On the Luroth semigroup of smooth Plane curves

在这个项目中，我们研究了正亏格的代数曲线的三重覆盖上的无基点铅笔。设X是亏格G的光滑代数曲线，它允许在亏格h&lt；大于或等于&gt；1的一般曲线C(在Brill-Noether意义下)上有三层覆盖。通过Castelnuovo-Severi不等式的简单应用，可以很容易地看出，除了基曲线C的拉回外，X上不存在次数小于或等于(g-3h)i的基点自由笔，从而完全确定了X的Luroth半群的相对较低的部分。因此，寻找X上不是由pi组成的铅笔的次数是很重要的。我们之所以研究h&gt；0和三张单的情形，原因如下：i)对于h=0和两张单的情形，X是超椭圆的，因此Luroth半群是众所周知的。ii)对于h=0和三张单的情形，G.Marten和F.-O…已经解决了这个问题对于h=0和n(&lt；大于或等于&gt；4)-Sheet的情形，最近，我们得到了一般的Coppens-Keem-Marten解。我们得到的主要定理如下：设X和C如上。如果g&lt；大于或等于&gt；(2[(3h+1)/]+1)([(3h+1)/]+1)，则X上存在任意次d&lt；大于或等于&gt；g-[(3h+1)/]-1的无基点铅笔，对于d&lt；大于或等于&gt；g-[(3h+1)/]，应用交理论的计数演算到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-[(3h+1)/]-1，由于很难证明W^1的不可约性，所以很难证明定理。最后，我们利用R.Miranda在amer.J.Math.107(1985)上发表的论文的结果和讨论，证明了我们定理中的次数界远不是Castelnuovo-Severi界。因此，它似乎不是最好的。事实上，观察循环三重覆盖，我们有：i)总的来说，Castelnuovo-Severi界是最好的。但它在特殊情况下可以改进。ii)对于循环三重覆盖的情况，我们将能够实质上改进我们的定理。较少

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