A strange fact about some cyclotimic integers
Marc Bogaerts
tests I have done until now give the same result:
If z is an algebraic integer of Q_p with Trace(z)=0 then all the
coefficients of the minimal polynomial are multiples of p except for the
first one (which is 1 of course).
An easy way to obtain such an algebraic number is to start with an
arbitrary a.n. x and take z = x - s(x), where s is a generator of the
(cyclic) Galois group of Q_p.
Is this result proven somewhere?
Achava Nakhash, the Loving Snake
wrote:
Not an answer to your interesting question, but you get all numbers of
trace 0 in this particular number field or in any cyclic extension,
where s is again the generator of the galois group. This is called
the additive form of Hilberts' Theorem 90. I don't know when it was
first proved.
Another way to get traceless algebraic integers is to regard
z = a_0 + a_1*(zeta) + ... + a_(p-2)*(zeta^(p-2))
Then tracelss is equivalent to p*a_0 = a_0 + a_1 + ... + a_(p-2)
I don't know what good this will do you, but it is a way of
determining traceless numbers in prime cyclotomic extensions.
Regards,
Achava
Achava Nakhash, the Loving Snake
Marc,
I don't know if it is proven somewhere or not. I suspect it is. It
is not hard. The answer popped into my head about 15 minutes ago. I
am too sleepy to give details, but here is the outline. As I
mentioned in my previous reply, if trace(z) = 0, then z = x - s(x) for
some x in the field. I can show it is sufficient to show that x - s
(s) must be divisible by pi, also known as 1 - zeta, the prime lying
over p, and which is invariant under the Galoins group, at least as an
ideal. Notice that 1 - s(1) = 0, and zeta^r - s(zeta^r) = zeta^r -
zeta^(2r) =
(zeta^r)( 1 - zeta^r). For all r, 1 - zeta^r is an associate of pi,
and hence divible by pi. It follows that z is divisible by pi. In
other words, any algebraic integer z with
trace(z) = 0 must be divisible by pi. Since pi is invariant under the
Galois group, all of the conjugates of z are also divisible by pi, and
so all of the elementary symmetric function in the conjugates of z
must be divisible by pi. But since these are also integers in the
usual sense, they must actually be divisible by p, QED.
The one gap in what I have written is not difficult, but I can barely
keep my eyes open at this point so I will stop and continue tomorrow.
Happy New Year.
Regards,
Achava
Achava Nakhash, the Loving Snake
I thought last night that I could prove that a traceless algebraic
integer in one of these cyclotomic fields must be x - s(x) for an
algebraic integer x, and the result really does follow from that.
Here is how far I have gotten. Suppose z = x - s(x), and z is an
algebraic integer in this field. Then we can write x = r/n, where r
is an algebraic integer in the field, and n is a rational integer.
This fact is true in any number field and is easy. Then
x - s(x) = r/n - s(r)/n = (r - s(r))/n, Here n must divide r - s(r)
because we know up front that z = (s - s(r))/n is an algebraic
integer. The numerator is clearly divisible by pi, so if (n, p) = 1,
z must be divisible by pi and so your conclusion holds. If n is
divisble by p, it is not clear to me what to do. So the only
possible cases that doesn't work is an algebraic integer z such that
z = x - s(x), and, writing x = r/n, we have n is divisble by p.
That is my current progress report. This path either leads to a proof
or to a counterexample
Regards,
Achava
Marc Bogaerts
> I don't know if it is proven somewhere or not. I suspect it is. It
> is not hard. The answer popped into my head about 15 minutes ago. I
> am too sleepy to give details, but here is the outline. As I
> mentioned in my previous reply, if trace(z) = 0, then z = x - s(x) for
> some x in the field.
Indeed, as in the proof of Hilbert's 90th theorem.
>I can show it is sufficient to show that x - s
> (s) must be divisible by pi, also known as 1 - zeta, the prime lying
> over p, and which is invariant under the Galoins group, at least as an
> ideal. Notice that 1 - s(1) = 0, and zeta^r - s(zeta^r) = zeta^r -
> zeta^(2r) =
> (zeta^r)( 1 - zeta^r).
