Lifting bases for (Z/pZ)^n to Z^n - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:45:35Z http://mathoverflow.net/feeds/question/1781 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/1781/lifting-bases-for-z-pzn-to-zn Lifting bases for (Z/pZ)^n to Z^n Andy Putman 2009-10-22T02:02:13Z 2009-10-22T05:21:58Z <p>The following question came up in my research. I suspect that it has a slick answer, but I can't seem to find it.</p> <p>Fix an integer n>=2 and a prime p. Define X(n) to be the set of primitive vectors in the Z-module Z^n and Y(n,p) to be the set of "lines" in the vector space (Z/pZ)^n (ie the spans of non-zero vectors). There is a natural surjective map f:X(n)-->Y(n,p) ("reduce mod p and take the span").</p> <p>Question : Does there exist a map g:Y(n,p)-->X(n) with the following two properties.</p> <ol> <li>f(g(L))=L for all L in Y(n,p).</li> <li>If {L_1,...,L_n} \subset Y(n,p) spans the vector space (Z/pZ)^n, then {g(L_1),...,g(L_n)} is a basis for the Z-module Z^n.</li> </ol> <p>Of course, I expect that the answer is no except in certain simple situations (for instance, it is yes for n=p=2), but I can't seem to find a proof.</p> <p>EDIT : Oops! I phrased the question incorrectly. Above is a corrected version.</p> http://mathoverflow.net/questions/1781/lifting-bases-for-z-pzn-to-zn/1787#1787 Answer by Steven Sivek for Lifting bases for (Z/pZ)^n to Z^n Steven Sivek 2009-10-22T02:40:30Z 2009-10-22T05:21:58Z <p>Here's a unified argument based on my comments to Scott's post that doesn't use quadratic reciprocity in any form. Suppose n=2 and p >= 5, and lift each line of slope i in Y(2,p) to a point (a<sub>i</sub>+pb<sub>i</sub>, ia<sub>i</sub>+pc<sub>i</sub>).</p> <p>Since each pair of lifts should give a basis of Z<sup>2</sup> and thus a matrix with determinant \pm 1, taking each pair from among i=1,2,k+2 (with 1 &lt;= k &lt;= p-3) gives us conditions</p> <p>a<sub>1</sub>a<sub>2</sub> = \pm 1 (mod p)</p> <p>k*a<sub>2</sub>a<sub>k+2</sub> = \pm 1 (mod p)</p> <p>(k+1)*a<sub>1</sub>a<sub>k+2</sub> = \pm 1 (mod p).</p> <p>Combining the first two gives ka<sub>2</sub><sup>2</sup>*a<sub>1</sub>a<sub>k+2</sub> = \pm 1, or a<sub>2</sub><sup>2</sup> = \pm(1+1/k) (mod p).</p> <p>But for k=1 this gives us a<sub>2</sub><sup>2</sup> = \pm 2, and for k=2 we get a<sub>2</sub><sup>2</sup> = \pm (1 + (p+1)/2) = \pm (p+3)/2, so either (p+3)/2 = 2 (mod p) or (p+3)/2 = -2 (mod p). These imply p=1 and p=7, respectively, so already the only possible solution is p=7. But if p=7 then k=3 gives a<sub>2</sub><sup>2</sup> = \pm 6, which is not \pm 2 (mod 7), so that doesn't work either. Thus a lift with n=2 can only possibly exist if p is 2 or 3.</p>