[JSH] A summary of James argument and the last two weeks
Arturo Magidin
Some readers may have experienced something akin to mental whiplash
over James's postings and claims over the past couple of days. This
post is my attempt at clarifying what he has done on those days for
those readers who are interested, and to clean up his presentation. In
addition, I will make some comments about his general argument, and
some challenges to it.
Obviously, this message will be long. Other than this introduction, I
will restrict myself to mathematics.
INTRODUCTION
============
This is the sort of post that David Libert used to excel at. His gift
for summarizing and referencing is sorely missed, though I do not
blame him for not posting them any more. I shall do my very best to
achieve something at least similar to his excellent posts, though I
doubt I will succeed very much.
The post will consist of 6 parts. In the first part, I will outline
the segment of James's general argument for FLT which includes the
point which has been questioned for over a year now. I will call the
challenged point "the divisibility conclusion" in this post.
In the second part, I will outline James's argument starting a couple
of weeks ago which he asserted would establish the divisibility
conclusion, by disproving the complete factorization theorem for
algebraic integers. I will also make comments as to the validity of
this argument. "Complete factorization theorem for algebraic integers"
refers to the theorem that states that a one variable polynomial of
degree n with integer coefficients may be factored into n linear
terms, each with algebraic integer coefficients. A proof may be found
in
http://www.math.umt.edu/~preprints/gauss.ps (Postscript format)
http://www.math.umt.edu/~preprints/gauss.pdf (PDF format)
or at
http://www.math.uwaterloo.ca/~dmckinnon/Papers/Gauss.pdf
In Part 3, I will give James's argument of the past two days, which
he aserts disproves the theorem I used to challenge his divisibility
conclusion. This has been the subject of his (mathematical) posts of
the last couple of days. I will also discuss the validity of this
argument.
In Part 4, I will present what I can see of James's argument for the
divisibility conclusion. This is the most tentative portion of this
post, since James has not been, IMHO, very clear in explaining just
what it is he is doing when he explains this.
In part 5, I will point out what difficulties persist in James's
argument for FLT with regards to his use of "objects," and what
happens in the portion after the divisibility conclusion.
Part 6 will be an appendix. It will contain James's explicit
calculations that disprove his divisiblity conclusion, as well as the
proof of the theorem I invoked to disprove it in the general case.
In the past two weeks in particular, James has attempted to disprove
theorems that have some (real or imagined) relation with the
criticisms of his proofs. James seems to think that if he can prove
those theorems wrong, this will immediately establish the correctness
of the entirety of his FLT argument. This, however, is not the
case. Even if he were to disprove the theorem that has been used to
prove his divisibilty conclusion false for most cases, he would still
have to establish that it is correct for all cases he needs it for. He
has not done so: he has produced explicit counterexamples to his own
divisibility conclusion, but it is worth noting that the specific
examples found would never in fact occur in the context of his whole
proof.
It would be possible to salvage his divisibility conclusion if he
could do the any of the following:
(1) Either prove the theorem I have invoked is wrong, AND establish
his divisibility conclusion for all values of m,f,j that may show up
in his FLT argument (it can be shown that certain values will
never show up: for example, m=0 will never occur, and f=2 may be
avoided). His current argument is clearly insufficient, since it
pretends to apply to all values of v, and yet explicitly fails for
some.
OR
(2) Prove that theorem I have invoked will be inapplicable for all
values of m,f,j that may show up in his FLT argument. In this case, that
would imply showing that, whenever m,f,j are values that ->can<- be
obtained by the outline of his FLT argument will result in a
polynomial which is not primitive and irreducible over Q. This,
however, seems harder to accomplish. For example, the case m=1, f=5,
j=1 results in an irreducible polynomial:
13825W^3 - 72W - 2
(irreducibility checked by the rational root theorem) though there
is, a priori, no reason why these numbers would fail to occur in his
argument for p=3.
PART 1 - THE ARGUMENT LEADING UP TO THE DIVISIBILITY CONCLUSION
===============================================================
Let f,j be positive integers, m a nonnegative integer, and define
v = -1 + mf^{2j}.
Let
f(W) = (v^3+1)W^3 - 3vW - 2.
For each fixed value of m, f, j, f(W) is a polynomial with integer
coefficients. Suppose that for each value of v we can find a
factorization of this polynomial into linear terms with algebraic
integer coefficients:
f(W) = (a1W + b1)(a2W + b2)(a3W + b3)
so that
a1*a2*a3 = v^3+1
a1*a2*b3 + a1*b2*a3 + b1*a2*a3 = 0
a1*b2*b3 + b1*a2*b3 + b1*b2*a3 = -3v
b1*b2*b3 = -2.
(Note that we must assume that v^3+1 is not zero for this to be a
factorization into linear terms. If v=-1, then f(W)=3W-2, so the
factorization must be f(W)=3W-2).
The "divisibility conclusion" of James (what I understand of the argument
will be presented in Part 4) is that, in this case, exactly one
of a1, a2, a3 will be coprime to f in the ring of algebraic
integers, and that the other two will each be multiples of f^j in the
ring of all algebraic integers.
Here, "r and s are coprime in the ring of algebraic integers" means
that there does not exist any algebraic integer x such that:
(a) x is not a unit in the ring of algebraic integers; and
(b) x divides r in the ring of algebraic integers: there exists an
algebraic integer t such that xt=r; and
(c) x divides s in the ring of algebraic integers: there exists an
algebraic integer u such that xu=s.
I will also call this property "there is a nonunit algebraic integer
divisor of both r and s."
Over the last year, I and others have challenged the divisibility
conclusion; I invoked and presented a proof of a result which, using
Galois Theory, shows that whenever the resulting polynomial is
irreducible over Q and primitive the divisibility conclusion will be
false; others presented explicit calculations that established what
a1, a2, a3 were, and proved that none of them were divisible by f^j;
in that case, we set j=(1/2) rather than an integer, but this does not
seem to present any difficulties with regards to the argument James
uses (he certainly has not objected to setting j=1/2).
