ORIGAMI CONSTRUCTIONS OF RINGS OF INTEGERS OF IMAGINARY QUADRATIC FIELDS

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1 #A34 INTEGERS 17 (017) ORIGAMI CONSTRUCTIONS OF RINGS OF INTEGERS OF IMAGINARY QUADRATIC FIELDS Jürgen Kritschgau Departent of Matheatics, Iowa State University, Aes, Iowa Adriana Salerno Departent of Matheatics, Bates College, Lewiston, Maine Received: 10/3/16, Revised: /14/17, Accepted: 7/5/17, Published: 8/7/17 Abstract In the aking of origai, one starts with a piece of paper, and through a series of folds along a given set of points one constructs coplicated three-diensional shapes Matheatically, one can think of the coplex nubers as representing the piece of paper, and the initial points and folds as a way to generate a subset of the coplex nubers Under certain constraints, this construction can give rise to a ring, which we call an origai ring We will talk about the basic construction of an origai ring and further extensions and iplications of these ideas in algebra and nuber theory, extending results of Buhler, etal In particular, in this paper we show that it is possible to obtain the ring of integers of an iaginary quadratic field through an origai construction 1 Introduction In origai, the artist uses intersections of folds as reference points to ake new folds This kind of construction can be extended to points on the coplex plane In [1], the authors define one such atheatical construction In this construction one can think of the coplex plane as representing the paper, and lines representing the folds The question they explored is: which points in the plane can be constructed through iterated intersections of lines, starting with a prescribed set of allowable angles and only the points 0 and 1? First, we say that the set S = {0, 1} is the set of initial points We fix a set U of angles, or directions, deterining which lines we can draw through the points in our set Thus, we can define the fold through the point p with angle u as the

2 INTEGERS: 17 (017) line given by L u (p) := {p + ru : r R} Notice that U can also be coprised of points (thinking of u U as defining a direction) on the unit circle, ie the circle group T Moreover, u and u define the sae line, so we can think of the directions as being in the quotient group T/{±1} Finally, if u and v in U deterine distinct folds, we say that I u,v (p, q) = L u (p) \ L v (q) is the unique point of intersection of the lines L u (p) and L v (q) We define R(U) to be the set of points obtained by iterated intersections I u,v (p, q), starting with the points in S and angles in U Alternatively, we ay define R(U) to be the sallest subset of C that contains 0 and 1 and I u,v (p, q) whenever it contains p and q, and u, v deterine distinct folds The ain theore of [1] is the following: Theore 1 If U is a subgroup of T/{±1}, and U of C 3, then R(U) is a subring Let U n denote the cyclic group of order n generated by e i /n (od {±1}) Then Buhler, et al, obtain the following corollary Theore Let n 3 If n is prie, then R(U n ) = Z[ n ] is the cyclotoic integer ring If n is not prie, then R(U n ) = Z[ n, 1 n ] In [4], Nedrenco explores whether R(U) is a subring of C even if U is not a group, and obtains a negative answer and soe necessary conditions for this to be true The ain result is that given the set of directions U = {1, e i, e i }, if 6 od then R(U) = Z + zz for soe z C Clearly, this will not always be a ring, as it will not always be closed under ultiplication In this paper, we explore the inverse proble, that is, given an interesting subring of C, can we obtain it via an origai construction? The answer is a rative in the case of the ring of integers of an iaginary quadratic field The next section of this paper delves deeper into the origai construction, in particular the intersection operator Soe properties in this section are crucial for understanding the proofs of our ain results This section also explores an exaple of an origai construction in ore depth, that of the Gaussian integers, since it illustrates the geoetric and algebraic approach through a very well known ring Finally, in Section 3, we state and prove our ain result

