A subtle question about principality

Dear Professor Kim,

I just have a few small questions. In lectures, we calculated the class group of , which has ring of algebraic integers . We then found that that (maximal ideal) ,

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

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(non-zero) is principal iff there exists s.t

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We had to consider if there was s.t

Since the general element of the ideal is , is it ‘more correct’ to consider if there was s.t

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

So

which was equivalent to the statement

or

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 in the ring of integers of an algebraic number field such that , then we are led to consider solutions to the equation

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

which is obviously more resrictive than

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.

 

Primitive elements

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 , we need to locate a linear combination with satisfying certain conditions spelled out in the proof of the Primitive Element Theorem. It might be , , , 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 . If you did this, yes it would be sufficient justification.

 

Norms, class groups, more, …

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 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 of integers inside an algebraic number field , all non-zero prime ideals are maximal. This fact is actually a bit tricky: it follows from the fact that is finite for any non-zero ideal 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 , very relevant are the prime factors of . This is because if

then

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

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

 

Rings of integers

Hi professor,

I was just wandering in the 2007 paper when it says from first principles to determine the ring of algebraic integers in , 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 and show that for any element that and are integers?

As regards with the previous question i think the was more like .
Thanks
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Reply:

In essence, yes. A general element of is of the form for . In that problem, you are expected to show that is an algebraic integer if and only if , 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 and could justify it, but then got awfully confused when presented with the same problem for specific .

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

 

Integers, class groups, multiplying ideals…

Many thanks, that wasn’t meant to sound like quite such a leading questions, I think I was in the midst of exam panic when I sent it! Sorry to fire off another list of questions, I’m fully aware that you must be inundated with emails at this time of year, so thank you again for being so prompt and clear in your responses!

1) In the 2008 exam qu1 part d: Is it possible to just calculate the minimal polynomial and see if the degree is the same as the extension?

2) In A Few Past Exam questions:

The discriminant in the second part is given as , no matter how many different methods I use to calculate this I don’t get the right answer! I was using the assumption that as a cubic we can use but this gives me ?

3) How do we multiply maximal ideals explicitly? For example in ‘A few past exam questions’ I can see how this is true for but do we then just deduce or is their some way of calculating this? I can also see that an alternative factorization could be as

but again don’t see how to explicitly calculate the second result.

4) In Integral Bases and Translations: The discriminant of B is given as , is this correct? My calculations gave me either or equivalently

5) In Few Class Groups:

How do we know the ring of algebraic integers is ? When calculating I get a possible algebraic integer with prime 2 (which is eliminated using Eisenstein) but am still left with prime 3 giving the possibility of algebraic integer

Kindest regards,

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Reply:

(1) Yes, it’s fine to do this. Another way is to use the proof of the primitive element theorem, as explained in `Primitive elements: an example.’

(2) You are right! That was a silly error on my part. Thankfully, it doesn’t affect the rest of the argument at all, so it went unnoticed.

(3) Multiplying explicitly isn’t too hard by just multiplying the generators. For example, in the case of and , we would get

But that isn’t how I obtained the formula you mention. It would have been a bit tricky to guess the generator just from the presentation above. What I actually did was factorize . Since , then only possibilities are

or

But it has to be the former since , so that .

(4) In this case, I think I’m right (surprise, surprise). Just follow the computation in that article using .

(5) I’m supposing you mean the problem where . The point is that for the translation , we get the irreducible polynomial

which is Eisenstein for the prime 3. Now follow the reasoning in `Integral bases and translations.

 

Norms of elements and principality

Dear Professor Kim,

Sorry to bombard you with these questions. I have come across a problem on your note ’some principle ideals’. When we factorize modulo 3 we get we then associate these factors with the ideals and respectively. When we compute the norm of we do so by calculating the determinant of the matrix and find that the norm is in fact 9, so is a principle ideal. However, we could just have easily used or and in each case I get a different answer for the determinant. Have I made an error or is there a canonical form of sort that I should be aware of?

Thank you for your time.

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Reply:

First of all, I presume your etc. are etc. All the elements you mention do indeed belong to the ideal and can be used as generators *when used together with the element 3*. Indeed they are all all evaluations at of polynomials that are congruent to mod . However, this does not mean they are generators on their own. Of course different elements in an ideal will have different norms in general. However, an element is a generator *by itself* (making into a principal ideal), exactly when . Of course such a need not exist. I haven’t calculated the norms of the elements you mention, but if their norms come out larger than 9, it merely says they are not generators (again, by themselves), while is.

A thorny point that comes out of this discussion is that if you had initially presented the ideal as , for example, then it might have been harder to see that it is principal.

 

More norms of ideals

Dear Professor Kim,

I am unsure of how to calculate the norm of in the ring , which is on Sheet 5 Question 4a.

