

本文研究了正亏格代数曲线的三重覆盖上的无基点束。设X是亏格为g的光滑代数曲线，它在亏格为h1的一般曲线C（Brill-Noether意义下）上有一个三层覆盖<greater than or equal>.设pi：X*C为覆盖投影。通过简单地应用Castelnuovo-Severi不等式，可以很容易地看出，除了从基曲线C的拉回之外，X上不存在次数小于或等于（g-3 h）/的基点自由束。因此，X的Luroth半群的一个相对低的部分完全由C的Luroth半群决定。因此，寻找X上不与π合成的铅笔的次数是很重要的。我们研究h &gt; 0和三层情形的原因是：i）对于h = 0和两层情形，X是超椭圆的，所以Luroth半群是众所周知的; ii）对于h = 0和三层情形，G.Martens和F. O ...更多信息 （iii）对于h = 0和n（<greater than or equal>4）-单的情形，Coppens-Keem-Martens最近解决了一般的，我们得到的主要定理是：定理设X和C如上。若g<greater than or equal>（2 [（3 h +1）/] +1）（[（3 h +1）/] +1），则X上存在任意次<greater than or equal>dg-[（3 h +1）/] -1的不与pi复合的基点自由束.对于dg<greater than or equal>-[（3 h +1）/]，将交理论的计数演算应用于Jacobian簇J（X）中的子簇W^r_（X），我们可以不那么困难地证明我们的定理.然而，对于d = g-[（3 h +1）/]-1，由于难以证明W^1_的不可约性，所以定理的证明比较困难。最后，我们利用R.米兰达在Amer.J.Math.107（1985）上发表的文章的结果和讨论，进一步推广了这一定理，并指出我们定理中的次数界与Castelnuovo-Severi界相差甚远.所以它似乎不是最好的。事实上，观察循环三重覆盖，我们有：i）一般来说，Castelnuovo-Severi界是最好的。ii）对于循环三重覆盖的情形，我们将能够实质性地改进我们的定理。少

In this project, we have studied base point free pencils on a triple coverin of an algebraic curve of positive genus. Let X be a smooth algebraic curve of genus g which admits a triple sheeted covering onto a general curve C (in the sense of Brill-Noether) of genus h <greater than or equal> 1. Letpi : X*C be the covering projection. By a simple application of the Castelnuovo-Severi inequality one can easily see that there does not exist a base point free pencil of degree less than or equal to (g-3h)/ on X other than the pull-backs from the base curve C.Hence, a relatively low part of the Luroth semigroup of X is determined that of C,completely. So it is important to seek the degrees of pencils on X which are not composed with pi. The reason why we studied the case h > 0 and triple sheeted is the following :i) For the case h = 0 and two sheeted, X is hyperelliptic, so the Luroth semigroup is well-known.ii) For the case h = 0 and three sheeted, it was already solved by G.Martens and F.-O … More .Schreyer.iii) For the case h = 0 and n (<greater than or equal> 4)-sheeted, recently, Coppens-Keem-Martens solved for generalThe main theorem we obtained is the following :THEOREM.Let X and C be as above. If g <greater than or equal> (2 [(3h+1)/] +1) ([(3h+1)/] +1), then there exists a base point free pencil of any degree d <greater than or equal> g-[(3h+1)/] -1 on X which is not composed with pi.For d <greater than or equal> g-[(3h+1)/], applying the enumerative calculus for the intersection theory to subvarieties W^r_ (X) in the Jacobian variety J (X), we can prove our theorem not so hard. However, for d = g-[(3h+1)/] -1, it was hard to prove Theorem bacause it was difficult to prove the irreducibility of W^1_. Finaly, we succeeded it by use of the results and discussion of R.Miranda's paper appearing in Amer.J.Math.107 (1985).It is noted that the bound of degree in our theorem is far from the Castelnuovo-Severi bound. So it seems not to be the best. In fact, observing cyclic triple covering, we have :i) In general, Castelnuovo-Severi bound is the best. But it can be improved for special cases.ii) For case of cyclic triple coverings, we will be able to improve our theorem substantially. Less