3 INTEGERS: 17 (017) 3 Properties of Origai Rings 1 The Intersection Operator Let U T, as before There are iportant properties of the I u,v (p, q) operator that are integral for us to prove our theore Let u, v U be two distinct angles Let p, q be points in R(U) Consider the pair of intersecting lines L u (p) and L v (q) In [1], it is shown that we can express I u,v (p, q) as I u,v (p, q) = u pv ūpv u v ūv + q vu qvu ūv u v [u, p] [v, q] = v + [u, v] [v, u] u, ( ) where [x, y] = xy xy Fro the algebraic closed for ( ) of the intersection operator, we can see by straightforward coputation that the following properties hold for for p, q, u, v C Syetry I u,v (p, q) = I v,u (q, p) Reduction I u,v (p, q) = I u,v (p, 0) + I v,u (q, 0) Linearity I u,v (p + q, 0) = I u,v (p, 0) + I u,v (q, 0) and ri u,v (p, 0) = I u,v (rp, 0) where r R Projection I u,v (p, 0) is a projection of p on the line {rv : r R} in the u direction Rotation For w T, wi u,v (p, q) = I wu,wv (wp, wq) An Illustrative Exaple Let S = {0, 1} be our set of initial points Now, let U = {1, e i /4, i} This is clearly not a group, since e i 4 i = e 3i 4 6 U, and so does not satisfy the hypotheses of Theore 1 Given a set of points, the set generated by intersections can be coputed in two ways; using the closed forula ( ), or geoetrically In Figure 1, we illustrate how to generate the first few points through line intersections Figure 1: Starting with {0, 1} and drawing lines with angles 0, /, /4, we see that we only generate two new points, 1 + i and i

4 INTEGERS: 17 (017) 4 In Figure, we show the di erent stages of the construction obtained by iterated intersections These figures were created by coding the forula for iterated intersections into Maple [] Figure : These graphs illustrate the sequence of point creation, where each successive graph shows all possible intersections fro the previous graph using U = {1, e i /4, i} as our set of allowable angles In the last graph, not all of the new points fit into the section of the plane we are picturing Notice that a pattern sees to eerge: the points constructed all have the for

5 INTEGERS: 17 (017) 5 a+bi where a, b Z This sees to indicate that this origai construction generates the Gaussian integers, a subring of C In fact, this is a special case of the ain result of [4], where z = i So, in fact, this suggests that it is possible to construct a subring of C without U being a group We prove in the next section that the ring of algebraic integers of an iaginary quadratic field can always be obtained through an origai construction 3 Constructing O(Q( p )) A natural question related to the previous section is: which subrings of C can be generated through an origai construction, that is, which subrings are origai rings? We have seen that the cyclotoic integers Z[ n ], where n is prie, are origai rings by Theore 1 Let < 0 be a square-free integer, so Q( p ) is an iaginary quadratic field Denote by O(Q( p )) the ring of algebraic integers in Q( p ) Recall that a coplex nuber is an algebraic integer if and only if it is the root of soe onic polynoial with coe cients in Z Then we have the following well-known theore (for details see, for exaple, [3, pg 15]) Theore 3 The set of algebraic integers in the quadratic field Q( p ) is {a + b p : a, b Z} if or 3 (od 4), p a + b : a, b Z, a b (od ) if 1 (od 4) And so, we can state our ain theore Theore 4 Let < 0 be a squarefree integer, and let = arg(1 + p ) Then O(Q( p )) = R(U) where 1 U = {1, i, e i }, if or 3 (od 4), and U = {1, e i, e i( ) }, if 1 (od 4) Notice that the Gaussian integers are a special case of Theore Proof of Theore 41 Let or 3 (od 4) and < 0 Let U = {1, i, e i } where is the principal arguent of 1 + p Lea 1 We have that I u,v (p, q) Z[ p ] whenever u, v U and p, q Z[ p ] Proof Since there are three possible directions, there are 3 Let p = a + b p and q = c + d p Then = 6 cases to consider

