Tag: geometry

The odd knights of the round table

Published January 28, 2010 by lievenlb

Here’s a tiny problem illustrating our limited knowledge of finite fields : “Imagine an infinite queue of Knights ${ K_1,K_2,K_3,\ldots } $, waiting to be seated at the unit-circular table. The master of ceremony (that is, you) must give Knights $K_a $ and $K_b $ a place at an odd root of unity, say $\omega_a $ and $\omega_b $, such that the seat at the odd root of unity $\omega_a \times \omega_b $ must be given to the Knight $K_{a \otimes b} $, where $a \otimes b $ is the Nim-multiplication of $a $ and $b $. Which place would you offer to Knight $K_{16} $, or Knight $K_n $, or, if you’re into ordinals, Knight $K_{\omega} $?”

What does this have to do with finite fields? Well, consider the simplest of all finite field $\mathbb{F}_2 = { 0,1 } $ and consider its algebraic closure $\overline{\mathbb{F}_2} $. Last year, we’ve run a series starting here, identifying the field $\overline{\mathbb{F}_2} $, following John H. Conway in ONAG, with the set of all ordinals smaller than $\omega^{\omega^{\omega}} $, given the Nim addition and multiplication. I know that ordinal numbers may be intimidating at first, so let’s just restrict to ordinary natural numbers for now. The Nim-addition of two numbers $n \oplus m $ can be calculated by writing the numbers n and m in binary form and add them without carrying. For example, $9 \oplus 1 = 1001+1 = 1000 = 8 $. Nim-multiplication is slightly more complicated and is best expressed using the so-called Fermat-powers $F_n = 2^{2^n} $. We then demand that $F_n \otimes m = F_n \times m $ whenever $m < F_n $ and $F_n \otimes F_n = \frac{3}{2}F_n $. Distributivity wrt. $\oplus $ can then be used to calculate arbitrary Nim-products. For example, $8 \otimes 3 = (4 \otimes 2) \otimes (2 \oplus 1) = (4 \otimes 3) \oplus (4 \otimes 2) = 12 \oplus 8 = 4 $. Conway’s remarkable result asserts that the ordinal numbers, equipped with Nim addition and multiplication, form an algebraically closed field of characteristic two. The closure $\overline{\mathbb{F}_2} $ is identified with the subfield of all ordinals smaller than $\omega^{\omega^{\omega}} $. For those of you who don’t feel like going transfinite, the subfield $~(\mathbb{N},\oplus,\otimes) $ is identified with the quadratic closure of $\mathbb{F}_2 $.

The connection between $\overline{\mathbb{F}_2} $ and the odd roots of unity has been advocated by Alain Connes in his talk before a general public at the IHES : “L’ange de la géométrie, le diable de l’algèbre et le corps à un élément” (the angel of geometry, the devil of algebra and the field with one element). He describes its content briefly in this YouTube-video

At first it was unclear to me which ‘coupling-problem’ Alain meant, but this has been clarified in his paper together with Caterina Consani Characteristic one, entropy and the absolute point. The non-zero elements of $\overline{\mathbb{F}_2} $ can be identified with the set of all odd roots of unity. For, if x is such a unit, it belongs to a finite subfield of the form $\mathbb{F}_{2^n} $ for some n, and, as the group of units of any finite field is cyclic, x is an element of order $2^n-1 $. Hence, $\mathbb{F}_{2^n}- { 0 } $ can be identified with the set of $2^n-1 $-roots of unity, with $e^{2 \pi i/n} $ corresponding to a generator of the unit-group. So, all elements of $\overline{\mathbb{F}_2} $ correspond to an odd root of unity. The observation that we get indeed all odd roots of unity may take you a couple of seconds (( If m is odd, then (2,m)=1 and so 2 is a unit in the finite cyclic group $~(\mathbb{Z}/m\mathbb{Z})^* $ whence $2^n = 1 (mod~m) $, so the m-roots of unity lie within those of order $2^n-1 $ )).