Over the last two weeks, James has attempted to establish the validity
of his divisibility conclusion/argument by trying to disprove two
theorems. More on the next parts.
PART 2. THE TWO ARGUMENTS FROM THE LAST TWO WEEKS
=================================================
Part 2a. The first stage. Apr 9 - Apr 22.
----------------------------------------
On Apr 9, James made the following post:
http://groups.google.com/groups?selm=3c65f87.0304090816.73a58257%40posting.google.com
In this post, he said:
"What you may also know as part of the story is my claim to
actually having suceeded to finding a short proof of Fermat's Last
Theorem, and that there are mathematicians who say otherwise.
"What you may not know is that these claims must rely on the claim
of ONE mathematician who said that he proved that given a polynomial
of degree n with integer coefficients, [there exists a] factorization
P(x) = (a1 x + b1)...(an x + bn)
[where] the a's and b's [may be chosen to] be algebraic integers."
I have edited James's actual statement, which he misstated the result
in question.
Formally, we have the following, which I will call the "complete
factorization theorem for A[x]", which A meaning the ring of all
algebraic integers:
THEOREM (Complete factorization for A[x]) Let P(x) be a polynomial
of degree n with algebraic integer coefficients. Then there exists
a factorization of P(x) into n (not necessarily monic) linear terms,
each with algebraic integer coefficients. In particular, this holds if
P(x) has integer coefficients.
A proof may be found in the links given in the introduction.
I noted the same day that there was no causal link between the
proofs that the divisibility conclusion was false and the complete
factorization theorem for A[x].
For the past two weeks, James continued on this general line of
argument. On Apr 14 2003, in the post
http://groups.google.com/groups?selm=3c65f87.0304141409.2b94be12%40posting.google.com
he presented his argument, in which he attempted to establish that the
complete factorization theorem for A[x] was false. There were some
corrections, but here is the basic line of argument:
Start with f(W) = (v^3+1)W^3 - 3vW + 1
All of these polynomials can be factored into linear terms with
algebraic integer coefficients. For the purpose of my exposition, it
will be easier to consider the dual polynomial
F(W) = W^3 - 3vW^3 + (v^3+1)
which is monic with integer coefficients. There is no difficulty in
doing this instead.
Factor F(W) = (W-r1)(W-r2)(W-r3),
where r1, r2, r3 are the three roots of the polynomial.
Use the same argument as before to deduce the divisibility conclusion:
for each value of v, exactly one of r1, r2, r3 will be coprime to f in
the ring of algebraic integers, and two will be divisible by f^j.
Show that this is the case for m=0, f<>3, j arbitrary, v=-1. In this
case, we have
F(W) = W^3 + 3W^2 = W^2(W+3),
so r1=r2=0, r3=3; Then r1, r2, are divisible by f^j, and r3 is
coprime to f.
Show that this is the case when v=1, with m=1, f=2, j=(1/2) (or
f=sqrt(2), j=1). In this case, we have
F(W) = W^3 -3W^2 + 2 =(W-1)(W^2-2W-2)
=(W-1)(W-(-1+sqrt(3)))(W-(1-sqrt(3))).
In this case, r1=1 is clearly coprime to 2, while 1+sqrt(3) and
1-sqrt(3) are both divisiors of 2:
(1+sqrt(3))(1-sqrt(3))(-1) = (1-3)(-1)=2.
Then consider the case v=4, that is m=1, f=5, j=1/2. Letting a1=-r1/sqrt(5),
a2=-r2/sqrt(5), and a3=-r3/sqrt(5), then each of a1, a2, a3 are roots
of
g(y) = 5y^6 - 144y^4 + 312 y^2 -169.
The claim then is that g(y) cannot be factored as the complete
factorization theorem promises, thereby disproving the complete
factorization theorem. He assures us that the conclusion that none of
a1, a2, a3 are algebraic integers depends on the complete
factorization theorem applied to this polynomial.
This last step is fallacious. Up to the calculation of g(y), the
complete factorization theorem has not been invoked. That none of a1,
a2, a3 are algebraic integers follows from the following two facts,
uncontested by James:
(1) The polynomial g(y) is primitive and irreducible over Q.
(2) THEOREM. Let f(x) be a primitive polynomial with integer
coefficients, which is irreducible over Q. Let r be a root of
f(x). Then r is an algebraic integer if and only if the leading
coefficient of f(x) is either 1 or -1.
At this point, we know for a fact that none of a1, a2, a3 are
algebraic integers. Since the divisibility conclusion of James asserts
that two of them must be algebraic integers (the ones that are
divisible by f^j=sqrt(5)), it follows that the divisibility conclusion
is incorrect, nd the complete factorization theorem is irrelevant.
Nonetheless, James said that the validity of his proof implied
the validity of his claim that g(y) could not be completely
factored in A[y]. He said: if g(y) can be completely factored, then my
argument is wrong.
On Tuesday, Apr 22, two posters gave an explicit factorization of
g(y). They were "Dot" and "Fred the Wonder Worm", who provided the
factorization in the following posts, respectively:
http://groups.google.com/groups?selm=210420032353503610%25dot%40at.dot.com
http://groups.google.com/groups?selm=b82u36%245aq%241%40spacebar.ucc.usyd.edu.au
James made a post agreeing to the factorization, apologizing for his
attacks on my honesty and mathematical abilities, and requesting that
people help him find where the error in his chain of reasoning was.
Part 2b. Apr 22-Today.
