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,

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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 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.

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