Assuming we succeed in fixing a one-to-one correspondence between the non-zero elements of $\overline{\mathbb{F}_2} $ and the odd roots of unity $\mu_{odd} $ respecting multiplication, how can we recover the addition on $\overline{\mathbb{F}_2} $? Well, here’s Alain’s coupling function, he ties up an element x of the algebraic closure to the element s(x)=x+1 (and as we are in characteristic two, this is an involution, so also the element tied up to x+1 is s(x+1)=(x+1)+1=x. The clue being that multiplication together with the coupling map s allows us to compute any sum of two elements as $x+y=x \times s(\frac{y}{x}) = x \times (\frac{y}{x}+1) $.
For example, all information about the finite field $\mathbb{F}_{2^4} $ is encoded in this identification with the 15-th roots of unity, together with the pairing s depicted as

Okay, we now have two identifications of the algebraic closure $\overline{\mathbb{F}_2} $ : the smaller ordinals equipped with Nim addition and Nim multiplication and the odd roots of unity with complex-multiplication and the Connes-coupling s. The question we started from asks for a general recipe to identify these two approaches.

To those of you who are convinced that finite fields (LOL, even characteristic two!) are objects far too trivial to bother thinking about : as far as I know, NOBODY knows how to do this explicitly, even restricting the ordinals to merely the natural numbers!

Please feel challenged! To get you started, I’ll show you how to place the first 15 Knights and give you a procedure (though far from explicit) to continue. Here’s the Nim-picture compatible with that above

To verify this, and to illustrate the general strategy, I’d better hand you the Nim-tables of the first 16 numbers. Here they are

It is known that the finite subfields of $~(\mathbb{N},\oplus,\otimes) $ are precisely the sets of numbers smaller than the Fermat-powers $F_n $. So, the first one is all numbers smaller than $F_1=4 $ (check!). The smallest generator of the multiplicative group (of order 3) is 2, so we take this to correspond to the unit-root $e^{2 \pi i/3} $. The next subfield are all numbers smaller than $F_2 = 16 $ and its multiplicative group has order 15. Now, choose the smallest integer k which generates this group, compatible with the condition that $k^{\otimes 5}=2 $. Verify that this number is 4 and that this forces the identification and coupling given above.

The next finite subfield would consist of all natural numbers smaller than $F_3=256 $. Hence, in this field we are looking for the smallest number k generating the multiplicative group of order 255 satisfying the extra condition that $k^{\otimes 17}=4 $ which would fix an identification at that level. Then, the next level would be all numbers smaller than $F_4=65536 $ and again we would like to find the smallest number generating the multiplicative group and such that the appropriate power is equal to the aforementioned k, etc. etc.

Can you give explicit (even inductive) formulae to achieve this? I guess even the problem of placing Knight 16 will give you a couple of hours to think about… (to be continued).

The artist and the mathematician

Published October 19, 2009 by lievenlb

Over the week-end I read The artist and the mathematician (subtitle : The story of Nicolas Bourbaki, the genius mathematician who never existed) by Amir D. Aczel.

Whereas the central character of the book should be Bourbaki, it focusses more on two of Bourbaki’s most colorful members, André Weil and Alexander Grothendieck, and the many stories and myths surrounding them.

The opening chapter (‘The Disappearance’) describes the Grothendieck’s early years (based on the excellent paper by Allyn Jackson Comme Appelé du Néant ) and his disappearance in the Pyrenees in the final years of last century. The next chapter (‘An Arrest in Finland’) recount the pre-WW2 years of Weil and the myth of his arrest in Finland and his near escape from execution (based on Weil’s memoires The Apprenticeship of a Mathematician). Chapter seven (‘The Café’) describes the first 10 proto-Bourbaki meetings following closely the study ‘A Parisian Café and Ten Proto-Bourbaki Meetings (1934-1935)‘ by Liliane Beaulieu. Etc. etc.

All the good ‘Bourbaki’-stories get a place in this book, not always historically correct. For example, on page 90 it is suggested that all of the following jokes were pulled at the Besse-conference, July 1935 : the baptizing of Nicolas, the writing of the Comptes-Rendus paper, the invention of the Bourbaki-daughter Betti and the printing of the wedding invitation card. In reality, all of these date from much later, the first two from the autumn of 1935, the final two no sooner than april 1939…

One thing I like about this book is the connection it makes with other disciplines, showing the influence of Bourbaki’s insistence on ‘structuralism’ in fields as different as philosophy, linguistics, anthropology and literary criticism. One example being Weil’s group-theoretic solution to the marriage-rules problem in tribes of Australian aborigines studied by Claude Lévi-Strauss, another the literary group Oulipo copying Bourbaki’s work-method.