---------------------
Within a few hours, James had apparently realized that there was no
logical connection between the complete factorization of g(y) and the
divisibility properties of the roots of the polynomial. He announced,
instead, that the explicit factorizations offered by Fred could be
used to disprove another theorem I used to claim the divisibilty
conclusion would be false whenever the polynomial f(w) was irreducible
(note that the two cases where the divisibility conclusion is true
occur with reducible polynomials).
Some people seem to have been confused by this quick turnaround,
thinking that James was saying that Fred's post vindicated his claim
that the complete factorization theorem was incorrect. This was not
the case, as is clear from James's comments in
http://groups.google.com/groups?selm=3c65f87.0304221514.7b559002%40posting.google.com
where he writes:
"It IS true that I haven't disproven your paper [i.e. the complete
factorization theorem for A[x]], and in fact, it's relevance to the
discussion seems to be at an end."
So James has (apparently) abandoned his attacks on the complete
factorization theorem, which was in any case never relevant to the
disproofs of the divisibility conclusion (as he was told since he
began his argument on April 9th).
Instead, James claims that the factorizations provided by Fred given
an example of a polynomial P(X) such that:
(1) P(X) is monic, primitive, has integer coefficients, and is
irreducible over Q; and
(2) P(X) has constant term divisible by 5; and
(3) At least one of the roots of P(X) is coprime to 5 in the ring
of all algebraic integers.
James is then challenging the following theorem; I will quote it here
in the most general form I presented last year, then I will quote a
special case of this result, which would be directly contradicted by
P(X), if James's claims were true.
The theorem in question, which for ease of reference I will call the
coprimeness/divisibility theorem, is:
THEOREM (Coprimeness/divisibility)
Let f(x) be a polynomial with integer coefficients. Suppose that
f(x) is of degree n, is primitive, and is irreducible over
Q. Suppose further that we are given a factorization
f(x) = (a1 x + b1) ... (an x + bn)
where a1,...,an,b1,...,bn are algebraic integers. Then for every
integer f and every positive rational number q,
(1) f^q is coprime to at least one ai in the ring of all
algebraic integers if and only if it is coprime to all
ai in the ring of all algebraic integers. f^q is
coprime to at least one bj in the ring of all algebraic
integers if and only if it is coprime to all bj
in the ring of all algebraic integers.
(2) f^q divides at least one ai in the ring of all algebraic
integers if and only if it divides all ai in the ring of
all algebraic integers. f^q divides at least one bj in
the ring of all algebraic integers if and only if it
divides all bj in the ring of all algebraic integers.
I offer a proof of the Coprimeness/divisibility Theorem in Part 6
below.
I have invoked this theorem to state the James's divisibility
conclusion must be false whenever f(W) is irreducible; that is, if for
the specific value of v (or m,f,j) used, the resulting polynomial is
irreducible over Q, then the conclusion must necessarily be false,
since it contradicts the theorem. (Note that f(W) is always primitive:
if v is odd, then -3v and 2 are coprime; if v is even, then 2 and
v^3+1 are coprime).
James is considering the following particular case of the above
polynomial, obtained when f is monic. I will call it the "factor
splitting" theorem, since it says that the prime factors of the
constant coefficient are then split and distributed among the roots.
THEOREM (Factor splitting)
Let f(x) be a polynomial with integer coefficients. Suppose that
f(x) is monic, primitive, and irreducible over Q. Let p be a
rational prime which divides the constant term of f(x). Then no
root of f(x) is coprime to p in the ring of all algebraic
integers. Equivalently, for every root of f(x) there exists an
algebraic integer r which is not a unit (in the ring of all
algebraic integers) and r divides both the root and p.
If James's claims for P(X) were true, then P(X), and p=5 would
constitute a counterexample to the factor splitting theorem. Since the
factor splitting theorem is a special case of the
coprimeness/divisibility theorem, it would disprove my challenge to
the divisibility conclusion (though it would not suffice to establish
the latter).
PART 3. THE CLAIMS ABOUT P(X)
=============================
Fred the Wonder Worm offered a factorization of g(y) in the post
http://groups.google.com/groups?selm=b82u36%245aq%241%40spacebar.ucc.usyd.edu.au
It is from this post that James extracts the polynomial
P(X). Explicitly,
Q(X) = X^6 - 1488 X^4 + 4176 X^2 - 5.
Three of the roots of Q(X) are given to be r1, r2, r3, which are
expressed as follows:
Let a be a root of this polynomial:
P(X)= x^12 + 6*x^11 - 181*x^10 - 960*x^9 + 7963*x^8 + 37678*x^7 - 117533*x^6 -
488570*x^5 + 686101*x^4 + 2232396*x^3 - 1726811*x^2 - 2931010*x + 987100
It is P(X) which will be the polynomial exhibited.
Define tr, r1, r2, r3 to be the following values:
tr = 176866667462880
r1 = (9862453210*a^11 + 54243492655*a^10 - 1762144506621*a^9 -
8336476474707*a^8 + 73634871906622*a^7 + 297005313003728*a^6 -
915848551626666*a^5 - 3051857657480243*a^4 + 3027640866778493*a^3 +
7747460630371967*a^2 + 403832110132962*a - 1198409518887700) / tr
r2 = (6573801085*a^11 + 29503072348*a^10 - 1208403339336*a^9 -
4366240041747*a^8 + 54222010610962*a^7 + 148750611762872*a^6 -
793269747747321*a^5 - 1487188701233408*a^4 + 3985921497989948*a^3 +
4839390026896655*a^2 - 5550606915303438*a - 5671414313897740) / tr
r3 = (6573801085*a^11 + 42808739587*a^10 - 1141875003141*a^9 -
6752351088912*a^8 + 44278396405132*a^7 + 248393754086744*a^6 -
459258252043281*a^5 - 2663297151231683*a^4 + 1065284831907473*a^3 +
7093998365144384*a^2 - 1778334433861968*a - 130135870547080) / tr
James then argues:
1. Since the constant term of P(x) is a multiple of 5, at least one
of its roots is not coprime to 5; let a be any such root, and we use it to
define r1, r2, r3.
