## Monthly Archives: May 2009

### Special office hour (ANT)

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

### Film

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 aboutThe 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.

### A subtle question about principality

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,

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

$|N(\alpha)| = N(I)$
—————————————————————–

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
———————————————————

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.

### 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?

———————————————————-

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.

### 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 $(3+(11)^1/2)$ 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

———————————————————–

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.

### 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 $Q[(103)^{1/2}]$, 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
—————————————————————————-

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 :-).

### 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 $3^5.17.19$, 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 $-4a^3-27b^2$ but this gives me $3^5.-13$?

3) How do we multiply maximal ideals explicitly? For example in ‘A few past exam questions’ I can see how this is true for $P_2P_5$ but do we then just deduce $P_2P_7$ or is their some way of calculating this? I can also see that an alternative factorization could be as

$(2)=P_2^2 (5+\alpha)=P_5P_7$

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 $4^4.(-p)^3$, is this correct? My calculations gave me either $4^2$ or equivalently $2^4.$

5) In Few Class Groups:

How do we know the ring of algebraic integers is $Z[\alpha]$? 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

$1/3(a_0+a_1x+a_2x^2)?$

Kindest regards,

—————————————————————

(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 $P_2=(2, \alpha)$ and $P_7=(7, \alpha-2)$, we would get

$P_2P_7=(14, 2\alpha-4, 7\alpha, \alpha^2-2\alpha).$

But that isn’t how I obtained the formula you mention. It would have been a bit tricky to guess the generator $(\alpha-2)$ just from the presentation above. What I actually did was factorize $(\alpha-2)$. Since $N(\alpha-2)=14$, then only possibilities are

$(\alpha-2)=P_2P_7$ or $(\alpha-2)=P_2P_7'.$

But it has to be the former since $(\alpha-2)\subset P_7$, so that $P_7|(\alpha-2)$.

(4) In this case, I think I’m right (surprise, surprise). Just follow the computation in that article using $N(\alpha)=-p$.

(5) I’m supposing you mean the problem where $\alpha=2^{1/3}$. The point is that for the translation $\beta=\alpha+1$, we get the irreducible polynomial

$(x-1)^3-2=x^3-3x^2+3x-3,$

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 $m(x)=x^3 + x - 1$ modulo 3 we get $(x+1)(x^2-x-1)$ we then associate these factors with the ideals $P_3$ and $P_9$ respectively. When we compute the norm of $x^2-x-1$ we do so by calculating the determinant of the matrix $L_{a^2-a-1}$ and find that the norm is in fact 9, so $P_9$ is a principle ideal. However, we could just have easily used $x^2+2x-1$ or $x^2+2x+2$ 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.

———————————————-

First of all, I presume your $x^2-x-1$ etc. are $a^2-a-1$ 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 $a$ of polynomials that are congruent to $x^2-x-1$ mod $3$. However, this does not mean they are generators on their own. Of course different elements in an ideal $I$ will have different norms in general. However, an element $b\in I$ is a generator *by itself* (making $I$ into a principal ideal), exactly when $|N(b)|=N(I)$. Of course such a $b$ 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 $a^2-a-1$ is.

A thorny point that comes out of this discussion is that if you had initially presented the ideal as $(3, a^2+2a-1)$, 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 $(2,2\sqrt{15})$ in the ring $Q(\sqrt{15})$, which is on Sheet 5 Question 4a.

I can see that this ideal can be written as $(2)$ so it will have norm 4. Also, in the ring $Z[\sqrt{15}]$ a general element looks like $n + m\sqrt{15},$ where $n,m$ belong to $latex Z$. So if we calculate the norm using the principle

$|(2)| = |Z[x]/(2)| = |Z[x]/(2(n + m\sqrt{15}))| = 4.$

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:

$Z[\sqrt{15}]/(2) = Z[x]/((x^2 - 15),2) = F_2[x]/(x^2 - 15)$
$= F_2[x]/(x^2 - 1) = F_2[x]/(x+1)(x-1) = F_2/(2)$

$|F_2/(2)| = 2$

Please tell me where I’m going wrong.

Many Thanks!
—————————————————-

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

$|(2)| = |Z[x]/(2)| = |Z[x]/(2(n + m\sqrt{15}))| = 4$

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

$|F_2[x]/(x+1)(x-1)|.$

Because the coefficients are in $F_2$, we have $x-1=x+1$. So

$F_2[x]/(x+1)(x-1)=F_2[x]/(x-1)^2\simeq F_2[t]/(t^2),$

from the isomorphism $F_2[x]\simeq F_2[t]$ that takes $x$ to $t+1$. It is easy to see that the $F_2$-vector space $F_2[t]/(t^2)$ has dimension 2 with basis $1, t$. Hence,

$|F_2[x]/(x+1)(x-1)|=|F_2[t]/(t^2) | =4.$

### Long email

I received a long last minute email. I’ll just copy it together with my answers without proper formatting or corrections because I’m also a bit tired.

