Dear Professor Kim,

Thank you for your reply. I have another question, this time regarding the group law on a cubic, C. I am happy with showing that the points of C form a group. However I am a little unsure of the ’simplified’ group law as described in the textbook. I believe that if we take O=(0,1,0) as the identity element we can treat the cubic as an affine curve. However as this corresponds to the ‘line at infinity’ in projective 2-space, what point (x,y) does it correspond to in affine 2-space?
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Reply:

IF C is the curve Y^2Z=X^3+axZ^2+bZ^3, in P^2, then it meets the line Z=0 in the point O=(0:1,0) as explained in the book. So we then express C as a union

C=\{O\}\cup C_0

where C_0=C\cap \{(X:Y:Z) | Z\neq 0\}. We then note that
\{(X:Y:Z) | Z\neq 0\}\simeq A^2 by the map

(X:Y:Z)\mapsto (X/Z,Y/Z).

Then C_0 goes to the curve in A^2 with equation

y^2=x^3+ax+b.

Of course the origin does not lie on this curve, because this affine curve corresponds exactly to the complement of the origin in C.

Does this answer your question? Let me know again if you’re still curious.

Here is a question from Dominque Miranda.

And here is my reply.

Dear Prof Kim.

In the Algebraic Geometry notes, we defined a Projective Algebraic Variety to be a subset of the projective space defined by the zeros of a homogeneous polynomial equation. Do we require the Algebraic set to be irreducible as well? and if so, why do we require this extra condition?

Thank you very much.

Acyr

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Because you know too much, you’re looking ahead already. We haven’t yet formally discussed the definition of a projective variety. Rather we’re just discussing examples. We’ll get to the issue of irreducibility and so on within a few weeks, and the question of the precisely correct definition.

As in previous years, I will maintain a blog for my two courses on Algebraic Geometry and Algebraic Number Theory. You can browse through some of the previous years’ entries for a general guide to usage. I feel it’s been considered a fairly helpful means of communication about the material covered, so please do take full advantage of it!

I will be in the fifth floor common room tomorrow from 4-5 PM to answer last minute questions.

Far be it from my intention to clutter the blog with film reviews, but maybe this is a good time, just to give all of you a bit of relief from studying. In any case, it will only be a short remark about`The Class (Entre les murs),’ describing a year’s work for a teacher at a school in northeastern Paris. I do try to keep up with films about education and this one has been highly acclaimed (Palme d’Or and all that), so I was glad to catch it on the plane last week. Reasonably pleasant to watch, sure enough, but the show left me with a question, which perhaps some of you can help me with if you’ve seen the film as well: Are such innocent and impressionable students as depicted there really supposed to be problematic in some way? Somehow, I couldn’t at all comprehend that the situation could be perceived as a difficult one. There were minor obstructions here and there, but everyone in class seemed quite communicative and engaged.

After asking around a bit, I thought to pose the question here for some feedback. So let me know.

Dear Professor Kim,

I just have a few small questions. In lectures, we calculated the class group of Q(\sqrt{10}), which has ring of algebraic integers Z[\sqrt{10}]. We then found that that (maximal ideal) P_2 = (2,\sqrt{10}), N(P_2) = 2.

After a bunch of calculations we had to see whether P_2 was principal or not. Using the result,

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I (non-zero) is principal iff there exists \alpha \in I s.t

|N(\alpha)| = N(I)
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We had to consider if there was n,m \in Z s.t

|N(n+m\sqrt{10})| = 2

Since the general element of the ideal  P_2 is 2n+m\sqrt{10}, is it ‘more correct’ to consider if there was n,m s.t

|N(2n+m\sqrt{10})| = 2

I know this doesn’t make a whole lot of difference, it’s just one of those things.

So

 n^2 - 10m^2 = \pm2

which was equivalent to the statement

n^2 = 2 \mod 5 or n^2 = 3 \mod 5

In general, do we consider any ‘modulo n’ so that the statement is simplified?

Many Thanks
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Reply:

You are absolutely right about the principality question. That is, when we have an ideal I in the ring of integers O_K of an algebraic number field K such that N(I)=n, then we are led to consider solutions to the equation

N(z)=\pm n

for various z. As we’ve seen in many examples, once z is expressed in terms of an integral basis, this becomes an equation in d=[K:Q] variables with integer coefficients to which we can consider solutions. Now, if there are no solutions with z\in O_K, then, a fortiori, there are no solutions z\in I and we can conclude that I is not principal. However, although I’m too lazy now to cook up an example, there are situations where there *is* a solution z\in O_K, but no solution z\in I. In the example you mention, the equation corresponding to z\in I is easily found to be

4n^2-10m^2=\pm 1

which is obviously more resrictive than

n^2-10m^2=\pm 1.