2. Argue that the definitions of r1, r2, r3, and the fact that r1,
r2, r3 are algebraic integers, implies that if a is a root of P(x)
which is not coprime to 5, then 5 must in fact divide a.
3. Since the constant term of P(X) is 987100 = 25*39484, there can be
no more than 2 roots which are not coprime to 5. Since P(X) is degree
12, that would contradict the factor splitting theorem.
Note that for the conclusion to (3) to hold, P(X) must be shown to be
irreducible. When I asked James to check, he replied that this was
irrelevant. I ->think<- that P(X) is meant to be irreducible over Q,
however, because presumably Fred got it by asking for a defining or
minimal polynomial. For the sake of argument, I agree to assume that
it is irreducible over the rationals, but leave open the possibility
that it is not.
There has been some problems with (1), as James said at first "some a
has a factor of 5", and then asserted that "has a factor" means "is a
multiple of", thus making it seem that he was assuming (2) rather than
proving it. This has led to some confusion on my part. I believe I can
clear up what James is actually doing here. I offer then my best
interpretation of what he does, given what he has written.
I believe the argument James attempted to write was as follows:
Since the constant term of Q(x) is a multiple of 5, at least one of
its roots is not coprime to 5. Let a be such a root. We will use a to
define r1, r2, r3.
Multiplying r1 by tr and collecting terms whose coefficient is a
multiple of 5 on the left hand side, and factoring out an a from the
right hand side, we have:
176866667462880r1-9862453210*a^11-54243492655*a^10+1198409518887700 =
=a(- 1762144506621*a^8 - 8336476474707*a^7 + 73634871906622*a^6
+ 297005313003728*a^5 - 915848551626666*a^4 - 3051857657480243*a^3
+ 3027640866778493*a^2 + 7747460630371967*a^1 + 403832110132962).
Now, the left hand side is clearly divisible by 5 in the ring of all
algebraic integers. Therefore, so is the right hand side. We cancel
out a 5 on the left hand side, and leave the division indicated on the
right hand side:
35373333492576r1-1972490642*a^11-10848698531*a^10+ 239681903777540=
=(a/5)(- 1762144506621*a^8 - 8336476474707*a^7 + 73634871906622*a^6
+ 297005313003728*a^5 - 915848551626666*a^4 - 3051857657480243*a^3
+ 3027640866778493*a^2 + 7747460630371967*a^1 + 403832110132962).
Now let x be any nonunit algebraic integer which is a common divisor
of r1, a, and 5. Then x divides the left hand side, since each term is
either an integer multiple of r1, of a, or of 5; therefore, it must
divide the right hand side. It cannot divide
(- 1762144506621*a^8 - 8336476474707*a^7 + 73634871906622*a^6
+ 297005313003728*a^5 - 915848551626666*a^4 - 3051857657480243*a^3
+ 3027640866778493*a^2 + 7747460630371967*a^1 + 403832110132962)
because to do that it would have to divide 403832110132962, which is
coprime to 5 hence to x.
James then proceeds the same way with r2: collecting he gets:
176866667462880r2-6573801085*a^11 - 4839390026896655*a^2 + 5671414313897740
= a(29503072348*a^9 - 1208403339336*a^8 - 4366240041747*a^7
+ 54222010610962*a^6 + 148750611762872*a^5 - 793269747747321*a^4
- 1487188701233408*a^3 + 3985921497989948*a^2 - 5550606915303438)
And then points out that both
a(- 1762144506621*a^8 - 8336476474707*a^7 + 73634871906622*a^6
+ 297005313003728*a^5 - 915848551626666*a^4 - 3051857657480243*a^3
+ 3027640866778493*a^2 + 7747460630371967*a^1 + 403832110132962).
(the right hand side from the expression for r1)
and then concludes that the difference between the last two terms,
5550606915303438a and 403832110132962a is equal to
5146774805170476a.
He then says this "must have a factor of 5", and so that a must be
divisible by 5. This would imply that at leats 10 roots of P(x) are
coprime to 5, and atl east one of them is not.
I do not see how he comes to the conclusion that the difference of the
last two terms of each expression must be a multiple of 5 from the
fact that they both are multiples of 5. Basically, he is saying that
if a*h_1(a) and a*h_2(a) are both multiples of 5, where each nonzero
coefficient of h_1 and of h_2 is coprime to 5, they both have nonzero
constant terms, and a is not coprime to 5, then a(h_1(0)-h_2(0)) must
be a multiple of 5, and this simply does not follow.
So I do not see how James arrives at the conclusion that
5146774805170476a must be a multiple of 5.
A slightly more promising route could be attempted by looking at
35373333492576r1-1972490642*a^11-10848698531*a^10+ 239681903777540=
=(a/5)(- 1762144506621*a^8 - 8336476474707*a^7 + 73634871906622*a^6
+ 297005313003728*a^5 - 915848551626666*a^4 - 3051857657480243*a^3
+ 3027640866778493*a^2 + 7747460630371967*a^1 + 403832110132962).
If x were a nontrivial common divisor of r1, a and 5, then it would
divide the left hand side. Then the right hand side would imply that
we have an algebraic integer of the form:
a*h(a), where h(x) is a polynomial with integer coefficients, were all
nonzero coefficients are coprime to 5 and h(0) is not 0; then a*h(a)
would necessarily have to be divisible by 5x, and we would then try to
conclude that 5 must divide a in that case. I tried constructing
some counterexamples with integers (replace 5 with a composite number
for such examples; we may do that, since in the ring of algebraic
integers, where a lives, 5 is not an irreducible anyway), but could
not after a few minutes.
