Monthly Archives: April 2009

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 y\sigma_3(y), is there a reason for this rather than y\sigma_2(y)?

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 p from the possible denominators of algebraic integers expressed in terms of a basis B=\{1,  \alpha, \alpha^2,  \ldots ,\alpha^d\} when the minimal polynomial for the algebraic integer \alpha is Eisenstein for the prime p.

Regarding the second question, I used y\sigma_3(y) because I tried it out and it worked. It’s a good exercise to try the argument with y\sigma_3(y) 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.

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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 |z| if |z - i| = |z - 2|.

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 |z-w| for any two complex numbers z and w 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

|z - i| = |z - 2|,

after which the answer should be easy.

Exam advice

Dr Kim

For those of us that unfortunately will not be able to make your revision classes for algebra 3, could you please tell us what you will be covering in those classes and could you please give any tips or advice for the upcoming exam.

Many thanks

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Reply:

It’s somewhat difficult to summarize the revision session because it was based on questions from the audience, rather than following a prepared plan. You can get a sense of the questions that came up by reviewing this blog. Review also the blog entries from around exam time last year. Pretty much all the advice I gave then is still relevant.

factorization

Dear Professor Kim,

Would you be able to go through question 3.b.i & 3.b.ii on the 2007 paper please? I know that the norm is 140 so the factors will have norms dividing (2^2.5.7)^r but I’m an unsure about what to do next and how to
lay out the answer.

Also, some help is needed with Q)4.b.iii on the 2007 paper. I’ve worked out a few norms using the fact that N(r – a) = m(r) but I don’t see anything useful.

Many Thanks!
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Reply:

Look at the document `A few past exam questions’ on the course webpage.

I strongly urge everyone to look at the material on the course webpage. I’ve received by now several questions on material that’s been discussed there, indicating that many people are not making full use of what’s already available.

Positive-definiteness

Dear Dr. Kim,

Could you show me how to determine whether it is positive definite for homework 8 question 1c please.
It will be very grateful if you can show me in your course blog.

Many thanks for your help.

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Reply:

The question takes the vector space V to be the span of \{(1,1,1)^t, (1,2,3)^t\} in C^3, and the Hermitian form to be

\langle v, w \rangle=v^t\bar{w}.

We need to see if \langle v,v\rangle >0 for all non-zero v=(v_1,v_2,v_3)^t \in V. But

\langle v, v\rangle =v^t\bar{v}=|v_1|^2+|v_2|^2+|v_3|^2>0

as long as some v_i is non-zero. So the form is positive definite. Notice that this form is actually defined and is positive definite on all of C^3. In general if you have a Hermitian form on a C-vectors space or a symmetric bilinear form on an R-vector space, if the original form is positive definite, then its restriction to any subspace is also positive definite, simply by exmaining the definition of positive-definiteness.

Minimal polynomials

Dr Kim,

Could you tell me where i am going wrong here:

Homework Sheet 3 – Question 4 part 2

a:=(1+\sqrt{8})/2

2a = 1 + \sqrt{8}

2a - 1 = \sqrt{8}

1/2(2a - 1) = \sqrt{2}

1/4(2a - 1)^2 = 2

a^2 - a - 7/4 = 0

Irreducible since Q(a)= Q(\sqrt{2}) has degree 2. Not an Algebraic Integer.

N=4

This is not the same answer as in the solutions.

Also could you explain briefly why Q(a) having degree 2 implies irreducibility. And that even if the answer is X^2 - X - 31/4 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 \alpha, we have

Q(\alpha)\simeq Q[x]/(m(x))

where m(x) is the minimal polynomial of \alpha. So the degree of m(x) is the same as the degree of the field extension. In the case of a, we see that m(x) must have degree 2. Since it divides x^2-x-7/4 and both are monic, we then must have m(x)=x^2-x-7/4. 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 Q(\alpha) of Q is d and f(x)\in Q[x] is a monic polynomial of degree d such that f(\alpha)=0, then f(x) must be the minimal polynomial and hence irreducible.

To get the N, you’ve already noted that b=2a satisfies the monic integral polynomial equation

(x-1)^2=8

and hence, is an algebraic integer.

I’ve being looking at finding integral bases and particularly at question 6 on the third homework sheet. I’m looking for a integral basis for Q(\sqrt{2} + \sqrt{3}) I understand that (\sqrt{2} + \sqrt{3})/2 will replace in to form an integral basis but I don’t understand how to get to it. It seems too long winded to try every possible value for \sum ab/2 until I find one which is a algebraic integer. What am I missing?
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Reply:

On the course webpage, check the document `An example of an integral basis.’

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 f(X)=X^3 - 2 is irreducible over Q

In the solutions it says irreducible by eisenstein p=2 but i thought p^2=4 and (-2)+4=2=0 \mod 2 ? 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. f(x) 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 f(x) 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 f(x) has degree 4. Then the possibilities for reducibility are:

(a) f(x)=g(x)h(x)

where g has degree 1 and h has dergee 3;

(b) g, h 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 2^2 does not divide the constant term, the criterion says x^3-2 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 x^4-2 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:

f(x) is irreducible if and only if f(ax+b) is irreducible for some constants a,b. Try writing out for yourself the elemenary proof of this. As an application, we checked, for example, the irreducibility of

f(x)=x^{p-1}+x^{p-2}+\cdots+x+1

for any prime p by checking the irreducibility of f(x+1) using Eisenstein’s criterion.

Degrees

Dr Kim,

A quick follow up question:

In certain solved examples of this type, we’ve assumed that Q(\sqrt{3},\sqrt{5}) has a degree 4, how do we know this already?

And in others we simply work out the degree of the minimum polynomial for Q[\sqrt{3}+\sqrt{5}] and prove it’s irreducible. Then it says by the tower theorem the degree of Q(\sqrt{3},\sqrt{5}) is less than or equal to 4?

It does seem that the degree of Q(\sqrt{3},\sqrt{5}) would be 4, but i’m not sure about the theory behind it.

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Reply:

Firstly, the fact that the degree of Q(\sqrt{3},\sqrt{5}) is less than or equal to 4 requires only the tower theorem since

[Q(\sqrt{3},\sqrt{5}):Q(\sqrt{3})] \leq 2

and

[Q(\sqrt{3}):Q]\leq 2.

So when you’ve proved that the minimal polynomial of \sqrt{3}+\sqrt{5} has degree 4, then you know that the subfield Q[\sqrt{3}+\sqrt{5}] \subset Q(\sqrt{3},\sqrt{5}) has degree *equal* to 4, showing that both have degree four and are the same field. This kind of argument is common.

For some more examples of the requisite argument, you can look at the document `Some irrationality’ on the course webpage.

Primitive elements

Dr Kim,

Concerning Primitive Elements, i have a few quieries about calculations. In the 2008 past paper, we have the example of finding the primitive element of

Q(\sqrt{3},\sqrt{5}).

The solutions begin with assuming this is \sqrt{3}+\sqrt{5}. Can we always do this? If not what method do we use to find it? Then we find the degree (of the field generated by the primitive element?) has to be 4 by the tower theorem. I understand this, but in another example if the degree was not the same as the field extension (ie.not 4) what do we do then/how do we define the subfield etc.?

Many thanks,
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Reply:

Look at the course webpage for the document `Primitive elements: an example.’ There, primitive elements are found using the method of the proof of the Primitive Element Theorem. (Note that some page references are now outdated, but you should be able to figure it out.)