Not quite, since s:zeta -> zeta^2 need not be a generator of the Galois
group (for p=7 zeta->zeta^3 is a generator but the square is not) but
anyway it is easy to see that zeta - s(zeta) is still a multiple of pi,
so the rest of the answer stands.
> For all r, 1 - zeta^r is an associate of pi,
> and hence divible by pi.
Because 1-zeta^r and pi have the same norm p (conjugates), (1-zeta^r)/pi
has norm 1 so is a unit.
> It follows that z is divisible by pi.
Because z lies in the primary ideal generated by pi.
> In
> other words, any algebraic integer z with
> trace(z) = 0 must be divisible by pi. Since pi is invariant under the
> Galois group, all of the conjugates of z are also divisible by pi, and
> so all of the elementary symmetric function in the conjugates of z
> must be divisible by pi. But since these are also integers in the
> usual sense, they must actually be divisible by p, QED.
>
> The one gap in what I have written is not difficult, but I can barely
> keep my eyes open at this point so I will stop and continue tomorrow.
> Happy New Year.
>
Thanks a lot for this beautilful expos�.
I was mainly interested in this result because it can give
Kronecker-Weber candidates for embeddings of integer algebraic
extensions.
Example:
Given the polynomial P(a)=a^4-a^3+121*a^2-171*a+4111, find a Cyclotomic
field that contains the roots of this polynomial.
if we eliminate the third degree monomial by a change of variables we
get that the third degree monomial in a becomes (-4*t-1) and
P(a+1/4) becomes a^4+965/8*a^2-885/8*a+1043405/256
so that 256*P(a/4+1/4) is a^4+1930*a^2-7080*a+1043405
All the coefficients except the head coefficient are multiples of 5 so
Q_5 is a candidate. Indeed the roots are (w is the 5th root of unity):
4*w+w^3-6*w^4, w+4*w^2-6*w^3, -6*w^2+4*w^3+w^4,-6*w+w^2+4*w^4
Greets,
Marc
victor_me...@yahoo.co.uk
> Let Q_p be the field of cyclotomics of prime degree p, then all the
> tests I have done until now give the same result:
>
> If z is an algebraic integer of Q_p with Trace(z)=0 then all the
> coefficients of the minimal polynomial are multiples of p except for the
> first one (which is 1 of course).
The conclusion holds iff Trace(z) is a multiple of p.
If p divides Trace(z) then pi = 1 - zeta divides z in Z[zeta].
This is the case for all conjugates of zeta, so pi divides all
non-leading coefficients of the minimum polynomial for z.
As pZ = Z intersect pi Z[zeta] then p divides all these coefficients.
Conversely if Trace(z) is not a multiple of p, then pi does not
divide z, nor any of its conjugates. As pi is prime in Z[zeta]
then pi does not divide the constant coefficient in the minimum
polynomial of f and so this coefficient is not divisible by p.
All this is "well-known" but whether any textbook author has
thought it worth writing down is another matter.
Marc Bogaerts
> I thought last night that I could prove that a traceless algebraic
> integer in one of these cyclotomic fields must be x - s(x) for an
> algebraic integer x, and the result really does follow from that.
> Here is how far I have gotten. Suppose z = x - s(x), and z is an
> algebraic integer in this field. Then we can write x = r/n, where r
> is an algebraic integer in the field, and n is a rational integer.
> This fact is true in any number field and is easy. Then
> x - s(x) = r/n - s(r)/n = (r - s(r))/n, Here n must divide r - s(r)
> because we know up front that z = (s - s(r))/n is an algebraic
> integer. The numerator is clearly divisible by pi, so if (n, p) = 1,
> z must be divisible by pi and so your conclusion holds. If n is
> divisble by p, it is not clear to me what to do. So the only
> possible cases that doesn't work is an algebraic integer z such that
> z = x - s(x), and, writing x = r/n, we have n is divisble by p.
>
There is an explicit construction of x from z here :
http://www.math.uiuc.edu/~r-ash/Algebra/Chapter7.pdf
page 11, Problems for Section 7.3 numbers 4-7
(x and z are interchanged here from your notation).
