Virtual Office Hours

> First, is there a way to calculate in Magma with some letters left
> undetermined?  For example, if I wanted the formula for the discriminant
> of an arbitrary cubic x^3+p*x+q, is there a way of defining p and q so
> that Magma will print out -4*p^3-27*q^2 without asking me to set values
> for p and q?

MAGMA doesn't make this so easy, though it can be done.   You create
a two variable polynomial ring S over the integers, then make a polynomial
ring over S in one variable x, and ask for the discriminant.  

> S := PolynomialRing(Integers(), 2);
> R := PolynomialRing(S);
> Discriminant(x^3+p*x+q);
-4*p^3 - 27*q^2

In PARI (type "gp" to run or click the PARI button on the web site)
you could just type "poldisc(x^3+p*x+q)".  

> Second, what is MECCAH?  And do we get an account through you or do we
> request it from somewhere else?

Yes, I will create an account for you.  Just write back with the login 
name you want (probably your fas login, but you can choose something  else).  

> I don't see how your argument works, considering that the double infinite
> cone (if it is what I think it is) is not convex.  Sure, you can conclude
> that there are two distinct points in the infinite cone whose difference 
> is
> in S, but it doesn't follow that the difference is in the infinite cone.
> Because of this, how do you know that each time you form a new cone, you
> have a new point?

 The following is a better hint:

Hint: Mimic the broof I gave in class of Blichfeld's lemma.  Namely,
consider the map \frac{1}{2}S --> V/L.  Since \frac{1}{2}S has volume
strictly greater than m*Vol(V/L) (consider the case of equal volume
separately), there must be a single point alpha in V/L that is the
image of m+1 distinct elements of \frac{1}{2}S.  This last conclusion
isn't obvious, but it's not too hard to see---If each element
of V/L is hit by at most m elements of \frac{1}{2}S, then since
the continuous map \frac{1}{2}S --> V/L preserves volume, the volume 
of \frac{1}{2}S is at most m*Vol(V/L), a contradiction.  (Cover V/L
by discs so that the inverse image of the disc in \frac{1}{2}S is
at most m copies of the disc, and note that \frac{1}{2}S is the union
of those inverse images, etc.).   Now we have P_1,...,P_{m+1} all
in \frac{1}{2}*S and all mapping to the same element of V/L.  As
before, show that 1/2(P_1-P_{m+1}), 1/2(P_2-P_{m+1}), etc. are all 
in L and all in S, and they are all different.  Conclude that the
intersection of S and L has m nonzero elements.    (To get 2*m distinct
points, probably take negatives and verify something.)

I don't know whether every ideal in an order is
generated by two elements.    

> This is somewhat of a rehash of the question earlier about "reducing"
> the set of generators of an ideal. I have a nice ideal 
> [
>     2,
>     a - 1,
>     b - 1,
>     a*b - a - b - 3
> ]
>  
> in Z[a,b] for which Magma hasn't realized that ab-a-b-3=(a-1)(b-1)-4.
> The ideal is not principal (Good b/c I don't want it to be) but I want
> to find a minimal generating set (specifically I want to see if it has 3
> or 2 generators). If I can cast the ideal to a module then I can use the
> MinimalBasis(M) command. However I can't seem to get the casting to work
> right:
>  
> >> Module(I);
>          ^
> Runtime error in 'Module': The ideal must be of a relative order
>  
> It seems to be because I need a RngOrdFracIdl to construct the module
> but I have a RngFunOrdIdl from the ideal command.
>  
> I also tried constructing the module directly:
>  
> >> M := sub;
>                   ^
> Runtime error in 'Module': Order must be over a maximal order
>  
> and even replacing O with OK (the ring of integers in K) still gives the
> same error. Moreover, it seems strange that I would need to cast into a
> module to accomplish this. Is there no MinimalBasis(I) command for
> ideals?

I don't know whether MAGMA can do this computation at all, because it requires
Groebner basis over rings (instead of fields), which are iffy in general.
Here's how to ask MAGMA for a minimal basis:
---------------
R := PolynomialRing(Integers(),2);
I := ideal;
MinimalBasis(I);
 
MinimalBasis(I);
               ^
Runtime error in 'MinimalBasis': Coefficient ring of argument 1 is not a field
---------------
However, as you can see MAGMA can't do it because we're working over a ring
instead of a field.  You could try it over a field, but then you lose the
interesting structure:
 
R := PolynomialRing(RationalField(),2);
I := ideal;
MinimalBasis(I);
[
    2
]
 
I'd be happy if you found some way to do this computation in MAGMA, but
my general impression is that for Groebner basis don't work so well over
integer rings.  Feel free to better mathematics by improving this.

