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Fox & Geese


The game of Fox and Geese is usually played on a cross-like
board. I learned about it from the second volume of the first edition of
Winning Ways
for your Mathematical Plays
which is now reprinted as number 3 of
the series. In the first edition, Elwyn Berlekamp,
John Conway and
Richard Guy claimed that the value of their
starting position (they play it on an 8×8 chess board with the Geese on
places a1,c1,e1 and g1 and the Fox at place e8) has exact value

1 +
1/on

where on is the class of all ordinal numbers so
1/on is by far the smallest infinitesimal number you can think
of. In this second edition which I bought a week ago, they write about
this :

We remained steadfast in that belief until we heard
objections from John Tromp. We then also received correspondence
from Jonathan Weldon, who seemed to prove to somewhat higher standards
of rigor that
“The value of Fox-and-Geese is 2 +
1/on”

Oops! But of course they try to talk themselves out
of it

Who was right? As often happens when good folks
disagree, the answer is “both!” because it turns out that the parties
are thinking of different things. The Winning Ways argument
supposed an indefinitely long board, while Welton more reasonably
considered the standard 8×8 checkerboard.

Anyway, let us be
happy that the matter is settled now and even more because they add an
enormous amount of new material on the game to this second edition (in
chapter 20; btw. if after yesterday you are still interested in the game of sprouts you might be interested in
chapter 17 of the same volume). Most of the calculations were done with
the combinatorial game suite program of Aaron
Siegel.

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SNORTgo

The
game of SNORT was invented by Simon Norton. The
rules of its SNORTgo-version are : black and white take turns in
putting a stone on a go-board such that no two stones of different
colour occupy neighbouring spots. In contrast to COLgo it is a
hot game meaning that many of its positions are non-numbers. The picture
on the left is a SNORTgo endgame of exact value

{{3|2}|-1}

and is therefore a fuzzy game meaning that the first player to
move has a winning strategy. Can you find the best move for white? and
for black?

If not, you can always consult the slides of the talk I gave this afternoon for the
WIS-seminar. Afterwards there was a game playing
afternoon but I felt too tired and went home early.

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Galois and the Brauer group

Last time we have seen that in order to classify all
non-commutative $l$-points one needs to control the finite
dimensional simple algebras having as their center a finite
dimensional field-extension of $l$. We have seen that the equivalence
classes of simple algebras with the same center $L$ form an Abelian
group, the
Brauer group. The calculation of Brauer groups
is best done using
Galois-cohomology. As an aside :
Evariste Galois was one of the more tragic figures in the history of
mathematics, he died at the age of 20 as a result of a duel. There is
a whole site the Evariste Galois archive dedicated to his
work.

But let us return to a simple algebra $T$ over the
field $L$ which we have seen to be of the form $M(k,S)$, full
matrices over a division algebra $S$. We know that the dimension of
$S$ over $L$ is a square, say $n^2$, and it can be shown that all
maximal commutative subfields of $S$ have dimension n over $L$.
In this way one can view a simple algebra as a bag containing all
sorts of degree n extensions of its center. All these maximal
subfields are also splitting fields for $S$, meaning that
if you tensor $S$ with one of them, say $M$, one obtains full nxn
matrices $M(n,M)$. Among this collection there is at least one
separable field but for a long time it was an open question
whether the collection of all maximal commutative subfields also
contains a Galois-extension of $L$. If this is the case, then
one could describe the division algebra $S$ as a crossed
product
. It was known for some time that there is always a simple
algebra $S’$ equivalent to $S$ which is a crossed product (usually
corresponding to a different number n’), that is, all elements of
the Brauer group can be represented by crossed products. It came as a
surprise when S.A. Amitsur in 1972 came up with examples of
non-crossed product division algebras, that is, division algebras $D$
such that none of its maximal commutative subfields is a Galois
extension of the center. His examples were generic
division algebras
$D(n)$. To define $D(n)$ take two generic
nxn matrices
, that is, nxn matrices A and B such that all its
entries are algebraically independent over $L$ and consider the
$L$-subalgebra generated by A and B in the full nxn matrixring over the
field $F$ generated by all entries of A and B. Somewhat surprisingly,
one can show that this subalgebra is a domain and inverting all its
central elements (which, again, is somewhat of a surprise that
there are lots of them apart from elements of $L$, the so called
central polynomials) one obtains the division algebra $D(n)$ with
center $F(n)$ which has trancendence degree n^2 1 over $L$. By the
way, it is still unknown (apart from some low n cases) whether $F(n)$
is purely trancendental over $L$. Now, utilising the generic
nature of $D(n)$, Amitsur was able to prove that when $L=Q$, the
field of rational numbers, $D(n)$ cannot be a crossed product unless
$n=2^s p_1…p_k$ with the p_i prime numbers and s at most 2. So, for
example $D(8)$ is not a crossed product.

One can then
ask whether any division algebra $S$, of dimension n^2 over $L$, is a
crossed whenever n is squarefree. Even teh simplest case, when n is a
prime number is not known unless p=2 or 3. This shows how little we do
know about finite dimensional division algebras : nobody knows
whether a division algebra of dimension 25 contains a maximal
cyclic subfield (the main problem in deciding this type of
problems is that we know so few methods to construct division
algebras; either they are constructed quite explicitly as a crossed
product or otherwise they are constructed by some generic construction
but then it is very hard to make explicit calculations with
them).

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