User talk:Ozob/Archive 5

smooth completion
Hi ozob, I would appreciate if you could check smooth completion for correctness. Tkuvho (talk) 13:44, 6 February 2012 (UTC)


 * There's nothing wrong that I can see. I touched up the lead a little, but it's just a change in presentation, not in content.  If you need a reference, I think all of this is in Hartshorne, chapter 4.  Ozob (talk) 23:35, 6 February 2012 (UTC)


 * Thanks. Comment at the AfD if you get a chance. Tkuvho (talk) 08:17, 7 February 2012 (UTC)

Group of rational points on the unit circle
Hi, I put some more work into the above recently. I'd like someone else to read it and see if it still makes sense. If you have a chance and feel like it, I'd appreciate you doing that. Thanks, Richard Peterson198.189.194.129 (talk) 21:52, 21 February 2012 (UTC)


 * Hi Richard,


 * I have some concerns about the article. First, let me get out of the way a minor concern: Italics for variables. Normally, single-letter Roman type variables like x, y, and G are italicized.  So are single-letter functions such as f.  (But Greek-letter variables such as &alpha; and multi-letter functions such as exp are usually left upright.)  The article doesn't consistently italicize x, y, t, u, G, H, and so on.


 * On to more interesting things. Under the heading "Group structure", one of the sentences is garbled. Currently it says:


 * If C4 denotes the cyclic subgroup with four elements generated by the point (0, 1) and Z is any infinite cyclic subgroup generated by a point of form elements generated by the point 0+i, and Z is any infinite cyclic subgroup generated by a point of form (a^2-b^2)/p + i2ab/p where $$ p = a^2 + b^2$$ and p is a prime of form 4k + 1, (and a, b are positive) then G is isomorphic to H = {C4} ⊕ Z ⊕ Z ⊕ ..., going on forever.


 * I think it ought to say:


 * If C4 denotes the cyclic subgroup with four elements generated by the point (0, 1), and Z is any infinite cyclic subgroup generated by a point of form (a^2-b^2)/p + i2ab/p where $$ p = a^2 + b^2$$ and p is a prime of form 4k + 1, (and a, b are positive) then G is isomorphic to H = {C4} ⊕ Z ⊕ Z ⊕ ..., going on forever.


 * That is, one phrase got repeated. But even without this phrase I don't think the sentence is clear.  While G is abstractly isomorphic to C4 plus an infinite number of copies of Z, it's confusing to identify all the infinite cycle subgroups of G by Z.  I would suggest giving these subgroups different names.  Maybe this sentence could become:


 * Let G2 denote the subgroup of G generated by the point (0, 1). G2 is a cyclic subgroup of order 4.  For an odd prime p, let Gp denote the subgroup of elements with denominator p or 1.  Gp is either empty or an infinite cyclic group.  If p &equiv; 1 (mod 4), then Fermat's theorem on sums of two squares says that p can be written as  for two positive integers a and b.  The point (a2 - b2)/p + (2ab/p)i is a generator of Gp.  If p &equiv; 3 (mod 4), then Gp is empty.  Furthermore, by factoring the denominators of an element of G, it can be shown that G is a direct sum of G2 and the Gp.  That is:
 * $$G \cong G_2 \oplus \bigoplus_{p \equiv 3 \bmod 4} G_p.$$


 * This might be too long (I tend to be too verbose) but I think it is much clearer: Each sentence expects less from the reader.


 * I have the feeling that there should be a relation between the groups Gp and heights. I don't really understand heights, so this is sort of a guess; but I think it's something like, "The subgroup generated by the points of (multiplicative) height less than or equal to M is the subgroup generated by Gp for p &le; M."  That sounds innocuous enough; maybe it's true?


 * Finally, here is an omission: The group of rational points on the circle is SO2(Q), the (definite) rank 2 special orthogonal group of the rational numbers. The group of rational points on the hyperbola is SO1,1(Q), the indefinite rank 2 special orthogonal group of the rational numbers.


 * Ozob (talk) 01:56, 22 February 2012 (UTC)
 * Thanks i've put in your sentence already. I think some things after it will need to be adjusted to the symbols in your sentence.The SO2(Q) is indeed an omission, i think SO2(R) is also, or was also, omitted from circle group. Whichever of the articles gets it in can have it mostly copied and pasted to the other.Thanks again, Rich198.189.194.129 (talk) 19:48, 22 February 2012 (UTC)
 * I know that by SO1,1(Q), you mean a group that preserves a quadratic form with 1,-1 sighnature, but that's about all i know about it.-Rich198.189.194.129 (talk) 20:42, 28 February 2012 (UTC)
 * Points on the unit circle have the form (cos &theta;, sin &theta;). Points on the unit hyperbola have the form (&plusmn;cosh &theta;, sinh &theta;).  If we want to hyperbolically rotate such a point through an angle &phi;, then we use the hyperbolic addition formulas cosh(&theta; + &phi;) = cosh(&theta;)cosh(&phi;) + sinh(&theta;)sinh(&phi;) and sinh(&theta; + &phi;) = cosh(&theta;)sinh(&phi;) + sinh(&theta;)cosh(&phi;).  We can write this as a matrix:

\begin{pmatrix} \cosh(\theta + \phi) \\ \sinh(\theta + \phi)\end{pmatrix} = \begin{pmatrix} \cosh(\phi) & \sinh(\phi) \\ \sinh(\phi) & \cosh(\phi) \end{pmatrix} \begin{pmatrix} \cosh(\theta) \\ \sinh(\theta) \end{pmatrix}. $$
 * SO1,1(R) consists of real 2&times;2 matrices with determinant 1 that preserve a signature (1,1)-form. That would be exactly the above matrices together with the matrices you get by replacing cosh(&phi;) by &minus;cosh(&phi;).  These matrices have determinant one by the analog of the Pythagorean theorem, i.e., the relation cosh(&phi;)2 &minus; sinh(&phi;)2 = 1.  SO1,1(Q) is of course the subgroup of such matrices having rational coefficients.  Ozob (talk) 02:34, 29 February 2012 (UTC)

On the (co-)homology of commutative rings
Hi Ozob. I was reading the wikipedia page Andre Quillen cohomology and I want to look at Quillen's paper On the (co-)homology of commutative rings. I checked on AMS but they don't seem to have it. Since you were the only one who edited that page, I thought you might know where to find it. Can you give me a link? Thanks in advance. Money is tight (talk) 12:10, 25 August 2012 (UTC)


 * If you do a Google Books search for "on the co-homology of commutative rings", it's the first hit (the title is the name of the conference, "Applications of categorical algebra"). Ozob (talk) 12:36, 25 August 2012 (UTC)


 * Thanks! Money is tight (talk) 03:37, 28 August 2012 (UTC)