If we take for u an element that satisfies two additional conditions,
namely that u is an algebraic integer and that Tr(u)=1 then x (z in the
text) is also an algebraic integer.
> That is my current progress report. This path either leads to a proof
> or to a counterexample
>
Have a nice day,
Marc
Achava Nakhash, the Loving Snake
>
> I was mainly interested in this result because it can give
> Kronecker-Weber candidates for embeddings of integer algebraic
> extensions.
>
> Example:
> Given the polynomial P(a)=a^4-a^3+121*a^2-171*a+4111, find a Cyclotomic
> field that contains the roots of this polynomial.
>
> if we eliminate the third degree monomial by a change of variables we
> get that the third degree monomial in a becomes (-4*t-1) and
> P(a+1/4) becomes a^4+965/8*a^2-885/8*a+1043405/256
> so that 256*P(a/4+1/4) is a^4+1930*a^2-7080*a+1043405
> All the coefficients except the head coefficient are multiples of 5 so
> Q_5 is a candidate. Indeed the roots are (w is the 5th root of unity):
> 4*w+w^3-6*w^4, w+4*w^2-6*w^3, -6*w^2+4*w^3+w^4,-6*w+w^2+4*w^4
>
> Greets,
> Marc
Hi Marc,
Thanks for the reply. Last night I woke up in the middle of the
night with an actual proof of your conjecture. Unfortunately, it was
gone in the morning, but I figured it out as I drove to work. Yes, I
am working on a Sunday.
The generaly algebraic integer in your field can be represented by
z = a_0 + a_1*zeta + ... + a_(p-2)*(zeta^(p-2)
Now trace(1) = p - 1, and trace(zeta^j) = -1 for any j such that j is
not divisible by p-1, Hence trace(z) = (p-1)*a_0 - a_1 - ... - a_
(p-2), so if trace(z) = 0, then
p*a_0 = a_0 + a_1 + ... + a_(p-2) = 0 =(mod pi),
where pi = 1 - zeta, of course, the generator of the unique prime
ideal lying over p.
But obviously zeta = 1 (mod pi), and of course 1 = 1 (mod pi), so for
our algebraic integer z,
z = a_0 + a_1 + ... + a_(p_2) = 0 (mod pi)
since trace(z) = 0.
Since z is divisible by pi, so are all of its conjugates, and thus so
are all of the elementary symmetric functions in z, and hence the
coefficients of the minimal polynomial of z. But if a rational
integer is divisible by pi, it is obviously also divisible by p.
Victor Meldrew adds the fact that your conclusion is true is trace(z)
is a multiple of p, and of course my proof works for this case as
well. He also states that the conclusion is false when trace(z) is
not divisible by p. My proof certainly shows that z is not divisible
by pi in this case, and that implies that the constant term of the
minimal polynomial is not divisible by p, so it works to show this as
well.
Thank you for posing such an interesting question, and thank you
Victor for giving us the full answer. On a personal note, I studied
algebraic number theory through the seventies, and I never saw this
result. Not only was it not in the textbooks, but none of the number
theorists with whom I spent time ever mentioned this. In our business
(I should say my former business) the term well-known means that it is
in the literature or in the folklore, but it does not mean what it
usually means in English, which is that a whole lot of people know it,
and you should know it if it is in your general area.
On the subject of language, and quite irrelevant to the thread, during
the seventies I studied class-field theory with a German graduate
student, so we looked at some books that were written in German. They
use the symbol Sp for trace. In English textbooks they generalize use
Tr. It turns out that the German word for trace is Spur. This
sounds kind of funny in English, because a spur is device you put on
the stirrups when riding a horse, and it gets the attention of the
horse better when you kick it than if you didn't have spurs. However,
the German pronunciation of Spur comes out in Enlish like shpoor, and
we do have a word spoor which is used when describing how animals can
track other elements by the smell they left behind. Of course what
the animal being hunted leaves behind is a trace of itself, so the two
seemingly different words trace and Spur are really closely related.
I don't know why I care about such things, but I do.
