Monthly Archives: May 2008

Workshop and 2201

Dear Prof. Kim,

with regard to the Non-Commutative Constructions in Arithmetic and Geometry workshop, please confirm the date and time since the e-mail says:

time: 7-8 June.

Also how formal is it.

secondly, please recommend me some books for MATH 2201, something that is more elementary or explicitly explains whats in the course. i have consulted the books you had recommended for the course but did not find the very helpful for me. i want something that gives me the roots of the topics studied.

Reply:

Yes, 7-8 June for the workshop is correct. You can see a precise schedule on the webpage. The level of formality is hard to describe precisely. There is a sense in which mathematics meetings are all rather informal, so it’s certainly no cause for concern. However, I should warn you that the lectures will be at a very high level. For serious students, I still think it’s good to come into contact with presentations by world-class researchers (this description obviously doesn’t include me) as early as possible. That’s why I issued the general invitations to students.

As far as linear algebra is concerned, there are two recommendations I can make:

1. Finite-dimensional vector spaces by Paul Halmos

This is a classic text that deals primarily with the *concepts* of linear algebra abstractly, and at a rather deep level.

2. Linear algebra in action by Harry Dym

This book is heavily computational and provides a very solid understanding of the important techniques in matrix algebra. It also look toward quite advanced work in analysis.

It could be better to move on to other things at this point to see really how linear algebra functions in higher level mathematics. That can often help consolidate your understanding of the basic material. The textbook `Algebra’ by Michael Artin is not about linear algebra, but contains a quick summary of the basics at the beginning. This is because he emphasizes throughout the text the examples from linear algebra, even when discussing groups, rings, fields, etc.

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Norm and trace via matrices

Hi sir.
We can calculate the norm of an element by considering the determinant of the linear map, but how, exactly, does one calculate the TRACE of an element using linear algebra methods? (-it is always good to have a last resort!)

Thanks.

Reply:

Here is a short note on this topic.

Last minute questions

I’ll be in the 5th floor common room tomorrow (15 May) from 3 to 4 PM to help you with them.

Hmm

I guess I should then also disavow concern with my image, serious or not….

New design

In keeping with my serious image 🙂 , I suppose I should disavow any concern with the mere appearance of this site. The reason for changing was that the previous format displayed the comments in very small type, making it a bit hard to read mathematics. Obviously, popular blogs are not constructed primarily to facilitate scientific communication.

Norm bounds

Hi,

I think I understand how to calculate some class groups. but what worries me is when in an exam we are rounding to calculate size of norms, as we don’t have a square number and are dividing by pi. Is there any guidance on whether we should round up or down in certain certain circumstances. And if say we calculate ideals with norms less than 7 instead of 6 is there anyway we will be able to tell our mistake?

Thanks for your help

Reply:

To be frank, I haven’t yet seen an example where this could cause a problem. But it might be useful to note that an overestimate for the norm bound is preferable to an underestimate, if there is any doubt. You see, with an overestimate, at the worst, you just have to find a few more relations or check a few more ideals for principality that you didn’t really have to. This doesn’t affect the final answer. But if you’ve made an estimate that’s too small, you may in fact have missed entirely a non-principal ideal representing an element in the class group not equivalent to any of those satisfying the (incorrect) bound.

If this paragraph is hard to understand, think about it a little bit. It is important that it make clear sense to you at this point.

3704 Sheet 6

Dear Sir,

In Problem Sheet 6, for k=Q(sq.root 10)

How did the last bit come about? Why is the class group finally {Ok,p2} Why is p3,p*3 and p5 dropped from the class group? The previous lecturer seemed to have
used mod 5 in his notation of the general element of Ok. Can you please give an answer using the methods from “A few class groups”. In particular can you please
give an answer with the same method as you did for Q(\sqrt{-33})? Is this possible?