But even if this claim is true, there remains a major problem: we know
that a is not coprime to 5; we also know that at least one root r1,
r2, r3 is not coprime to 5 (since the constant coefficient of Q(x) is
also a multiple of 5). But it does not follow that there must be a
nontrivial common divisor of that r, a, and 5. In integers, even
though 6, 10, and 15 are such that any two have a nontrivial divisor,
yet there is no nontrivial divisor to all three.
So frankly, I am confused about just how James's arguments are
supposed to contradict the factor splitting theorem.
PART 4. THE ARGUMENT LEADING TO THE DIVISIBILITY CONCLUSION
===========================================================
As noted, this is the part that I find the most difficult to analyze.
I will use the simpler polynomial James used over the last week,
taking the dual as I did above.
We start with
f(W) = W^3 - 3vW^2 + (v^3+1)
= W^3 - 3W^2(-1+mf^{2j}) + ((-1+mf^{2j})^3+1)
= (W+a1)(W+a2)(W+a3).
Then, expanding the second line and writing as if it were a polynomial
in m, we have:
(W+a1)(W+a2)(W+a3) = f^{6j}m^3 - 3f^{4j}m^2 + 3f^{2j}(1-W^2)m + W^3.
Now James seems to argue thusly: the expression on the right is a
polynomial in m. Some terms on the left will have factors of m, and
the rest will have to divide the "constant term" W^3. The same is true
for the term W on each summand. Some of the ai must have factors of m,
since their product is equal to m(3f^{2j} - 3f^{4j}m + f^{6j}m^2); 1
must be of the form w+c where w divides m and c dividies the constant
term. From this we deduce that if any ai has a factor of m, then it
must be a multiple of f^j; and since at most two of a1, a2, a3 may be
a multiple of f^j (since f^{2j} divides v^3+1, but f^{3j} does not),
then it follows that exactly two of them are mulitples of f^j, and one
of them is coprime to f.
I cannot understand the final part of his argument in the
page. Writing 1 as a sum w+c is of course possible, but it is not
unique and seems to be a complete mess. I do not see how one "sees"
that from this it follows that any ai that has a factor of m must be a
multiple of f^j.
Previouse expositions by James began by considering the case m=0, so
v=-1, and the polynomial becomes
W^3 +3W^2 = W^2(W+3). The argument then seemed to be that since
a1=a2=0, those are the a's that "have factors of m", and since a3=3,
a3 is a "factor of the constant term", and that this will hold for all
values of m; the conclusion is then that any factor of (W+a3) will
necessarily divide THE CONSTANT TERM OF f(W) AS A POLYNOMIAL IN W,
namely v^3+1. There is far too much confusion for me to figure out
what is going on here.
It seems hardly surprising that this argument leads to contradictions
and problems. I note that the exposition on the webpage is not the
same as what was given by James last week, showing that this argument
is still in a fluid state.
PART 5. OTHER UNRELATED PROBLEMS WITH JAMES'S PROOF
==================================================
James seems to think that if he can only prove the theorem wrong, he
will have established the truth of his proof. He knows that if the
theorem is right, then he is wrong. He seems to think the converse
also holds: that if the theorem is wrong, then he is right. This is
simply not the case.
In addition to the problem of the divisibility conclusion, James
currently uses "objects" in his argument The thid line on his page,
http://groups.msn.com/AmateurMath/fltproof.msnw
says "All that follows is in an object ring."
The hyperlink at "object" leads to a page that contains the following:
"OBJECTS are members of commutative rings where any unit and its
multiplicative inverse are units in all possible commutative rings in
which either and all integers are members, where no member is a factor
of an object for which it is not a factor in all possible commutative
rings that include all integers in which it and that object are
members."
I note that this is a change from last week, when the words "and all
integers" were not in the text.
The first problem is that the definition is incoherent as written. The
second is that the definition of object uses the term object in its
conclusions. Therefore, this is not a definition; absent a definition,
his entire "argument" is nonsense as written.
James has said in some posts that "objects" are a class which includes
all algebraic integers, but is stritly larger than it; and yet
includes no "fractions".
It is impossible for a subring of the algebraic numbers to properly
include all algebraic integers and have no fractions. If a/b is an
element which is in this ring but is not an algebraic integer, with a
and b algebraic integers, coprime, then let x and y be algebraic
integers such that xa+yb = 1; then
x(a/b) + y = (xa+yb)/b = 1/b is an element of the ring, and so it
contains fractions.
In addition, if we take objects as a given definition, and the
statement above to be a description of some of their properties, then
these "objects" cannot include any algebraic integer which is not an
integer. For example, sqrt(2) cannot be an object: if it were, then so
would 2sqrt(2).
But 2 divides 2sqrt(2) in Z[sqrt(2)]; yet in Z[sqrt(8)], which
contains both 2 and 2*sqrt(2), 2 does not divide 2*sqrt(2). The exact
same argument works for any algebraic integer r which is not an
integer: 2 divides 2r in the ring Z[r], but does not divide 2r in the
ring Z[2r].
So there are large inconsistencies between the claimed properties of
objects. This already demolishes James's proof.
Finally, I do not believe anybody has ventured beyond the divisibility
conclusion; that is, I do not think there is anybody who has checkd
James's exposition beyond that point, assuming it were true. Last time
I checked, it descended into incoherence and no contradiction was
apparent. So even if he were to fix the problem with "objects" and
were to establish the divisibility conclusion, it is far from a given
that the rest of the argument does indeed hold up.
PART 6. THE EXPLICIT DISPROOF OF THE DIVISIBILITY CONCLUSION,
AND A PROOF OF THE COPRIMENESS/DIVISIBILITY THEOREM.
=============================================================
Part 6a. Explicit disproof
--------------------------
James considers the following polynomial
F(W) = (v^3+1)W^3 - 3vW + 1.