Regards,
Achava
Regards,
Marc Bogaerts
> On 31 Dec 2009, 14:59, Marc Bogaerts <mbg-dot-ni...@gmail.com> wrote:
>> Let Q_p be the field of cyclotomics of prime degree p, then all the
>> tests I have done until now give the same result:
>>
>> If z is an algebraic integer of Q_p with Trace(z)=0 then all the
>> coefficients of the minimal polynomial are multiples of p except for the
>> first one (which is 1 of course).
>
> The conclusion holds iff Trace(z) is a multiple of p.
> If p divides Trace(z) then pi = 1 - zeta divides z in Z[zeta].
> This is the case for all conjugates of zeta, so pi divides all
> non-leading coefficients of the minimum polynomial for z.
> As pZ = Z intersect pi Z[zeta] then p divides all these coefficients.
>
> Conversely if Trace(z) is not a multiple of p, then pi does not
> divide z, nor any of its conjugates. As pi is prime in Z[zeta]
> then pi does not divide the constant coefficient in the minimum
> polynomial of f and so this coefficient is not divisible by p.
>
maximal ideal generated by pi.
Thanks.
Marc Bogaerts
> Victor Meldrew adds the fact that your conclusion is true is trace(z)
> is a multiple of p, and of course my proof works for this case as
> well. He also states that the conclusion is false when trace(z) is
> not divisible by p. My proof certainly shows that z is not divisible
> by pi in this case, and that implies that the constant term of the
> minimal polynomial is not divisible by p, so it works to show this as
> well.
>
To summarize:
1) The algebraic integers with Tr(x)=0 mod p form a vector space of
dimension p-2.
2) zeta, zeta^2,..zeta^(p-1) form a basis for the ring of the algebraic
integers.
3) The sum of these elements is -1 (cf. the cyclotomic polynomial).
4) If a1, a2, .., a3 are the coefficients of x in this basis then
Tr(x)=Sum(ai). So the set of a.i. with trace 0 mod p form an ideal.
5) The trace of a rational integer n is n*(p-1).
So the only rational integers with trace 0 mod p are the multiples of p.
6) if t=Tr(x) then Tr(x+t) is t+t*(p-1) = t*p.
7) Trace(zeta)=-1 (see 2) so Trace(1-zeta^r)=p for all integer r.
8) 1-zeta, 1-zeta^2, .., 1-zeta^(p-1) span the ideal of the elements
with trace 0 mod p, so this ideal is a primary ideal generated by 1-zeta.
9) If x is an algebraic integer then y=x+Tr(x) lies in this ideal so the
quotient by this ideal is the Galois Field F_p, it is thus a maximal ideal.
10) If x lies in this ideal then its minimal polynomial is mapped into
zero by the quotient (x mod p) so its coefficients are 0 mod p.
> Thank you for posing such an interesting question, and thank you
> Victor for giving us the full answer. On a personal note, I studied
> algebraic number theory through the seventies, and I never saw this
> result. Not only was it not in the textbooks, but none of the number
> theorists with whom I spent time ever mentioned this. In our business
> (I should say my former business) the term well-known means that it is
> in the literature or in the folklore, but it does not mean what it
> usually means in English, which is that a whole lot of people know it,
> and you should know it if it is in your general area.
>
> On the subject of language, and quite irrelevant to the thread, during
> the seventies I studied class-field theory with a German graduate
> student, so we looked at some books that were written in German. They
> use the symbol Sp for trace. In English textbooks they generalize use
> Tr. It turns out that the German word for trace is Spur. This
> sounds kind of funny in English, because a spur is device you put on
> the stirrups when riding a horse, and it gets the attention of the
> horse better when you kick it than if you didn't have spurs. However,
> the German pronunciation of Spur comes out in Enlish like shpoor, and
> we do have a word spoor which is used when describing how animals can
> track other elements by the smell they left behind. Of course what
> the animal being hunted leaves behind is a trace of itself, so the two
> seemingly different words trace and Spur are really closely related.
>
> I don't know why I care about such things, but I do.
>
I find this extremely funny. In Dutch we also have the word "spoor"
which means trace but also spur.