Reply:

I hope you can see that the methods pretty much the same for all the examples. In that problem, there is a list

O_K,P_2, P_3, P_3', (2), P_5, P_2P_3, P_2P_3'

of ideal of norm less than 7, and every element in the class group is equivalent to one of these. We know that P_2^2=(2), and hence,

P_2^{-1} \sim P_2

The relation

(\sqrt{10})=P_2P_5

shows that

P_5 \sim  P_2^{-1} \sim P_2

(2-\sqrt{10}) =  P_2P_3

then gives

P_2P_3\sim  O_K

and

P_3 \sim P_2^{-1} \sim  P_2

Also,

P_3'\sim P_3^{-1} \sim P_2^{-1} \sim P_2

and finally,

P_3'P_2 \sim P_2^2\sim   O_K

So it turns out that

O_K, P_2

represent all classes in the class group.

One sees thereby that the class group is either trivial (if P_2 is principal), or Z/2 (if P_2 is not principal). At this point, one checks that there is no element of O_K with norm 2, using the \mod 5 argument, and hence, that P_2 cannot be principal.

Addendum to previous post

Thank you for your response. What confused me was, I was taking l to equal 8 as opposed to 2 because it is the largest natural number whose square divides the discriminant. So in the future am I right in saying that I will look for the largest prime whose square divides the discriminant?

Reply:

In essence, yes. The reason for doing this is contained inside the `first trick’ of lecture 13.

Integral bases and translation

Hello Professor,

Here is another Problem I’m having, It’s from the 2004 exam paper, Q2 ii) :

f(X) = X^4 – p. Let *alpha* be the zero of f.

By making the change of variable g(X) = f(X+1) or otherwise, Show that if p= 3 mod 4 then

{1, *alpha*, *alpha*^2, *alpha*^3}

is an integral basis in Q(*alpha*).

I have calculated the absolute value discriminant of the basis {1, *alpha*, *alpha*^2, *alpha*^3} to equal 64p^3 using N(f'(*alpha*)).

I’m not too sure what method to use, could you perhaps point me in the right direction.

Reply:

The discriminant is incorrect. Be careful of the degree of the number field. In any case, here is a solution.

More class group questions

Dear Sir,

In reference to your material posted online for “A few class group”,

In the 2nd example where m(x)= x^3+4x-2. How did you factorise m(x) over F7? How did you do it? Why did u choose to divide m(x) with (x+5) initially? Is there a general way for obtaining the factors of m(x)? Is it by trial and error?

Also from the 3rd example, I am really confused by the sentence “Thus, we see that {e,[P2],[P3],[P2P3]} is an order 4 subgroup of CLk in which each element has order 2.” What does this mean? What is e? and why do we consider a subgroup of order 4? Where did the P7 go? How did you derive the inequalities for |Clk|?

After the 3rd example I seem to be really confused in determining if Pi is a principal or not. Can you please help in giving a general way to decide if each prime is a principle idea? Does it differ for every question? Is the prime a principal as long as the N(pi)=(pi)?

Thank you. Thanks for holding up this blog too!!!

Reply:

Regarding factorization of polynomials in F_p[x], there are some algorithms that improve the efficiency, but I do it by trial and error up to a cubic. This is because in these cases, a polynomial is reducible if and only if it has a root. It’s not so hard to plug in the numbers in small fields like F_7 to check for roots. That’s how I obtained the factor x+5.

In the third example, e refers to the identity element in the class group, that is, the class of the unit ideal O_K. In the computations of the previous paragraph, we showed that the set of \{e, [P_2], [P_3], [P_2P_3]\} really has order 4, and is also closed under the group law and taking inverses. For example, P_2^2=e.

Similar observations show that in fact, all elements have order 2. Thus, this set is a subgroup of Cl_K isomorphic to Z/2\times Z/2.
This is where 4\leq |Cl_K| comes from. But from the upper bound of 7, we had removed P_5, and hence, arrived at an upper bound of 6.

The group theory that follows then allows us to reach the conclusion. Because of that group theory, we can circumvent entirely a consideration of P_7.

As for principality, the main fact we use is that I is principal if and only if there is an element a\in I such that |N(a)|=N(I).