6 INTEGERS: 17 (017) 6 1 I 1,i (p, q) = c + b p I 1,e i (p, q) = b + c d + b p 3 I i,1 (p, q) = a + d p 4 I i,e i (p, q) = a + (a c + d) p 5 I e i,1(p, q) = a b + d + d p 6 I e i,i(p, q) = c + ( a + b + c) p In other words, if p, q Z[ p ], then so is I u,v (p, q) All of these can be obtained fro straightforward coputations using equation ( ) For exaple, I 1,i (p, q) = [1, a + bp ] i + [i, c + dp ] = c + b p [1, i] [i, 1] This concludes the proof for the closure of the intersection operator In other words, as long as our set of known points starts with eleents in Z[ p ], then the intersections will also be in Z[ p ] We can also express this clai as R(U) Z[ p ] It reains to be shown that any eleent in Z[ p ] is also an eleent in R(U) Lea The set Z[ p ] R(U) Proof Let a + b p be an eleent in Z[ p ] We want to show that it can be constructed fro starting with {0, 1} and the given set U We can reduce the proble by showing that given points {n+k p, n+1+k p } we can construct n 1 + k p, n + + k p, n + (k ± 1) p and n (k ± 1) p In Figure 3 we give an illustration of this step of the proof In essence, the following is the induction step to a double induction on the real and iaginary coponents of an arbitrary integer we are constructing We prove that for any two adjacent points in the construction, we can construct points that are adjacent in every direction Since our initial set of points is {0, 1}, we can construct any eleent of the for a + b p We will now construct the desired points using the appropriate reference points Constructing n + + k p Consider I i,1 (I 1,e i (I e i,i(n + k p, n k p ), n k p ), n k p )

7 INTEGERS: 17 (017) 7 Figure 3: Given two points next to each other, we want to show that we can generate all the points iediately around the This results in ore points next to each other, upon which we can repeat the process Notice that we can evaluate this expression using the six cases enuerated in Lea 1 In particular, we apply case (6) first to get I i,1 (I 1,e i (n (k + 1) p, n k p ), n k p ) Next, we apply case () to get I i,1 (n (k + 1) p, n k p ) Finally, we use case (3) to get n + + k p Constructing n 1 + k p Consider I i,1 (I 1,e i (I i,e i (n + k p, n k p ), n + k p ), n + k p ) First, we apply case (4) to get I i,1 (I 1,e i (n + (k 1) p, n + k p ), n + k p ) Next, we apply case () to get I i,1 (n 1 + (k 1) p, n) Finally, we apply case (3) to get n 1 + k p Constructing n + (k + 1) p and n (k + 1) p Consider I e i,i(n + k p, n k p ) Using case (6) we get n (k + 1) p Now consider I i,1 (n + k p, n (k + 1) p ) Using case (3) we get n + (k + 1) p Constructing n + (k 1) p and n (k 1) p Consider I i,e i (n + k p, n k p ) Using case (4) we get n+(k 1) p Consider I i,1 (n+1+k p, n+(k 1) p ) Using case (3) we get n (k 1) p Thus we have shown that Z[ p ] R(U), copleting the proof

8 INTEGERS: 17 (017) 8 3 Proof of Theore 4 The proof for Theore 4 eploys the sae strategy as the proof for 41, with a subtle di erence given by the slightly di erent structure of the ring Let 1 (od 4) and < 0 Let U = {1, e i, e i( ) } where is the principal arguent of 1 + p Lea 3 We have that I u,v (p, q) O(Q( p )) where u, v U and p, q O(Q( p )) Proof Again, there are six cases to consider Let p = a+bp and q = c+dp, where a b od and c d od The cases are as follows 1 I 1,e i (p, q) = b + c d + b p I 1,e i( )(p, q) = c + d b + b p 3 I e i,1(p, q) = a b + d + d p 4 I e i,e i( )(p, q) = (a b+c+d)+(b a+c+d)p 5 I e i( ),1(p, q) = a + b d + d p 6 I e i( ),e i (p, q) = (a+b+c d)+(a+b c+d)p All of these cases can be obtained, again, by straightforward coputation using ( ), and are left as an exercise Notice that (1), (), (3), and (5) all are clearly in the ring of algebraic integers The only additional fact to show is that (4) and (6) are as well But it s easy to see that a b + c + d b a + c + d od, since a b od And siilarly a + b + c d a + b c + d od, because c d od This concludes the proof for the closure of the intersection operator In other words, as long as our set of known points contains only eleents in O(Q( p )), then the intersections will also be in O(Q( p )) That is, R(U) O(Q( p )) It reains to be shown that any eleent in O(Q( p )) is also an eleent in R(U) Lea 4 The set O(Q( p )) R(U) Proof Let a + bp be an eleent in O(Q( p )) As before, we can reduce the proble to one of double induction This is done by showing that given points p n + k, n + + kp

