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Geometry
Volume 2013 (2013), Article ID 369420, 6 pages
http://dx.doi.org/10.1155/2013/369420
Research Article

Galois Group at Each Point for Some Self-Dual Curves

1Graduate School of Science and Technology, Niigata University, Niigata 950-2181, Japan
2Department of Mathematics, Faculty of Science, Niigata University, Niigata 950-2181, Japan

Received 10 October 2012; Accepted 21 December 2012

Academic Editor: Michel Planat

Copyright © 2013 Hiroyuki Hayashi and Hisao Yoshihara. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We study the Galois group defined by a point projection for plane curve. First, we present a sufficient condition that the group is primitive and then determine the structure at each point for some self-dual curves.

1. Introduction

This study is a continuation of [14], and so forth. In general, it is not easy to determine the Galois group at every point for plane curve, in particular for curve with singular point. When we determine the structure of , it is important to know whether it is primitive or not. However, there are not so many results which are useful for our purpose (cf. [5]). In this paper we give a geometrical criterion and then determine the group at each point for some self-dual curves.

Let be an algebraically closed field of characteristic zero. We fix it as the ground field of our discussions. Let be an irreducible plane curve of degree () and the rational function field of . Let be a set of homogeneous coordinates on and put . Let be the defining equation of and put , where .

1.1. Galois Group

Let be the resolution of singularities of . For a point , let be the dual line in the dual space of corresponding to . We define the morphism by where is the point in corresponding to the line , which passes through and if . In case , the line is the tangent line to the branch of at . Clearly, we have and a field extension , where denotes the multiplicity of at . In case , we understand . We put ; if there is no fear of confusion, we simply denote it by . Since the extension depends only on , we denote by , that is, we have . Let be the Galois closure of and the Galois group .

Definition 1. We call the Galois group at for . In case is a Galois extension, the point is said to be a Galois point.

In case is the field of complex numbers, is isomorphic to the monodromy group of the covering [6, 7].

1.2. Self-Dual Curve

Definition 2. A point is said to be a cusp of if it is a singular point and consists of a single point. Furthermore, if is a blow-up and is a nonsingular point of the proper transform of , the point is said to be a simple cusp.

Denote by the dual curve of .

Definition 3. If is projectively equivalent to , then is said to be a self-dual curve.

Suppose is smooth. Then, is self-dual if and only if . However, if has a singular point, the condition that is self-dual becomes complicated. The following proposition has been known (cf. [8]).

Proposition 4. Suppose is one of the following curves:(1)has just one singular point; (2) is rational and has only simple cusps as singular points.Then, is a self-dual curve if and only if is projectively equivalent to the curve defined by .

Example 5. It seems that only a few self-dual curves have been known. Here, we present some of them(I): the curve defined by ;(II): the curve defined by (cf. [9]);(III): the curve defined by (cf. [10]).For the curve , if , then and are not simple cusps and has no flex. The curve has two cusps and , where is not a simple cusp. The curve has three cusps: , and and the normalization is an elliptic curve. It is easy to find the dual curve of ; however, in the other curves we need some consideration, for the details, see [9, 10].

Remark 6. Let be the rational map giving the dual of , that is, where is the defining equation of . In the case where , the map turns out to be a quadratic transformation of :
We use the following notation: (i): the cyclic group of order ;(ii): the symmetric group of degree ;(iii): the intersection number of two curves and at ;(iv): the line passing through and ;(v): a line passing through ;(vi): the tangent line to at .

2. Statement of Results

We need some preparations before stating the results. A curve means a nonsingular projective algebraic curve. Let and be curves and a surjective morphism, which we call a covering for short. We denote by the ramification index of at . If there is no fear of confusion, we simply denote it by .

Definition 7. Let be the covering above. If there exists a curve and coverings and such that , and , then is said to be decomposable and an intermediate covering. If such a curve does not exist, then is said to be indecomposable (cf. [11]).

