Re: Question 4 on midterm
William Stein
> William,
> On question 4, you ask to "show how to find n in Z and
> alpha in OK" and to "Explain your steps you use". Does
> that mean I have to actually find them? n is easy to obtain,
> by following the procedure to get alpha is trickier. I
> have to solve a big chinese remainder theorem with ideals
> that are generated by two elements. Not something I want
> to do by hand. Is it enough just to set up the Chinese
> Remainder Problem and state the solution will give you alpha...
Ok, do the following (correctly!) and I'll give full credit:
(1) clearly explain how you "would" find n and alpha (i.e., explain
an algorithm) -- this should be easy, since I basically explained this in
class.
(2) actually compute an n and alpha using any method at all (e.g., asking
sage). show your code.
(3) prove that the n and alpha in (2) are right, e.g., by doing some
sort of calculation, e.g., verify that alpha is in the ideal and
conversely that each generator of the ideal is in (n,alpha).
Reasonable?
-- William
William Stein
> For verifying that the ideals are equal do you want us to show the actual
> relations i.e. x,y \in O_k such that n*x + \alpha*y = g, for g in (2a+2, a-11, -5/2*a-17/2)
> or does it suffice to use the "in" command in sage?
Don't just use the "in" command. Since both are ideals, all you have to
observe is that n in (2a+2, a-11, -5/2*a-17/2) and g in (2a+2,
a-11, -5/2*a-17/2) ,
and conversely that each of 2a+2, a-11, -5/2*a-17/2 are in (n, g).
By "show", it's enough to just write, e.g., 2a+2 in terms of n, g in any way.
William