9 INTEGERS: 17 (017) 9 Figure 4: Given two points next to each other, we want to show that we can generate all the points iediately around the Notice that this can be iterated to generate all points The graph on the left illustrates the starting points The graph on the right shows the points whose construction is su cient to prove the theore we can construct n + k p, n kp, n (k 1)p Figure 4 is an illustration of the points we want to construct We will now show how to construct the desired points Constructing n+1+(k 1)p Consider p n + k I e i,e, n + + kp i( ) By applying case (4) fro above, we see that p n + k I e i,e, n + + kp i( ) and n (k + 1)p = n (k 1)p Constructing n+1+(k+1)p Consider p n + k I e i( ),e, n + + kp i By applying case (6) fro above, we see that p n + k I e i( ),e, n + + kp i = n (k + 1)p

10 INTEGERS: 17 (017) 10 Constructing n +kp Consider p n (k + 1) I e i,1 I 1,e i( ), n + kp, n + kp By applying case (1) fro above, we can reduce the previous expression to p n 1 + (k + 1) I e i,1, n + kp We further reduce the expression using case (5) fro above The result is p n 1 + (k + 1) I e i,1, n + kp = n + kp Constructing n+4+kp Consider p n (k + 1) I e i( ),1 I 1,e, n + + kp, n + + kp By applying case (5) fro above, we can reduce the previous expression to p n (k + 1) I e i( ),1, n + + kp We further reduce the expression using case (1) fro above The result is p n (k + 1) I e i( ),1, n + + kp = n kp By induction, given our initial set of points {0, 1} and the proofs above, we see that p a + b : a, b Z, a b (od ) R(U), copleting the proof 33 Soe Final Illustrations and Rearks In Figure 5 we see that when we use U = {1, i, e i } as our angle set, then the origai ring grows into the first and third quadrants, and bleeds into the others As discussed, R(U) = Z[i] In Figure 6 we see that when we use U = {1, e i arg(1+p 3), e i( arg(1+p 3)) } as our angle set, then the origai ring grows along the real line, and slowly bleeds into

11 INTEGERS: 17 (017) 11 the rest of the coplex plane This is an illustration of the second case of Theore 4 In this case, R(U) = O(Q( p 3)) Of course, R(U) is assued to be closed under the intersection operator, so the growth pattern doesn t atter in an abstract sense However, coputationally, it eans that the nuber of steps it takes to construct a point is not related to that point s odulus In fact, we get an entirely di erent easure of distance if we only consider the nuber of steps it takes to generate a point One possible additional exploration is, given ore general starting angles and points, to describe the dynaics of the iterative process These exaples also serve to illustrate the progression of the iterative process, which was coded into Maple [] to produce the graphs Figure 5: This graph depicts the first 5 generations of origai points using U = {1, i, e i 4 } Figure 6: This graph depicts the first 5 generations of origai points using U = {1, e i arg(1+p 3), e i( arg(1+p 3)) }

12 INTEGERS: 17 (017) 1 Acknowledgeents This work was part of Kritschgau s senior Honors Thesis at Bates College, advised by Salerno, and we thank the Bates Matheatics Departent, in particular Peter Wong, for useful feedback and discussions throughout the thesis process We would also like to thank the rest of the Honors coittee, Paela Harris and Matthew Coté, for the careful reading of the thesis and their insightful questions Finally, we would like to thank the 015 Suer Undergraduate Applied Matheatics Institute (SUAMI) at Carnegie Mellon University for inspiring the research proble We also thank the referee for suggestions that largely iproved the iages and presentation References [1] J Buhler, S Butler, W de Launey, and R Graha, Origai rings, J Aust Math Soc 9 (01), [] Maple 17 Maplesoft, a division of Waterloo Maple Inc, Waterloo, Ontario [3] D A Marcus, Nuber Fields Springer Verlag, New York, 1997 [4] D Nedrenco, On origai rings, arxiv: v1, (015)