Another interesting part is Aczel’s analysis of Bourbaki’s end. In the late 50ties, Grothendieck tried to convince his fellow Bourbakis to redo their work on the foundations of mathematics, changing these from set theory to category theory. He failed as others felt that the foundations had already been laid and there was no going back. Grothendieck left, and Bourbaki would gradually decline following its refusal to accept new methods. In Grothendieck’s own words (in “Promenade” 63, n. 78, as translated by Aczel) :

“Additionally, since the 1950s, the idea of structure has become passé, superseded by the influx of new ‘categorical’ methods in certain of the most dynamical areas of mathematics, such as topology or algebraic geometry. (Thus, the notion of ‘topos’ refuses to enter into the ‘Bourbaki sack’ os structures, decidedly already too full!) In making this decision, in full cognizance, not to engage in this revision, Bourbaki has itself renounced its initial ambition, which has been to furnish both the foundations and the basic language for all of modern mathematics.”

Finally, it is interesting to watch Aczel’s own transformation throughout the book, from slavishly copying the existing Weil-myths and pranks at the beginning of the book, to the following harsh criticism on Weil, towards the end (p. 209) :

“From other information in his autobiography, one gets the distinct impression that Weil was infatuated with the childish pranks of ‘inventing’ a person who never existed, creating for him false papers and a false identity, complete with a daughter, Betti, who even gets married, parents and relatives, and membership in a nonexistent Academy of Sciences of the nonexistent nation of Polvedia (sic).
Weil was so taken with these activities that he even listed, as his only honor by the time of his death ‘Member, Poldevian Academy of Sciences’. It seems that Weil could simply not go beyond these games: he could not grasp the deep significance and power of the organization he helped found. He was too close, and thus unable to see the great achievements Bourbaki was producing and to acknowledge and promote these achievements. Bourbaki changed the way we do mathematics, but Weil really saw only the pranks and the creation of a nonexistent person.”

Judging from my own reluctance to continue with the series on the Bourbaki code, an overdose reading about Weil’s life appears to have this effect on people…

Grothendieck’s functor of points

Published September 29, 2009 by lievenlb

A comment-thread well worth following while on vacation was Algebraic Geometry without Prime Ideals at the Secret Blogging Seminar. Peter Woit became lyric about it :

My nomination for the all-time highest quality discussion ever held in a blog comment section goes to the comments on this posting at Secret Blogging Seminar, where several of the best (relatively)-young algebraic geometers in the business discuss the foundations of the subject and how it should be taught.

I follow far too few comment-sections to make such a definite statement, but found the contributions by James Borger and David Ben-Zvi of exceptional high quality. They made a case for using Grothendieck’s ‘functor of points’ approach in teaching algebraic geometry instead of the ‘usual’ approach via prime spectra and their structure sheaves.

The text below was written on december 15th of last year, but never posted. As far as I recall it was meant to be part two of the ‘Brave New Geometries’-series starting with the Mumford’s treasure map post. Anyway, it may perhaps serve someone unfamiliar with Grothendieck’s functorial approach to make the first few timid steps in that directions.

Allyn Jackson’s beautiful account of Grothendieck’s life “Comme Appele du Neant, part II” (the first part of the paper can be found here) contains this gem :

“One striking characteristic of Grothendieck’s
mode of thinking is that it seemed to rely so little
on examples. This can be seen in the legend of the
so-called “Grothendieck prime”.

In a mathematical
conversation, someone suggested to Grothendieck
that they should consider a particular prime number.
“You mean an actual number?” Grothendieck
asked. The other person replied, yes, an actual
prime number. Grothendieck suggested, “All right,
take 57.”

But Grothendieck must have known that 57 is not
prime, right? Absolutely not, said David Mumford
of Brown University. “He doesn’t think concretely.””

We have seen before how Mumford’s doodles allow us to depict all ‘points’ of the affine scheme $\mathbf{spec}(\mathbb{Z}[x]) $, that is, all prime ideals of the integral polynomial ring $\mathbb{Z}[x] $.
Perhaps not too surprising, in view of the above story, Alexander Grothendieck pushed the view that one should consider all ideals, rather than just the primes. He achieved this by associating the ‘functor of points’ to an affine scheme.