Definition 8. Let be the covering above and all the ramification points for . Put . The covering is said to be an s-covering over if there exists no ramification point in except . The is said to be an s-covering if it is an s-covering over each .

Definition 9. With the same notation as in Definition 8, we call (or, simply (,…,)) the ramification data for .

We give several sufficient conditions that is indecomposable. Some of them will not be used later in this paper.

Proposition 10. Let be the covering above and . If one of the following conditions is satisfied, then is indecomposable.(1)For some , is prime and .(2).(3) is a rational curve, is an s-covering except over and is prime for each , where .

Proposition 11. With the same notation as in Proposition 10, if is an s-covering and satisfies one of the following conditions, then is indecomposable.(1) is a rational curve, , , and is prime for each . (2) is a rational curve and is prime for each .(3) is a rational curve and is prime for each .
Hereafter, we follow the notation in Section 1. By taking a suitable projective change of coordinates, we can assume the projection center is without changing the structure of . Putting , we have and . Put , where and let    be the roots of . Then, we can consider as a permutation subgroup of . Note that is a transitive subgroup of . Hence, is a primitive group if and only if the isotropy subgroup of an element of is a maximal subgroup of .

Theorem 12. The group is primitive if and only if is indecomposable. In particular, if is a prime number, then is primitive for .

Definition 13. Assume is a smooth point or a cusp. A line is said to be a simple -tangent line to if the following conditions are satisfied:(1)if (resp., ), then (resp., ), where and ;(2)the curves and have normal crossings except at .Sometimes we call a simple -tangent for short.

Note that a simple -tangent yields an s-covering over .

Lemma 14. We have the following assertions for .(1)If each line has normal crossings with or is a simple -tangent line to such that is a prime number, then is primitive (cf. [5, Lemma ]).(2)If there exists a simple -tangent line , then contains a transposition.

The following lemma is well known.

Lemma 15. If a permutation group is primitive and contains a transposition, then it is a full symmetric group.

Combining the results above, we get the following corollary.

Corollary 16. If the covering is one of the coverings in Propositions 10 or 11 and is an s-covering over with for some , then is a full symmetric group. In particular, if each line has normal crossings with or is a simple -tangent, then is a full symmetric group.

Corollary 16 implies [2, Theorem 1 and ]. Now we can state the structure of as follows.

Theorem 17. For the curves in Example 5, the Galois groups are as follows, where indicates the trivial group(I)the case (see Table 1);(II)the case (see Table 2);(III)the case (see Table 3).

tab1
Table 1
tab2
Table 2
tab3
Table 3

Remark 18. For the curves in Theorem 17, is a Galois point if and only if is a cyclic group. However, the same assertion does not hold true in general, see, for example, [3].

3. Proofs

First, we prove Propositions 10 and 11.

Claim 1. Suppose and a ramification point satisfy the following conditions:(1) is an s-covering over ;(2) is prime.If there exists an intermediate covering , then is unramified at .

Proof. Suppose is ramified at . Then, since is prime, we have , hence is not a branch point for . Then, there will appear another ramification point for in . This is a contradiction.

The proof of Proposition 10 is as follows. Suppose is decomposable and there exists a covering as in Definition 7. First, we prove the assertion (1). Since is prime, is unramified at by Claim 1. Hence, we have . Since there exists at least two points in , we have , which contradicts the assumption. Next we prove (2). Clearly and are ramified at and , respectively. Put . Then, since , consists of one or two points. In the former case, consists of two points, on the other hand in the latter case consists of one point, where . In each case we infer the inequality , which is a contradiction. We go to the proof of (3). Then, by Claim 1, is not a branch point for . Thus, is the only branch point for . Then, by Hurwitz’s formula, we have , where is the genus of , is the degree of , and . Since , this inequality implies , which is a contradiction.