In our case, the latter already has no solution, so we don’t need to consider the more refined equation.

Regarding your second question, firstly, the equation is not *equivalent* to the congruence equation but just implies it. So if the congruence equation has no solution, neither does the original, which is how we used it. Now, I don’t quite understand your final question, but perhaps I should remark that considering congruences is a standard way of investigating solutions to quadratic equations. In fact, it is useful for *any* Diophantine equation. However, a rather deep theorem says that for quadratic equations, sufficiently many congruence equations completely determine whether or not the original equation has rational solutions.

Dear Dr Kim,

There is a post on your blog regarding finding primitive elements. Your advice was to look at the Primitive elements: an example document. Can I assume this would be an acceptable answer in the exam and, would just stating that the method is the same as the one used for proving the Primitive Element Theorem be sufficient justification or do we need to provide further explanation?

Thanks in advance,

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Reply:

Of course the method is acceptable, but I don’t understand what you mean by `sufficient justification.’

Let’s remind ourselves what method we are speaking of:

To find a primitive element in Q(\alpha, \beta), we need to locate a linear combination \alpha + c\beta with c\in Q satisfying certain conditions spelled out in the proof of the Primitive Element Theorem. It might be \alpha+\beta, \alpha-\beta, \alpha+(1/2)\beta, etc. depending on the situation, even though the result tends to be rather simple in the examples that have come up. To use the method of the theorem would mean checking that the conditions are satified for some specific c. If you did this, yes it would be sufficient justification.

Sorry to bombard you with my problems professor, but i was attempting problem sheet 6 in a bid to understand how to calculate class groups properly, and have no real problem with it up until the point where we start to deduce which of the prime ideals are principal. When i say prime ideals i hope i’m right in calling the curly p with subscript of a prime number that. If the norm of a general element is a prime we say that the prime ideal is maximal right? I was attempting question one for root 11 and the one step i seem to have difficulty with is when we calculate the other norms of (3+(11)^1/2) and above, i have a vague understanding of what we do where we assign the norms to the prime ideals depending on what they are. But this step in general seems to allude me whenever i attempt these questions. So if you could shed some light on this step or could just guide me to the theorems or lemmas that would help with this area that would be extremely helpful. Thanks

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Reply:

I’m sorry to say this so close to the exam, but some of your questions are a bit worrisome. For example, the question `If the norm of a general element is a prime we say that the prime ideal is maximal right?’. It’s hard to make out what you mean. A correct statement is that inside the ring O_K of integers inside an algebraic number field K, all non-zero prime ideals are maximal. This fact is actually a bit tricky: it follows from the fact that O_K/I is finite for any non-zero ideal I and that any finite integral domain is a field. I hope you’re not confused about the *definition* of prime and maximal ideals, which just come from general algebra.

Let me guess a bit at what the confusion might be. When factorizing an ideal I, very relevant are the prime factors of N(I). This is because if

I=P_1P_2\cdots P_k

then

N(I)=N(P_1)N(P_2)\cdots N(P_k).

We know in fact that each N(P_i) will be a prime power factor of N(I). This allows us to look for prime ideal factors of I.

I hope you’ve already thoroughly reviewed the online notes. Chapter 4 is the relevant part for this material.

Hi professor,

I was just wandering in the 2007 paper when it says from first principles to determine the ring of algebraic integers in Q[(103)^{1/2}], what this actually means, hope it doesn’t sound like a dumb question. Do we just bear in mind the definition of an algebraic integer and produce a basis for Q[(103)^{1/2}] and show that for any element a + b[(103)^{1/2}] that a and b are integers?

As regards with the previous question i think the  nZ^d was more like n(Z^d).
Thanks
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Reply:

In essence, yes. A general element of Q[(103)^{1/2}] is of the form a+b(103)^{1/2} for a,b\in Q. In that problem, you are expected to show that a+b(103)^{1/2} is an algebraic integer if and only if a,b\in Z, using just the definition. The context of the problem might help you to understand what is expected from the solution: I noticed at some point that there were students who knew the (important!) formula for the ring of integers in general Q[\sqrt{d}] and could justify it, but then got awfully confused when presented with the same problem for specific d.

By the way, to defend my notation nZ^d, note that there are two different ways to insert brackets, both leading to the same subgroups of Z^d. Hence, it’s OK to omit them :-) .

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