Consider an arbitrary affine integral scheme $X $ with coordinate ring $\mathbb{Z}[X] = \mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k) $, then any ringmorphism
$\phi~:~\mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k) \rightarrow R $
is determined by an n-tuple of elements $~(r_1,\ldots,r_n) = (\phi(t_1),\ldots,\phi(t_n)) $ from $R $ which must satisfy the polynomial relations $f_i(r_1,\ldots,r_n)=0 $. Thus, Grothendieck argued, one can consider $~(r_1,\ldots,r_n) $ an an ‘$R $-point’ of $X $ and all such tuples form a set $h_X(R) $ called the set of $R $-points of $X $. But then we have a functor

$h_X~:~\mathbf{commutative rings} \rightarrow \mathbf{sets} \qquad R \mapsto h_X(R)=Rings(\mathbb{Z}[t_1,\ldots,t_n]/(f_1,\ldots,f_k),R) $

So, what is this mysterious functor in the special case of interest to us, that is when $X = \mathbf{spec}(\mathbb{Z}[x]) $?
Well, in that case there are no relations to be satisfied so any ringmorphism $\mathbb{Z}[x] \rightarrow R $ is fully determined by the image of $x $ which can be any element $r \in R $. That is, $Ring(\mathbb{Z}[x],R) = R $ and therefore Grothendieck’s functor of points
$h_{\mathbf{spec}(\mathbb{Z}[x]} $ is nothing but the forgetful functor.

But, surely the forgetful functor cannot give us interesting extra information on Mumford’s drawing?
Well, have a look at the slightly extended drawing below :

What are these ‘smudgy’ lines and ‘spiky’ points? Well, before we come to those let us consider the easier case of identifying the $R $-points in case $R $ is a domain. Then, for any $r \in R $, the inverse image of the zero prime ideal of $R $ under the ringmap $\phi_r~:~\mathbb{Z}[x] \rightarrow R $ must be a prime ideal of $\mathbb{Z}[x] $, that is, something visible in Mumford’s drawing. Let’s consider a few easy cases :

For starters, what are the $\mathbb{Z} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $? Any natural number $n \in \mathbb{Z} $ determines the surjective ringmorphism $\phi_n~:~\mathbb{Z}[x] \rightarrow \mathbb{Z} $ identifying $\mathbb{Z} $ with the quotient $\mathbb{Z}[x]/(x-n) $, identifying the ‘arithmetic line’ $\mathbf{spec}(\mathbb{Z}) = { (2),(3),(5),\ldots,(p),\ldots, (0) } $ with the horizontal line in $\mathbf{spec}(\mathbb{Z}[x]) $ corresponding to the principal ideal $~(x-n) $ (such as the indicated line $~(x) $).

When $\mathbb{Q} $ are the rational numbers, then $\lambda = \frac{m}{n} $ with $m,n $ coprime integers, in which case we have $\phi_{\lambda}^{-1}(0) = (nx-m) $, hence we get again an horizontal line in $\mathbf{spec}(\mathbb{Z}[x]) $. For $ \overline{\mathbb{Q}} $, the algebraic closure of $\mathbb{Q} $ we have for any $\lambda $ that $\phi_{\lambda}^{-1}(0) = (f(x)) $ where $f(x) $ is a minimal integral polynomial for which $\lambda $ is a root.
But what happens when $K = \mathbb{C} $ and $\lambda $ is a trancendental number? Well, in that case the ringmorphism $\phi_{\lambda}~:~\mathbb{Z}[x] \rightarrow \mathbb{C} $ is injective and therefore $\phi_{\lambda}^{-1}(0) = (0) $ so we get the whole arithmetic plane!

In the case of a finite field $\mathbb{F}_{p^n} $ we have seen that there are ‘fat’ points in the arithmetic plane, corresponding to maximal ideals $~(p,f(x)) $ (with $f(x) $ a polynomial of degree $n $ which remains irreducible over $\mathbb{F}_p $), having $\mathbb{F}_{p^n} $ as their residue field. But these are not the only $\mathbb{F}_{p^n} $-points. For, take any element $\lambda \in \mathbb{F}_{p^n} $, then the map $\phi_{\lambda} $ takes $\mathbb{Z}[x] $ to the subfield of $\mathbb{F}_{p^n} $ generated by $\lambda $. That is, the $\mathbb{F}_{p^n} $-points of $\mathbf{spec}(\mathbb{Z}[x]) $ consists of all fat points with residue field $\mathbb{F}_{p^n} $, together with slightly slimmer points having as their residue field $\mathbb{F}_{p^m} $ where $m $ is a divisor of $n $. In all, there are precisely $p^n $ (that is, the number of elements of $\mathbb{F}_{p^n} $) such points, as could be expected.

Things become quickly more interesting when we consider $R $-points for rings containing nilpotent elements.