What are some key results in algebraic number theory and how can they be proven?

[Diophantus]

Hi all,

I have been studying the Pisot-Vijayaraghavan numbers recently however I have little background in the basic theory of algebraic numbers. There are a number of standard results which the texts give without proof and which I haven't yet been able to prove myself.

Here are a few that are bugging me:

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k), then a(1), a(2), ... ,a(k) are distict.

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k), then a^n = a(1)^n is an algebraic integer with conjugates a(2)^n , ... , a(k)^n.

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k) and let P(x) be a monic polynomial with rational integer coefficients, then P(a) =/= 0 implies P(a(i)) =/= 0 for all i = 2 , ... , k.

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k) and let P(x) be a monic polynomial with rational integer coefficients, then P(a) is an algebraic integer with conjugates P(a(2)) , ... , P(a(k)).

Any hints on how to prove the above would be very useful.

Regards.

 

[Hurkyl]

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k), then a(1), a(2), ... ,a(k) are distict.

The conjugates of a are the roots of the minimal polynomial f of a. Assume f has a repeated root. Derive a contradiction. Derivatives will be useful. (Do you know f is irreducible?)

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k) and let P(x) be a monic polynomial with rational integer coefficients, then P(a) =/= 0 implies P(a(i)) =/= 0 for all i = 2 , ... , k.

Let f be the minimal polynomial of a. If P(a(i)) = 0, does that tell you something?



At the moment, the only way I see for sure to prove the other two involves conjugation. Are you familiar with that? Or any Galois theory? (at least one of the above is also easy to prove via conjugation)

 

[Diophantus]

Hi Hurkyl,

Thanks for the reply. I wanted to reply sooner myself but been really busy. I've managed to prove the results you helped me with now. As soon as I proved some basic lemmas such as f irreducible over Q they were not too dificult.

At the moment, the only way I see for sure to prove the other two involves conjugation. Are you familiar with that? Or any Galois theory? (at least one of the above is also easy to prove via conjugation)

Alas I am not familiar with conjugation or Galois theory other than a vague notion of what they are. I am certainly very willing to learn the necessary tools though if you are willing to give me a few pointers.

Many regards and thanks again for your previous guidance

Diophantus

 

[Hurkyl]

I'm certainly not a substitute for a textbook!

In this context, the basic "picture" is that all of the roots of an irreducible polynomial over Q "look" the same. Conjugation makes this more explicit, by actually permuting the roots around. (Complex conjugation is the most famous example)

Let a = a(1) be an algebraic integer with conjugates a(2) , ... , a(k), then a^n = a(1)^n is an algebraic integer with conjugates a(2)^n , ... , a(k)^n.

For example, the geometric idea is that all of the a(i) should look the same... so all of the a(j)^n should also look the same. Algebraically, any conjugation must send a(1) to some a(i). Therefore, it sends a(1)^n to a(i)^n. (And all i occur) Thus, the conjugates of a(1)^n are the a(j)^n.

 

[Diophantus]

Is this the same as conjugation actions in group theory? As luck has it I think my next algebra lecture is due to cover that.

 

[Diophantus]

Hmmm maybe not. Conjugation seems to mean a plethora of different things.

What sort of textbook should I be looking for? It would be even better if anyone could recommend some titles.

Regards

 

[Hurkyl]

I'm sort of hoping someone better qualified could chime on in this. My expectation is that an introductory book on algebraic number theory would be very likely to discuss them, as would any abstract algebra text that talked much about fields.

(So look ahead in your algebra book to see what it has to say about fields. )

 

[mathwonk]

well i am not quite up on this but i agree with hurkyl. in general if a is an element of a galois extension, and s1,...,sn are the elements of the galois group, then the conjugates of a are thye elements of the orbit of a by s1,...,sn, i.e. the elemenmts s1(a),...,sn(a).

since automorphisms commute with polynomialsl, it follwos that the orbit of p(a), where p is a polynomial over the abse ring or field, is the oprbit of p(a) under the si, which is the same as the si applied to the orbit of a.

mumble mumble,...

 

[mathwonk]

the fact that a polynom,ial, applied to an algebraic integer is a gain an algebraic integer is analogous to the afct from field theory that the set of algebrasic elements over as field forms a field. i.e. algebraic integers are elements not algebraic, but integral over sa ring, and the integral elements also form a ring, as follows from the criterion using the cayley hamilton theorem argument, that an element is integrakl iff it generates a finite module over the base ring,. so this topic is the analogous theory for rings of the galois theory for fields.

 

[mathwonk]

the standard text on alg number theory is by neukirch, but it is pricy.

 

[mathwonk]

the standard construction is to form the product of, all (X-si(a)) for all galois elements si, and argue that this product is symmetric wrt the si, hence belongs to the base field. this shows that there exists a polynomial over the base field whose roots are the conjugates of a.

 

[mathwonk]

you have tod efine con jugates to prove they are distinct. i.e. if they are defiend as the orbit under the gaklois they are distinct by definition. if they are defiend as the full set of roots, with multiplicities of the minimal polynomial of one of them, then you need to know the minimal polynomial has no repeated roots. this is always true for any irredudible polynomial over a field of characteristic zero, like Q, but not in general fields.

 

[DeadWolfe]

Damn guy, you really got to work on your typing.