He agrees this polynomial can always be factored into linear terms
with algebraic integer coefficients, and that his argument leading to
the divisibility condition would hold for this polynomial, regardless
of the truth of either the complete factorization or the
coprimeness/divisibility theorems, and that the divisibility
conclusion will hold in the ring of all algebraic integers.
When v=4, that is, m=1, f=5, j=(1/2), we have
F(W) = 65W^3 -12W + 1 = (a1W+1)(a2W+1)(a3W+1),
where a1, a2, a3 are minus the roots of the dual polynomial
X^3-12X^2+65.
The divisibility conclusion is that exactly two of a1, a2, a3 are
divisible by sqrt(5) in the ring of all algebraic integers. That is,
exactly two of a1/sqrt(5), a2/sqrt(5), and a3/sqrt(5) are algebraic
integers. It is equivalent to take r1/sqrt(5), r2/sqrt(5), r3/sqrt(5),
the roots of X^3-12X^2+65. We do so.
Let r be any of r1, r2, r3. Then
r^3 - 12r^2 + 65 = 0; dividing by 5sqrt(5), we have
(r/sqrt(5))^3 - (12/sqrt(5))(r/sqrt(5))^2 + (13/sqrt(5)) = 0.
That is, r/sqrt(5) is a root of
x^3 - (12/sqrt(5))x^2 + (13/sqrt(5)) = 0.
Mutliplying by 5 we have that they are roots of
5x^3 - 12sqrt(5)x^2 + 13sqrt(5) = 0;
Therefore, they all satisfy
5x^3 = sqrt(5)(12x^2 -13).
Squaring both sides, they all satisfy
25x^6 = 5(144x^4 - 312 x^2 + 169)
Cancelling 5 and collecting, they are all roots of:
5x^6 - 144x^4 + 312 x^2 - 169 = 0.
James agrees to this, and to the fact that this polynomial is
irreducible over Q. Therefore, we may conclude that no root of this
polynomial is an algebraic integer. Since each of r1/sqrt(5),
r2/sqrt(5), r3/sqrt(5) are roots, none of them are algebraic
integers. Therefore, neither are a1/sqrt(5), a2/sqrt(5),
a3/sqrt(5). This proves that the divisiblity conclusion is incorrect.
Part 6b. The coprimeness/divisibility theorem.
---------------------------------------------
I will require the following result from Galois Theory:
THEOREM. Let f(x) be an irreducible polynomial with integer
coefficients, and let K be any Galois extension of Q which contains
all the roots of f(x). If r and s are two (not necessarily distinct)
roots of f(x) in K, then there exists a Galois automorphism g:K->K
which fixes Q pointwise, and such that g(r)=g(s).
This theorem is false if f(x) is not irreducible, and it is the reason
why the Coprimeness/divisibility theorem only holds when f(x) is
irreducible.
THEOREM. Let K be any Galois extension of Q, and let g be any Galois
automorphism g:K->K which fixes Q. Then there is an automorphism of
the algebraic numbers which, restricted to K, equals g.
Now the proof.
Lemma 1. Let a be an algebraic integer, and let g be any Galois
automorphism of the field of all algebraic numbers over Q; then g(a)
is an algebraic integer.
Proof. If f(x) is monic with integer coefficients and f(a)=0, then
0 = g(0) = g(f(a)) = f(g(a)), so g(a) is also an algebraic
integer. QED
Lemma 2. Let a and b be algebraic integers, g a Galois automorphism of
the field of algebraic numbers. Then a divides b in the ring
of algebraic integers if and only if g(a) divides g(b). a is a unit in
the ring of algebraic integers if and only if g(a) is a unit.
Proof. a divides b if and only if there exists an algebraic integer x
such that ax=b; if and only if g(ax)=g(b); if and only if
g(a)g(x)=g(b); if and only if g(a) divides g(b) (since g(x) is an
algebraic integer if and only if x is).
a is a unit if and only if it divides 1, if and only if g(a) divides
g(1)=1, if and only if g(a) is a unit. QED
Lemma 3. Let f(x) be a polynomial with integer coefficients, r a root,
g a Galois automorphism of the field of all algebraic numbers. Then
g(r) is a root of f(x).
Proof. Same as Lemma 1. QED
Lemma 4. Let a and b be algebraic integers, g a Galois automorphism of
the field of algebraic numbers. Then a and b are coprime
in the ring of algebraic integers if and only if g(a) and g(b) are
coprime in that ring.
Proof. x divides both a and b if and only if g(x) divides both g(a)
and g(b); and x is a unit if and only if g(x) is a unit. QED.
Lemma 5. Let f(x) be a polynomial of degree n with integer
coefficients, which is primitive. Suppose we can factor f(x) as a
product of linear terms with algebraic integer coefficients
f(x)= (a1x+b1)...(anx+bn).
Then for each pair i,j=1,...,n there exists a Galois automorphism of the
field of algebraic numbers, and an algebraic integer unit u, such that
g(ai)=uaj; g(bi)=ubj.
Proof. Since f(x) is primitive, we must have that the gcd of ai and bi
is 1 for each i=1,...,n.
Pick i and j. Let g be the Galois automorphism which sends the root
-bi/ai to the root -bj/aj. Then
g(bi/ai) = bj/aj; so g(bi)*aj = bj*g(ai). Therefore, bj divides
g(bi)aj, but is coprime to aj; so bj divides g(bi). Symmetrically, aj
divides g(ai).
Doing the same with g^{-1}, which maps bj/aj to ai/bi, we conclude
that ai divides g^{-1}(aj); and bi divides g^{-1}(bj). Applying g to
both of these we get that g(ai) divides aj, g(bi) divides bj. Since aj
divides g(ai), we know there is a unit u such that g(ai)=u*aj;
likewise, there is a unit v such that g(bi)=v*bj. It only remains to
show that u=v. But
aj/bj = g(ai/bi) = u*aj/(v*bj), so u/v = 1; therefore u=v, as
desired. QED.