> In problem number 6 it appears that I is the entire ring of integers
> because if x = sqrt(5) then
> 2 + x \in I so x(2 + x) = 5 + 2x \in I.  Also 5 in I so 2 + x - 5 = -3 
> + x \in I.  And so 2*(-3 + x) + 5 + 2x = -1 \in I so 1 in I.
>
> Did you mean it to be this way?  Seems kinda odd to do the CRT with one 
> of the ideals being the entire ring.

No, I made a mistake by making I the unit ideal.  However, I'm going
to let the problem stand as is, so as not to cause too much confusion, 
and because the rest of the problem still makes sense (it's just
much easier).  Please feel free to make up an additional non-unit
ideal I_2 and do the problem for that ideal too.

On MAGMA, if a function has an "optional parameter," how do you change 
the default value?
 
If the function is called "f" and the parameter is called param and
you want to set it to 1, you do this:
 
   f(normal arguments : param := 1);

> Dear Professor Stein,
>   In question 3c of set 2, do you mean "nonzero prime ideal" instead of
> "nonzero ideal?"

Yes, I meant nonzero PRIME ideal.

> 1.  I was checking my answer on finding the discriminant of Q(sqrt(-3)) 
> on MAGMA, and it said it was -12.  I'm pretty sure it's -3 (cf p. 413 of 
> Artin).  I typed in
>
> P:=PolynomialRing(RationalField());
> N:=NumberField(x^2+3);
> Discriminant(N);
>
> and it said "-12"
>

That's the discriminant of the number field, which is defined in MAGMA
to be the discriminant of the polynomial that defines the number field.
You want the discriminant of the maximal order:

O := MaximalOrder(N);
Discriminant(O);

etc.

> 2.  If one has a number field L over a numberfield K, and one writes 
> L=K(\alpha) for some \alpha whose minimal polynomial is f(x), (which we 
> can always do), then I know that its true that f(x) cannot have multiple 
> roots.  I can't remember how to prove this though; it's just an 
> elementary thing from Galois theory.  Do you remember?

You should prove that if f had a double root then f would be reducible. 
Hint: gcd(f,f') would not be 1 if there were is a double root.

> 3.  On p. 50 of Swinnerton-Dyer, he says that the homeomorphism induced 
> by an element a in the number field k is measure preserving on V_k/ k.  
> I don't see how this follows; there are certainly homeomorphisms that 
> are not measure preserving (this was a recurring theme on my homework 
> for real analysis last semester).

Let's assume the results stated in Appendix A1.3.  Then every invariants
measure on V_k/k is a scalar multiple of the Haar measure.  If a is in k,
and mu is a Haar measure on V_k/k, then S|-->mu(a*S) is another measure
on V_k/k, so it is a fixed scalar multiple ||a|| of mu.   Because V_k/k 
has finite volume and a is an automorphism (hence surjective), we have 
that mu(a*V_k/k) = mu(V_k/k), so the scalar multiple ||a|| is 1.  This
implies that multiplication by a preserves the measure, as claimed on 
page 50.
  

On page 4 of Swinnerton-Dyer, there is a comment about embeddings of k
into C. He claims that there are [k:Q]=n such embeddings. I just don't
see what these embeddings are. It seems like the natural embeddings are
those given by elements of the Galois group of k over Q. What am I
missing?
 
Remember that k need not be Galois.  For example, if k=Q(2^(1/3)), then there
are no nontrivial elements of the automorphism group of k over Q.
 
Here's how to understand what Swinnerton-Dyer is talking about.  By the primitive
element theorem, there exists a single element z in k such that k=Q(z).  The
map Q[x] --> k got by sending x to z induces an isomorphism
                Q[x]/(f) --> k
where f is the minimal polynomial of z.  So instead let's view k as being
Q[x]/(f), and ask, "in how many ways can I embed Q[x]/(f) into the complex
numbers?"   To give an embedding of Q[x]/(f) into C is the same as giving
a homomorphism from Q[x] to C that sends f to 0.  This is the same as giving
the image of x in C, with the single requirement that the image of x be a
root of f.    Thus the embeddings of Q[x]/(f) into C are in bijection with
the roots of f.  Since deg(f) = [k:Q], the assertion follows.