MK

—————————————————————
Because of lack of time, I will be brief with the answers:

1. Everything is OK except for the statement about the JCF and
the linearly independent eigenvector. A basis for the eigenspace
can be taken as 1. There is just one Jordan block. This should tell you
what the JCF is. (Obviously can’#t be zero).

2. You are right.

3. Right again.

4. The definition as given is correct. For example, might have
f=m. There must have been a misprint in the mark scheme.

5. Yes, these are the same. More precisely, Ker(A)=Ker(-A) for
obvious reasons.

6. One way of thinking about C is as R^2 with some multiplication.
Anyways, the dimension is 2 for the reason you say as well. The dimension
of any field F as a vector space over itself is one. The dimension of
R as a Q-vector space is infinite.

7. See what you can do with L^{98}v.

8. Yes to both questions.

9. This is somewhat tricky. Look at example 125 on the 3704 lecture notes for the general idea.

10. You could, but it might be quicker to take a general symmetric
matrix and show that it can’t work.

11. This is a bit complicated to explain by email in a short amount of
time.

> Dear Prof Kim,
>
> Sorry for the late email. I have quite a few questions on 2201 course
> material. I will be very grateful if you could clear my queries.
>
>
> COURSE SUMMARY AND LECTURE NOTES:
> 1) Summary sheet on course mentions ‘vector space of polynomials with
> complex coefficients of degree at most 5’ Would this be {i, ix, i(x^2),
> i(x^3), i(x^4), i(x^5)} So on D(differentiation):VtoV we get min
> polynomial=ch polynomial=x^5 and so JCF is the zero matrix? And so D has
> only 1 linearly independent eigenvector (i.e. 0)?
>
> 2) Lecture notes page 43: In the proof for diagonalisation theorem, 3
> lines before the end of proof, ‘Hence by the inductive hypothesis, there
> is an orthonormal basis’ shouldn’t this be orthogonal basis?
>
> 3) Page 47 proof of sylvesters law of inertia: In the new basis ci =
> bi/sqrt(q(bi)) and not bi/sqrt(q(bi,bi))?
>
> Also further in the proof ‘q(u)=x1^2+…+xr^2>0’ shouldn’t this be >=0 as
> only then further on U intersection W = {0}
>
> 4) Lecture notes page 25: Definition of minimal polynomial second
> condition on f(T)=0 and f=/0 then deg(f)>=deg(m). Shouldnt deg(f) be
> strictly > deg(m)as suggested in the 2008 mark scheme.
>
> 5) Definition of t-th generalised eigenspace is V(t)(la)=ker((la.Id-T)^t)
> (page27) but in primary decomposition theorem (page28) we use this def as
> V(t)(la)=ker((T-la.Id)^t). Would this not change signs? Are they the same
> because for e.g. kernel implies T(v)=0 so T(-v)=0 too.
>
> 6) What is the dimension of C as an R-vector space? Is this 2 because
> basis is {1, i}?
> What is the basis of C as a C-vector space?
> What is the basis of Q as a Q-vector space?
> What is the basis of R as a Q-vector space?
>
> 7) Suppose there exists a vector v such
> that (L)^100(v)=0,(L)^99(v)=/0. Prove that there exists a
> vector w such that (L)^2(w)=0 and (L)(w)=/0.
> What would be the steps for this proof?
>
>
> HOMEWORK SHEETS:
> 8) Sheet 2 qu7 – Do we only do one long division to get ans?
> Sheet 2 qu9 – Not on syllabus?
>
> 9) How to do sheet 3 qu7?
>
> 10) Sheet 6 qu4 – Can we do this by finding all sym bilinear forms on
> field 2 (8 matrices) and showing none of their quadratic forms are xy?
>
> 11) Sheet 7 qu4 – How do I do part a? Part b is simple. For part c, do we
> explicitly show this for the 9 different possibilities. Most of these
> already in required form (i.e. q00, q01, q10, q11, etc..)
>
>
> Sorry for so many questions. I don’t live near university. Otherwise I
> would have come in to see you. If its easier for you to reply over a
> telephone conversation, please call me on 07971530295.
>
> Thank you very much for your time.
>
>