Next we prove Proposition 11. In each case we use the reduction to absurdity, that is, suppose is decomposable. So we use the notation . In the case (I), by Claim 3, is unramified at . Since and are rational, from Hurwitz’s formula, we infer that is ramified with the index . Then, since there exists no ramification points in except , must branch at and . However, there exists an unramified point in , this is a contradiction. Therefore, is indecomposable. In the case (II), by Claim 1, is unramified at for . Since is rational, by Hurwitz’s formula, we have a contradiction. In the case (III) similarly, by Claim 1, is unramified at every point; however, since is rational, must be an identity, which is a contradiction. This completes the proof of Proposition 11.

The proof of Theorem 12 is as follows: suppose is not primitive and let be the isotropy group of in . Then, there exists a subgroup of such that . Let be the nonsingular model of the intermediate field which corresponds to by the Galois correspondence. Then, there exist the coverings and such that . Thus, is decomposable. The converse assertion is clear from the Galois correspondence.

The proof of Lemma 14 is simple. In view of Definition 13, we see that the assertion (1) is another expression of (3) in Proposition 11. The assertion (2) may be well known (cf. [7]).

Now we proceed to the proof of Theorem 17. The structure of depends on the covering and depends on the position of . We prove by examining the cases where lies on the tangent line to at the cusp or at the flex. Hereafter, we assume is the curve in Theorem 17. Since is a self-dual curve and has only cusps as the singularity, the following remark is clear.

Remark 19. Suppose a line satisfies the following conditions:(1) does not pass through any cusp;(2) is not the tangent line to at the flex.Then, is a simple -tangent line to or and have normal crossings.

Proof of the Case (I). Assume . It has the following property.

Claim 2. The tangent line is and , which does not lie on . In case , has one flex at . On the other hand, in case , has no flex.

Proof. Calculating the Hessian of (cf. [12]), we infer readily the assertions.

If , or , then can be determined directly. In fact, if , then consider the affine part of , that is, the affine defining equation is . Then, putting , we get , hence . The other case is similarly determined. If , then consider the affine part , we get . Putting , we get , hence . As we have seen above, these points are Galois ones.

Next we treat the case . First, we prove the subcase . Since is a self-dual curve and has no flex, we see that, if a line passes through neither nor , then it has normal crossings with or it is a simple -tangent line to . Furthermore, by Hurwitz’s formula, we see there exists a simple -tangent. Then, by (1) in Proposition 11 and Lemma 15, we have . Next we prove the subcase . Then, is a flex (resp., cusp) and the tangent line at does not meet except at . If a line does not pass through , then it has normal crossings with or it is a simple -tangent line to . By (2) in Proposition 11 and Lemma 15, we have . The proof of the case is the same.

Now we prove the case where and . If and , then has two ramification points and such that , and . Thus, is not an s-covering. If passes through neither nor , then is a simple -tangent to or has normal crossings with . By (3) in Proposition 10, is indecomposable. Since there exists a simple -tangent , we conclude . In case and or , is an s-covering and and , hence by (2) in Proposition 10, is primitive and there exists a simple -tangent line , thus we conclude . In view of Remark 19, we conclude easily from the similar argument that when .

Proof of the Case (II). Assume . It has the following property.

Claim 3. The is and , which does not lie on . Furthermore, and . The has one flex of order , that is, and does not pass though .

Proof. The last assertion is checked by Hurwitz’s formula and the others are simple.

Remark 20. The coordinates of the flex are computed as .

Clearly, if or , then . If , then , hence is primitive. We divide the proof into three cases:(1);(2); (3) is the other point.

In any case, by Hurwitz’s formula, we infer that there exists at least one simple -tangent line passing through , hence . Then consider the case . If , then has ramification points and such that and . Thus, is not an s-covering. Consider for the most special case . We infer from Hurwitz’s formula that the ramification data is . By (3) in Proposition 10, we have . There are several cases of position of which yield different ramification data; however, it is easy to see that there exists such that . Then from Propositions 10 or 11, we conclude .

Proof of the Case (III). Assume . It has the following property. There exists a projective transformation such that and , or so that interchanges .