> Regards,
> Achava
>
> Regards,
Achava Nakhash, the Loving Snake
> Achava Nakhash, the Loving Snake wrote:
>
> > Victor Meldrew adds the fact that your conclusion is true is trace(z)
> > is a multiple of p, and of course my proof works for this case as
> > well. He also states that the conclusion is false when trace(z) is
> > not divisible by p. My proof certainly shows that z is not divisible
> > by pi in this case, and that implies that the constant term of the
> > minimal polynomial is not divisible by p, so it works to show this as
> > well.
>
> To summarize:
>
> 1) The algebraic integers with Tr(x)=0 mod p form a vector space of
> dimension p-2.
> 2) zeta, zeta^2,..zeta^(p-1) form a basis for the ring of the algebraic
> integers.
> 3) The sum of these elements is -1 (cf. the cyclotomic polynomial).
> 4) If a1, a2, .., a3 are the coefficients of x in this basis then
> Tr(x)=Sum(ai). So the set of a.i. with trace 0 mod p form an ideal.
> 5) The trace of a rational integer n is n*(p-1).
> So the only rational integers with trace 0 mod p are the multiples of p.
> 6) if t=Tr(x) then Tr(x+t) is t+t*(p-1) = t*p.
> 7) Trace(zeta)=-1 (see 2) so Trace(1-zeta^r)=p for all integer r.
> 8) 1-zeta, 1-zeta^2, .., 1-zeta^(p-1) span the ideal of the elements
> with trace 0 mod p, so this ideal is a primary ideal generated by 1-zeta.
Just a little language nitpick. In English an ideal generated by a
single element is called a principal ideal, not a primary ideal.
There is something called a primary ideal, namely an ideal P such that
if xy is in P, then either x or some power of y is in P, so it is a
weakening of the concept of prime ideal. There is a theorem called
the Laskeer-Noether theorem that every ideal of a Noetherian ring has
a primary decomposition, that is, can be written as an intersection of
finitely many primary ideals. To me the most intesting thing about
this theorem is that Lasker is Emmanuel Lasker who became the world
champion of chess for many years.
> 9) If x is an algebraic integer then y=x+Tr(x) lies in this ideal so the
> quotient by this ideal is the Galois Field F_p, it is thus a maximal ideal.
> 10) If x lies in this ideal then its minimal polynomial is mapped into
> zero by the quotient (x mod p) so its coefficients are 0 mod p.
Regards,
Achava
Achava Nakhash, the Loving Snake
> Achava Nakhash, the Loving Snake wrote:
I was too sleepy to deal with the mathematics when I responded earlier
today. Now I am getting to that.
> To summarize:
>
> 1) The algebraic integers with Tr(x)=0 mod p form a vector space of
> dimension p-2.
A vector space over what field? Certainly not over Q, since you can
multiply any a.i. to 1 with an element of Q, and Tr(1) is not 0.
Perhaps you mean a module over the integers? That is certainly true.
> 2) zeta, zeta^2,..zeta^(p-1) form a basis for the ring of the algebraic
> integers.
I believe this is right. I used 1 instead of zeta^(p-1), but I can't
see this making any difference.
> 3) The sum of these elements is -1 (cf. thecyclotomic polynomial).
Yes.
> 4) If a1, a2, .., a3 are the coefficients of x in this basis then
> Tr(x)=Sum(ai). So the set of a.i. with trace 0 mod p form an ideal.
I believe Tr(x) = - Sum(a_i). It is true that this set is an ideal.
I don't see how this follows from what you wrote. It is easy to see
that the trace of any multiple of pi is divisible by p, since all of
the conjugates are divisible by pi. To show the converse you also
need to use the fact that the powers of zeta (not divisible by p) are
congruent to 1 mod pi, so x = Sum(a_i) (mod pi). Thus if Tr(x) = 0,
Sum(a_i) = - Sum(a_i) (mod pi),
> 5) The trace of a rational integer n is n*(p-1)
> So the only rational integers with trace 0 mod p are the multiples of p.
Yes.
> 6) if t=Tr(x) then Tr(x+t) is t+t*(p-1) = t*p.
Yes.
> 7) Trace(zeta)=-1 (see 2) so Trace(1-zeta^r)=p for all integer r.