Lemma 6. Let f be an integer, q a positive rational number. Let x be
an algebraic integer, g a Galois automorphism of the field of
algebraic numbers. Then f^q divides x if and only it divides g(x); f^q
is coprime to x if and only if it is coprime to g(x).
Proof. f^q is a root of x^q-f = 0. Therefore, g(f^q) is also a root of
this polynomial; but the roots of this polynomial are exactly f^q,
zf^q,...,z^{q-1}f^q, where z is a primitive q-th root of unity.
f^q divides x if and only if g(f^q) divides g(x) if and only if
z^i(f^q) divides g(x); but z^i is a unit, so z^i(f^q) divides g(x) if
and only if f^q divides g(x).
The same is true if we replace "divides" with "is coprime to." QED
Proof of the Coprimeness/Divisibility Theorem:
Clearly, if the clause about all the ai and all the bj holds, then the
clause about at least one holds.
Conversely, suppose that f^q divides some ai; let j be arbitrary,
j=1,...,n. Let g be a Galois automorphism of the field of all
algebraic numbers which sends bi/ai to bj/aj. Then g(ai)=uaj, u a
unit. f^q divides ai if and only if f^q divides g(ai)=uaj if and only
if it divides aj. So f^q divides aj. Same argument holds if f^q
divides bj; or if we replace "divides" with "is coprime to." QED
======================================================================
"Why do you take so much trouble to expose such a reasoner as
Mr. Smith? I answer as a deceased friend of mine used to answer
on like occasions - A man's capacity is no measure of his power
to do mischief. Mr. Smith has untiring energy, which does
something; self-evident honesty of conviction, which does more;
and a long purse, which does most of all. He has made at least
ten publications, full of figures few readers can critize. A great
many people are staggered to this extend, that they imagine there
must be the indefinite "something" in the mysterious "all this".
They are brought to the point of suspicion that the mathematicians
ought not to treat "all this" with such undisguised contempt,
at least."
-- "A Budget of Paradoxes", Vol. 2 p. 129 by Augustus de Morgan
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Jason Pawloski
David C. Ullrich
Pawloski) wrote:
>>http://www.math.umt.edu/~preprints/gauss.ps
>
>This URL does not work.
Sure it works! Although I confess I don't quite see how the
text of the document, "404 Not Found The requested
URL /~preprints/gauss.ps was not found on this server."
proves what Arturo claims it proves...
You might find
http://www.math.umt.edu/~magidin/preprints/gauss.ps
more interesting.
>Jason
******************
David C. Ullrich
Randy Poe
Try these:
http://www.matem.unam.mx/~magidin/preprints/gauss.pdf
http://www.matem.unam.mx/~magidin/preprints/gauss.ps
- Randy
Arturo Magidin
Jason Pawloski <jpaw...@nemesissoftware.com> wrote:
>>http://www.math.umt.edu/~preprints/gauss.ps
>
>This URL does not work.
Oops.
http://www.math.umt.edu/~magidin/preprints/gauss.ps
http://www.math.umt.edu/~magidin/preprints/gauss.pdf
I tried to be very careful with the post URLs, since I have to copy
them by hands; and which one do I mess up?
======================================================================
"It's not denial. I'm just very selective about
what I accept as reality."
--- Calvin ("Calvin and Hobbes")
======================================================================
Arturo Magidin
mag...@math.berkeley.edu
Virgil
mag...@math.berkeley.edu (Arturo Magidin) wrote:
> Oops.
>
> http://www.math.umt.edu/~magidin/preprints/gauss.ps
> http://www.math.umt.edu/~magidin/preprints/gauss.pdf
>
> I tried to be very careful with the post URLs, since I have to copy
> them by hands; and which one do I mess up?
>
> ======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
> ======================================================================
>
> Arturo Magidin
> mag...@math.berkeley.edu
I always use a copy and paste operation, especially with
URLs, since my eyes aren't as good as they used to be.
Doesn't your operating system allow this?
Arturo Magidin
Virgil <vmh...@attbi.com> wrote:
>I always use a copy and paste operation, especially with
>URLs, since my eyes aren't as good as they used to be.
>
>Doesn't your operating system allow this?
I do not know how to do a cut and paste on either SuSE (the platform I
am using now), or on Red Hat, the platform I was using before. I have
never been able to do it, either within the same desktop or from one
desktop to another.
Of course, that would not have helped with my error in giving the URL
for the manuscript; that was just me being stupid.
And my eyes have never been what they used to be... (very near
sighted, with quite a bit of astigmatism thrown in)
W. Dale Hall
Arturo Magidin wrote:
> In article <vmhjr2-22D52B....@netnews.attbi.com>,
> Virgil <vmh...@attbi.com> wrote:
>
>
>>I always use a copy and paste operation, especially with
>>URLs, since my eyes aren't as good as they used to be.
>>
>>Doesn't your operating system allow this?
>
>
> I do not know how to do a cut and paste on either SuSE (the platform I
> am using now), or on Red Hat, the platform I was using before. I have
> never been able to do it, either within the same desktop or from one
> desktop to another.
>
Actually, pasting (no cutting required) from any window to any window in
many Linux systems (indeed, many Unix systems of all stripes) is close
to trivial:
If you are using an X-windows interface (X11, Gnome, KDE), you can
highlight by holding left button down as you drag the cursor from one
end of the text to the other (if it's a URL embedded in HTML that's
being viewed with a browser, you have to be careful not to select the
URL & jump across the hyperlink); then paste by middle-clicking in your
window of choice.
