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,

 

Renumberings and algebraic integers

Hi,

Many thanks for the revision class last week. I just have a few more specific questions I wondered if you would be able to help me with.

1) You often refer to ‘Theorem 107′ in solutions however this doesn’t seem to correlate with the notes I have, is there any chance you could clarify what this theorem says and how it is applied?

2) In ‘An example of an integral basis’ you select , is there a reason for this rather than ?

Many thanks for your help.

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Reply:

I’m sorry. When I revised the notes, I forgot to update the references on the supplementary material. Most of the time, I do think it’s a good exercise to locate the correct reference somewhere near the incorrect one. (This is not just an excuse for me to be lazy!) In the case of the old theorem 107, this is theorem 111 in the current notes. It allows us to rule out primes from the possible denominators of algebraic integers expressed in terms of a basis when the minimal polynomial for the algebraic integer is Eisenstein for the prime .

Regarding the second question, I used because I tried it out and it worked. It’s a good exercise to try the argument with instead and see if it’s just as good. It shouldn’t be immediately obvious until you’ve followed the pattern of the given proof a good deal of the way. Let me know how it goes.

 

Geometry of complex numbers

Dear professor,

How has your Easter been?

I was doing some revision and I came across a question I would like to ask you. Please enlighten me on it.

Find the least value of if

Thank you for your time.

Hope you had a good holiday.

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Reply:

The Easter break went by quite quickly. I was mostly working at the Institut Des Hautes Etudes Scientifiques in France, but my family got to roam around Paris a bit and attended some Easter events at the Notre Dame Cathedral .

Regarding your question, note that for any two complex numbers and is just the distance between them regarded as points in the plane. In particular, you should be able to draw the line (why?) defined by the condition

after which the answer should be easy.

 

Minimal polynomials

Dr Kim,

Could you tell me where i am going wrong here:

Homework Sheet 3 – Question 4 part 2

Irreducible since has degree 2. Not an Algebraic Integer.

This is not the same answer as in the solutions.

Also could you explain briefly why having degree 2 implies irreducibility. And that even if the answer is as in the solutions, N=2 rather than 4? (I would think that multiplying by 2 does not give a monic polynomial with integer coefficients or would 2 do this?)

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Reply:

For the minimal polynomial, your answer is correct. The solutions contain a misprint.

For any algebraic number , we have

where is the minimal polynomial of . So the degree of is the same as the degree of the field extension. In the case of , we see that must have degree 2. Since it divides and both are monic, we then must have . In particular, the polynomial is irreducible. Of course in this case, it’s easy to see directly that it has no rational roots, and hence is irreducible. But if you think about it, even in general, if you know that the degree of the field extension of is and is a monic polynomial of degree such that , then must be the minimal polynomial and hence irreducible.

To get the , you’ve already noted that satisfies the monic integral polynomial equation

and hence, is an algebraic integer.

 

Checking irreducibility

Dr. Kim

Would you mind telling me the different methods i can use to prove polynomial irreducibility over Q. I know Eisenstein’s Criterion and the brute force approach of finding no zeros that divide the constant (gauss?).

Also i was confused about this. Homework Sheet 2, question 2:

Show is irreducible over Q

In the solutions it says irreducible by eisenstein but i thought and ? Am i correct or have i misunderstood the criterion. Instead is it ok to show +-1,+-2 are not zeros of the polynomial?

Thanks for your help, i thought i would ask these basic questions now as not to waste anyone’s time at our revision class.

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Reply:

I won’t explain in much detail and the general problem can be quite complex, although there are various algorithms for dealing with it.

The facts you refer to are:

1. has a linear factor if and only if it has a rational root. Look at the online notes lecture 1 for the proof and for a review of how to check for rational roots. A corollary of this is that a polynomial of degree two or three is irreducible if and only if it has no roots. (This statement, by the way, is true over any field.)

2. Suppose has degree 4. Then the possibilities for reducibility are:

(a)

where has degree 1 and has dergee 3;

(b) both have degree two. To consider this possibility is slightly more tedious, but still manageable by hand.

To consider the general possibilities in this manner for polynomials of degree five or more becomes quite unwieldy.

But in special situations, one can use

3. Eisenstein’s criterion. Perhaps you have misunderstood the statement of the criterion. In the example you mention, since the prime 2 divides all non-leading coefficients and does not divide the constant term, the criterion says is irreducible. Since this polynomial has degree 3, you are right that one could simply check that it has no rational roots to conclude irreducibility. But Eisenstein’s criterion is already useful for since we don’t have to go through the tedious check to rule out degree two factors.