Claim 4. The flexes of are , and , hence the tangent lines to at them are , and , respectively. On the other hand, the tangent lines to at , and are , and , respectively. There exist just three points satisfying the following conditions:(1);(2)if does not pass through any cusp, then and have normal crossings or there exist two points satisfying .Such is an intersection , where , indeed , and . Therefore, if , then there exists a line passing through such that is a simple -tangent line to .

Proof. Making use of the results in [10] and observing the self-duality of , we can check the assertions by direct computations.

Now let us begin the proof. If , then , hence is primitive. The lines and yield the ramification points of order three of , hence we infer from Hurwitz’s formula that there exists such that . Thus, we get . For or , using the projective transformation above, we see .

Next consider the case . Then we have , hence is primitive. Using Hurwitz’s formula or the self-duality of , we see that there exists a simple -tangent line to , thus we have .

Finally, we consider the remaining case .

Claim 5. Let be the number of ramification points with index . Then we have , where . In particular, if , then , furthermore; and .

Proof. The former assertion is clear from Claim 4 and Hurwitz’s formula. The proof of the latter assertion is as follows: observing Claim 4, we infer that, if , then is unique , which is the intersections of the three lines and (Figure 1). Similarly observing Claim 4, we infer that if , then or . In this case, we have , hence .

369420.fig.001
Figure 1

Claim 6. If is an s-covering, then is indecomposable.

Proof. By Claim 5 the ramification index is , or . Suppose is decomposable. Then, or . By Claim 1, is unramified at , where or . By Claim 5, we have . As we have seen in the proof of Proposition 10, cannot be ramified at only one point. Thus, we have . If , then the proof is clear by (3) in Proposition 11. If , then . In case , is ramified at and . Since , this cannot occur. In case , is ramified at and with ; however, these do not satisfy Hurwitz’s formula. If , then and from Claim 4 and Hurwitz’s formula we infer that the ramification data is . Suppose is decomposable. Then, or . If , then is ramified at . However, since , this case cannot occur. Then, we have . We see easily that is ramified at with . However, this does not satisfy Hurwitz’s formula. Therefore, is indecomposable.

Now we resume the proof. We prove by examining the cases:(i);(ii);(iii), , and ;(iv) is the point not appearing in the above case.

By Claims 5 and 6, the proof is complete for (i) and (iv). So let us treat the case (ii). By Claim 6, is primitive. However, there exists no simple -tangent line. Take and consider the affine part . The defining equation is . Putting and , we get . Here, we consider the Galois group obtained by the special value . By the aid of a software, for example, PARI, we see that the polynomial in is irreducible and the Galois group of this polynomial is . Let be the roots of with respect to . Note that    is regular near and are the roots of . We can find satisfying the conditions: (resp., ) is a generator of the minimal splitting field of (resp., ) over (resp., ). Suppose the degree of is less than . Then, so is , which is a contradiction. Hence we have , thus we conclude . The proofs of the other two cases and are almost the same.

The proof of the case (iii) is as follows: here we notice that if , then is not an s-covering. First we consider the special case where is in some , for example, Then the ramification data is and . Suppose is decomposable. Then, by Claim 1, is unramified at and . Namely, is ramified at just two points. Then, the ramification data of is or , where or , respectively. However, it is easy to see that this is impossible considering and , so is indecomposable. Since there exist , we conclude . On the other hand, if is not in for each , then, by (3) in Proposition 10, is indecomposable. Since there exists a simple -tangent, we have .

Thus, we complete all the proofs.

Remark 21. In the list of Theorem 17 only two kinds of group appear. Of course, other kinds will appear in other examples, for example, let us take the Fermat quartic . Then, there exist 12 points such that is the dihedral group of order 8 (cf. [13]).

Problem. Concerning the Galois groups for , full symmetric group degenerates into the cyclic group. How does the symmetric group degenerate for various curves?

Acknowledgments

The authors would like to express their thanks to Oka for teaching the example of self-dual curve . They thank also the referee(s) for carefully reading the paper and giving the suitable suggestions for improvements.

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