Except when r is divisible by p, in which case Tr(zeta^r) = Tr(0) = 0.
> 8) 1-zeta, 1-zeta^2, .., 1-zeta^(p-1) span the ideal of the elements
> with trace 0 mod p, so this ideal is a primary ideal generated by 1-zeta.
As mentioned in the previous post, these are called principal ideas in
English. The fact is true.
> 9) If x is an algebraic integer then y=x+Tr(x) lies in this ideal so the
> quotient by this ideal is the Galois Field F_p, it is thus a maximal ideal.
Yes. Not the usual way it is derived. It is easy to see that the
ring of algebraic integers modulo pi = (1 - zeta) has exactly p
elements and that is sufficient.
> 10) If x lies in this ideal then its minimal polynomial is mapped into
> zero by the quotient (x mod p) so its coefficients are 0 mod p.
I don't follow this. Why is this true? It seems to me it has to
mapped to a power of x, since the lead coefficient must be 1 and that
is not divisible by p. However x and all its conjugates must be
divisible by pi. Hence all the elementary symmetric functions in the
conjugates of x are divisible by pi (this is easy to see by looking
at what they are, but it follows from more general considerations as
well) and as these are the coefficients of the minimal polynomial
except for the term of highest degree, they must all be divisible by
pi and hence by p.
Regards,
Achava
Marc Bogaerts
I realized afterwards but I'm a bit dyslexic so words starting with the
same consonants are prone to be confused by me :).
Remark, the ideal in question is a prime ideal also.
> There is something called a primary ideal, namely an ideal P such that
> if xy is in P, then either x or some power of y is in P, so it is a
> weakening of the concept of prime ideal. There is a theorem called
> the Laskeer-Noether theorem that every ideal of a Noetherian ring has
> a primary decomposition, that is, can be written as an intersection of
> finitely many primary ideals. To me the most intesting thing about
> this theorem is that Lasker is Emmanuel Lasker who became the world
> champion of chess for many years.
>
>> 9) If x is an algebraic integer then y=x+Tr(x) lies in this ideal so the
>> quotient by this ideal is the Galois Field F_p, it is thus a maximal ideal.
>> 10) If x lies in this ideal then its minimal polynomial is mapped into
>> zero by the quotient (x mod p) so its coefficients are 0 mod p.
>
Have a nice day,
Marc
Marc Bogaerts
> On Jan 5, 4:04 am, Marc Bogaerts <mbg-dot-ni...@gmail.com> wrote:
>> Achava Nakhash, the Loving Snake wrote:
>
> I was too sleepy to deal with the mathematics when I responded earlier
> today. Now I am getting to that.
>
>
>> To summarize:
>>
>> 1) The algebraic integers with Tr(x)=0 mod p form a vector space of
>> dimension p-2.
>
> A vector space over what field? Certainly not over Q, since you can
> multiply any a.i. to 1 with an element of Q, and Tr(1) is not 0.
> Perhaps you mean a module over the integers? That is certainly true.
>
>> 2) zeta, zeta^2,..zeta^(p-1) form a basis for the ring of the algebraic
>> integers.
>
That's the basis used by the GAP programming language (I think they call
it a Zumbroich base).
> I believe this is right. I used 1 instead of zeta^(p-1), but I can't
> see this making any difference.
>
>> 3) The sum of these elements is -1 (cf. thecyclotomic polynomial).
>
> Yes.
>
>> 4) If a1, a2, .., a3 are the coefficients of x in this basis then
>> Tr(x)=Sum(ai). So the set of a.i. with trace 0 mod p form an ideal.
>
> I believe Tr(x) = - Sum(a_i).
Sorry, I was wrong there again.
>
> It is true that this set is an ideal.
> I don't see how this follows from what you wrote.
Multiplying an such an element with zeta permutes the coefficients, so
does the multiplication with powers of zeta and integer combinations of
them.