4. It is often convenient to use a change of variable:

is irreducible if and only if is irreducible for some constants . Try writing out for yourself the elemenary proof of this. As an application, we checked, for example, the irreducibility of

for any prime by checking the irreducibility of using Eisenstein’s criterion.

 

Computing orders mod 23

Hi Sir,

Sorry to disturb you again. I was working out the order of 3 mod 23, which i found to be 11. I did this the following way:

I was wondering if there was a simpler method to find the order instead of doing it using this way?

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Reply:

It’s hard to give a general answer for small moduli, where the improvement over brute force calculation tends to be rather incremental. It might be somewhat helpful to analyze the *possible orders*. In the case at hand, the non-zero elements modulo 23 under multplication form a group of order . Thus, the possible orders of elements are 1,2, 11, and 22. Since 3 clearly doesn’t have order 1 or 2, the possibilities are 11 and 22. But I don’t see an obvious way to rule out the possibility of 22 without calculating as you’ve done. On the other hand, if any element satisfies

then we can immediately conclude that the order is 22.

 

Primitive roots of 1 modulo a prime

Hi Sir,

I wanted to make sure what i was doing was right. The question says find the primitive roots of the primes 5 and 7.

For 5

so i obtained 1,2,3,4.

For 7

so obtained 1,2,3,4,5,6.

Is this the correct way to do this? I was getting confused between primitive roots and residue classes. Residue classes are just the numbers co-prime to O(5-1)=O(4), which is 3,1 and for O(7-1)=O(6) is 5,1.

Am i right or on the wrong track?

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Reply: You are getting several notions mixed up. If you fix a modulus , then the
residue classes modulo are represented by all the possible remainders after dividing by , and hence, are

For example, the residue classes modulo 6 are

On the other hand, the residue classes that are *invertible* modulo are represented by the classes where is coprime to . For example, the residue classes that are invertible modulo 6 are

If happens to be a prime, then all non-zero residue classes are invertible modulo

Fermat’s little theorem says that if is a prime, then for any non-zero residue class , we have

So *all* the non-zero residue classes are -th roots of 1 modulo . However, the *primitive* roots of 1 are those for which

but

for any . That is, these are the residue classes that have exact order with respect to multiplication. For example, modulo 5, is a 4-th root of 1, but definitely not a primitive fourth root. Similarly,

but also

So fails as well to be a primitive 4-th root of 1. However, you can check that and *are* primitive 4-th roots of 1.

Now go back to your investigation of the primitive 6-th roots of 1 modulo 7.

By the way, your question indicates that you should read the definitions and examples in the notes more carefully several times.

 

root of unity

Dear Dr Kim

I know you must be recieving many emails of this sort at this moment in time, so i’ll try and keep this as brief as possible: just a quick (easy) question,

When trying to calculate the roots of a polynomial say

I know it can be split into its linear factors where = cube root of 5

and .

How do you calculate the value of ?

many thanks
sorry again for the inconvenience

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Reply:

Importantly, the correct value is not . This is the complex number on the unit circle at an angle of with the positive real axis. , on the other hand, is a real number. , and are the three solutions to the equation

For this reason, whenever is a solution to an equation of the form

so are and .

As for the value, it depends on the problem. Sometimes, is a perfectly legitimate final expression for a number. On the other hand, you can write it in terms of radicals as

 

Question on automorphism groups

Hello Prof Kim

I got a question about groups and ring ( algebra 4). Can you explain to me what it means to “describe explicitly” for n = 1, 2,3, ….?

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Reply:

In many parts of mathematics, one uses the notation to denote the automorphisms of , where is a set usually with some extra structure. Thus, consists of maps

with the property that

(1) has an inverse;

(2) is compatible with whatever structure has, if any.

So when you’re being asked to describe, say, explicitly, you are being asked to describe all invertible maps

that are compatible with the group structure. One point you shouldn’t get confused by is that isl *always* a group under composition, whatever structure has. If is itself a group, will in general be some other group.

Note that if we had left out the group structure and just considered the set , then the automorphisms would be isomorphic to (‘essentially the same as’) , the symmetric group on three letters. But with the group structure, you need to be more careful. Not all permutations will preserve the group structure. Think about it a bit and ask again if it’s still confusing.