It's easy to get the hang of it, by opening an xterm window and typing
stuff and playing with the highlight & middle-click operation. As long
as you don't do something like "rm -fR *" as root (KIDS, DON'T TRY THIS
AT HOME), you're probably safe. Otherwise, it also works within an
emacs session. I don't do vi, so I wouldn't know about that one.
> Of course, that would not have helped with my error in giving the URL
> for the manuscript; that was just me being stupid.
>
> And my eyes have never been what they used to be... (very near
> sighted, with quite a bit of astigmatism thrown in)
>
> ======================================================================
> "It's not denial. I'm just very selective about
> what I accept as reality."
> --- Calvin ("Calvin and Hobbes")
> ======================================================================
>
> Arturo Magidin
> mag...@math.berkeley.edu
>
Dale.
Arturo Magidin
>> I do not know how to do a cut and paste on either SuSE (the platform I
>> am using now), or on Red Hat, the platform I was using before. I have
>> never been able to do it, either within the same desktop or from one
>> desktop to another.
>>
>
>Actually, pasting (no cutting required) from any window to any window in
>many Linux systems (indeed, many Unix systems of all stripes) is close
>to trivial:
>
>If you are using an X-windows interface (X11, Gnome, KDE), you can
>highlight by holding left button down as you drag the cursor from one
>end of the text to the other (if it's a URL embedded in HTML that's
>being viewed with a browser, you have to be careful not to select the
>URL & jump across the hyperlink); then paste by middle-clicking in your
>window of choice.
Ah; I cannot do that from one desktop to another (the highlight
disappears when I left/right click to change desktops), but apparently
I can do it from within the same desktop. My old mouse only had two
buttons, so maybe that's why I could never do it there.
Thanks!
Wayne Brown
> It's easy to get the hang of it, by opening an xterm window and typing
> stuff and playing with the highlight & middle-click operation. As long
> as you don't do something like "rm -fR *" as root (KIDS, DON'T TRY THIS
> AT HOME), you're probably safe. Otherwise, it also works within an
> emacs session. I don't do vi, so I wouldn't know about that one.
It works just fine in vi as well.
--
Wayne Brown | "When your tail's in a crack, you improvise
fwb...@bellsouth.net | if you're good enough. Otherwise you give
| your pelt to the trapper."
"e^(i*pi) = -1" -- Euler | -- John Myers Myers, "Silverlock"
Wayne Brown
> Ah; I cannot do that from one desktop to another (the highlight
> disappears when I left/right click to change desktops), but apparently
> I can do it from within the same desktop. My old mouse only had two
> buttons, so maybe that's why I could never do it there.
With two-button mice, the middle-click operation is usually simulated
by clicking the right and left mouse buttons simultaneously.
Dik T. Winter
> W. Dale Hall <dh...@farir.com> wrote:
> > I don't do vi, so I wouldn't know about that one.
>
> It works just fine in vi as well.
Yup, because it is a feature of X. Pushing the middle button will enter
high-lighted text in the pointed window at the cursor position. So in
vi when you are in insert mode it will enter the text; when you are not
it will do something completely different. But it is easy under X to
have high-lighted text that gets normal when you perform some
innocious action. I think that is the problem Arturo has with his
"desktop switch". I have seen it occur to myself on occasion. But
I am not privy to the innards of X to explain further. I think I
prefer the copy/paste strategy with explicit keys (although that does
not work with vi on my home machine either...).
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
Andrzej Kolowski
> Some readers may have experienced something akin to mental whiplash
> over James's postings and claims over the past couple of days. This
> post is my attempt at clarifying what he has done on those days for
> those readers who are interested, and to clean up his presentation. In
> addition, I will make some comments about his general argument, and
> some challenges to it.
>
[snip]
Arturo,
I snipped out your excellent summary only to save space - I have
printed it and will study it in more detail.
You have gone to great lengths to be fair and non-judgemental,
though I am sure JSH would not agree. However I think it is a
mistake this time to devote further effort to this. In the past
month JSH has been largely confused regarding the previous arguments,
and confused again regarding the effect of FTWW's factorization.
The confusion, partly the result of a poor memory and partly of
course due to an intense blinding desire to prevail, is pathetic. In
his desperate attempts to prove you wrong he has completely ignored
the fact that his definition of "objects" is invalid and that his
current argument based on this definition is therefore also useless.
It seems as though he cannot completely resign himself to the
counterexamples to his previous attempt using algebraic integers,
although now he has some time ago conceded the validity of every
link in the chain regarding the counterexamples. It is very
possible that he has forgotten this.
I would argue that you are giving too much attention to
the incompetent low-level mathematics of a confused and troubled
man. JSH perhaps does not realize that his "work" is quite possibly among
the best-known current mathematical writing in the world, probably more
widely read than the typical paper in Annals of Mathematics. He complains
that mathematicians ignore and suppress what he has done, as if we
were all engaged in one massive conspiracy or sheeplike herd, when in
fact he has been given more chances to vent his theories than any published
mathematical author has ever had. If there were genuine merit there
someone would notice it. He does not deserve the publicity.
And your considerable mathematical powers would be better spent
on your own original work. You must know that your own career is not
being advanced by this activity.
That said, I appreciate what you have done: not because it
successfully refutes JSH, but because I have learned a fair
amount of algebraic number theory in a painless way by observing
your handling of innumerable problems, proofs, and counterexamples
along the way.
Andrzej
>
> ======================================================================
> "Why do you take so much trouble to expose such a reasoner as
> Mr. Smith? I answer as a deceased friend of mine used to answer
> on like occasions - A man's capacity is no measure of his power
> to do mischief. Mr. Smith has untiring energy, which does
> something; self-evident honesty of conviction, which does more;
> and a long purse, which does most of all. He has made at least
> ten publications, full of figures few readers can critize. A great
> many people are staggered to this extend, that they imagine there
^t