I can see that this ideal can be written as so it will have norm 4. Also, in the ring a general element looks like where belong to $ latex Z$. So if we calculate the norm using the principle

From what I understand this is the reasoning you give in ‘Some remarks on factorization’.

However, if we use the method which given further down that sheet I get:

Please tell me where I’m going wrong.

Many Thanks!
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Reply:

first of all, I hope you can see that the line

above doesn’t make too much sense. The second displayed equation is almost right, except an error occurs when computing

Because the coefficients are in , we have . So

from the isomorphism that takes to . It is easy to see that the -vector space has dimension 2 with basis . Hence,

 

Question on discriminants

Hi Professor Kim hope you’re doing well. I just had a quick question to do with the notes on lecture 12 on the pdf file, in lecture 12 theres a ‘proof for the theorem’, I was just wandering is this an alternative proof for theorem 106? Or is it part of the same proof but just looking at the Norm of the derivative of the minimal polynomial?

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Reply:

I’m sorry. That portion of the notes is a bit disorganized. You are right that it is just an alternative proof. Thanks for pointing this out.

 

3074 exam questions

Hi Sir,

I have a few questions regarding the final exam:

1/. Will this years exam follow the same format of the test you set last year? i.e a stronger basis on method rather than proof.

2/. When looking at last years exam, I was unable to find the proof of question 1 part d in the notes. Where can I find the proof?

3/. Is it essential to learn the proof of dedekinds theorem for the exam?

4/. In last years exam, question 5, i.e calculating the class group, you said the class group was less than or equal to 6 where is this result from i.e the last paragraph of the answers.

Kindest Regards,
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Reply:

1. It’s difficult for me to give any precise answer to this question. I will just say that the overall distribution of types of problem will not differ too much from previous years. But you shuold remember my repeated warning that the notion of similarity in relation to a collection of problems needs to be understood rather generally, as happens when you have a sophisticated understanding of the material.

2. You should look at the article on primitive elements on the course webpage.

3. I can’t answer such a specific question. Dedekind’s Prime Factorization Theorem is certainly on the list of important theorems that I put in the summary.

4. For this question first I’ll ask you to read the previous portion of the proof very carefully. You can ask again if you’re completely stuck.

 

More norms

Dear Prof.,

I am a student doing your 3704 course. Please kindly answer some questions that i got. Thanks in advance

1) I noticed this pattern in some of the calculations, is this correct?

where is a root of the minimal polynomial; and are constants.

2) I am following the solution of 2008 past paper question 2. Is it true that is an algebraic integer if its norm and trace are integers.

3) When distinguishing between two fields, and with basis and , we check that if is in or not. Does this simply mean if is square-free in the rational?
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Reply:

The general nature of your questions indicates you should review the course more thoroughly from the beginning. I will try to indicate just a few points where your questions are ill-formed and/or worrisome.

(1) When you say and are `constants,’ it’s very unclear what this means. All of the quantities are `constant’ in the usual sense that they are numbers. Also, when you say ` is a root of the minimal polynomial’ what minimal polynomial are you referring to?

One general fact about norms of numbers that follows immediately from the definitions (that is, you should know how to prove it easily at this point) is that

In particular, . Of course, should be a *natural number* here, that is, . When you refer to as a constant, you might mean that it is rational. In that case, if the norm is being taken inside a number field of degree , then . Therefore, if is rational and is any element of , then

Once again, you should have a very clear understanding of how this equality comes about starting from the definition. The worst possible approach to this material is to try to pick up on patterns of equalities that occur in the notes and then vaguely guess which variations on them are admissible. You may not be doing this, but this strategy is unfortunately common, so I’m giving a strong warning against it whenever I can. Note that in this discussion, whether is the root of some specific minimal polynomial or is irrelevant. Try to understand for yourself why that phrase occurred in your question. (Of course any algebraic number is a root of its *own* minimal polynomial.)

(2) This question makes sense. The fact is that the norm and trace of any algebraic integer are integers. However, the converse is false in general. On the other hand, if and is a quadratic field, then is an algebraic integer if and only if it norm and trace are integers. Prove this last statement as an exercise.

(3) The word `square-free’ for an integer means that is not divisible by the square of any prime. A rational number written in reduced form is square-free if both and are square-free integers. A rational number is a square if there is a rational number such that . Be careful not to confuse these elementary concepts. Once again, the way you formulated the question is itself a bit worrisome. For example, the sentence that begins `Does this simply mean…’ is very bad in many ways.

Regarding the motivating problem, if and are isomorphic, then for any choice of bases and , we would have that is a square. In particular, if is not a square for some bases and , then and are *not* isomorphic. One should be aware that the converse is false: being a square certainly does not imply that and are isomorphic. Examples illustrating all this are contained in the notes.