> It is easy to see
> that the trace of any multiple of pi is divisible by p, since all of
> the conjugates are divisible by pi. To show the converse you also
> need to use the fact that the powers of zeta (not divisible by p) are
> congruent to 1 mod pi, so x = Sum(a_i) (mod pi). Thus if Tr(x) = 0,
> Sum(a_i) = - Sum(a_i) (mod pi),
>
>> 5) The trace of a rational integer n is n*(p-1)
>> So the only rational integers with trace 0 mod p are the multiples of p.
>
> Yes.
>
>> 6) if t=Tr(x) then Tr(x+t) is t+t*(p-1) = t*p.
>
> Yes.
>
>> 7) Trace(zeta)=-1 (see 2) so Trace(1-zeta^r)=p for all integer r.
>
> Except when r is divisible by p, in which case Tr(zeta^r) = Tr(0) = 0.
>
Sorry, wrong again.
>> 8) 1-zeta, 1-zeta^2, .., 1-zeta^(p-1) span the ideal of the elements
>> with trace 0 mod p, so this ideal is a primary ideal generated by 1-zeta.
>
> As mentioned in the previous post, these are called principal ideas in
> English. The fact is true.
>
Ooops dyslexia again !!!!
>> 9) If x is an algebraic integer then y=x+Tr(x) lies in this ideal so the
>> quotient by this ideal is the Galois Field F_p, it is thus a maximal ideal.
>
> Yes. Not the usual way it is derived. It is easy to see that the
> ring of algebraic integers modulo pi = (1 - zeta) has exactly p
> elements and that is sufficient.
>
>> 10) If x lies in this ideal then its minimal polynomial is mapped into
>> zero by the quotient (x mod p) so its coefficients are 0 mod p.
>
> I don't follow this. Why is this true? It seems to me it has to
> mapped to a power of x, since the lead coefficient must be 1 and that
> is not divisible by p. However x and all its conjugates must be
> divisible by pi. Hence all the elementary symmetric functions in the
> conjugates of x are divisible by pi (this is easy to see by looking
> at what they are, but it follows from more general considerations as
> well) and as these are the coefficients of the minimal polynomial
> except for the term of highest degree, they must all be divisible by
> pi and hence by p.
>
Forget what I said here, I was a bit quick, thinking that a mapping from
the polynomials over the ring of a. integers to polynomials over F_p by
taking the mod p of the coefficients had any meaning.
Your reasoning is completely correct.
I have the case in mind where p is not a prime but a power of a prime,
in which case the statement doesn't hold.
Indeed X^3-3*X+1 splits completely in C_9 having roots (I write z
instead of zeta) z^2++z^7, z^4+z^5 and -z^2-z^4-z^5-z^7.
But some cases exist where the statement applies, like
x=z^2+z^4+2*z^5-z^7 having minimal polynomial
X^6-9X^4+81X^2-81X+27
Have a fine day.
Achava Nakhash, the Loving Snake
> > It is true that this set is an ideal.
> > I don't see how this follows from what you wrote.
Marc said,
> Multiplying an such an element with zeta permutes the coefficients, so
> does the multiplication with powers of zeta and integer combinations of
> them.
This is not quite correct. For instance, using your z in place of
zeta,
(z^(p-1))*z = 1 = -z - z^2 - ... - z^(p-1) using your Zumbroich basis,
and Tr(1) = p-1. So the trace is not the same, but the traces differ
by a multiple of p, so they are the same mod p. Using your x with
coefficients a_1, ..., a_(p-1), then Tr(x*z) - Tr(x) is a multiple of
p. Using the additivity of trace, we see that the set of algebraic
integers in this ring whose trace is constant mod p forms an ideal.
To me this is rather remarkable, particularly since it comes from a
simple and utterly unenlightening calculation. I always suspect that
there is some sort of algebraic situation underlying the resulst, and
of course the challenge is to find it. At any rate, you still get a
result that allows you to successfully complete this step.
> x=z^2+z^4+2*z^5-z^7 having minimal polynomial
> X^6-9X^4+81X^2-81X+27
It looks like I snipped too much! Oh well. I haven't even thought
about the prime power case. I don't know much about cyclotomic fields
that are not from p'th roots of unity where p is an odd prime. Perhas
I shall let this thread be my excure for looking into them.
Regards,
Achava