# Henselianity in the language of rings, with Franziska Jahnke

I would like to write about a new paper ([AJ15]) which Franziska Jahnke and I have written and put on the arXiv. It’s called Henselianity in the language of rings. In it we investigate the relationship between the following four properties of a field $K$:

(h) $K$ is henselian, i.e. $K$ admits a nontrivial henselian valuation,

(eh) $K$ is elementarily henselian, i.e. every $L\equiv K$ is henselian,

(def) $K$ admits a definable nontrivial henselian valuation, and

($\emptyset$-def) $K$ admits a $\emptyset$-definable nontrivial henselian valuation.

# Puzzle corner: Solution to Sylvy’s puzzle #4

Cross-reference to puzzle page on my website

In this post I want to discuss the solution to Puzzle #4. Let’s recall the puzzle: it was to find an uncountable chain in $(\mathcal{P}(\mathbb{N}),\subseteq)$.

# Puzzle corner: Sylvy’s puzzle #5

Cross-reference to puzzle page on my website

Here is my next puzzle! Sorry that it’s been greatly delayed.

I mentioned previously the problem of constructing the midpoint of a given circle, using only ruler and compass. One student solved it nearly immediately (congratulations!), so here I offer a more general problem.

#### Sylvy’s puzzle #5

Given a parabola in the plane, construct its vertex using straightedge-and-compass construction (sometimes called ruler-and-compass’ construction).

# Sylvy’s Weekly Puzzle #3, Solution (part II)

Cross-reference to puzzle page on my website

In this post I want to carry on discussing the solution to puzzle #3. Recall that $\mathcal{S}$ denotes the world that Sonic explores in the Special Stage’. We are thinking of $\mathcal{S}$ as a topological space, upto topological equivalence.

We have already taken a look at the cases of the sphere and the torus, and we had shown that $\mathcal{S}$  cannot be a sphere and that $\mathcal{S}$ could be a torus. In fact we saw the map’ of the level: so it really IS a torus! But the question is: what can we say about $\mathcal{S}$ using only the topological information that sonic can gather?’

The final part of puzzle #3 asked us to see if $\mathcal{S}$ could be a Klein Bottle. A Klein bottle is the topological space obtained by taking a square and gluing together the opposite edges, as in the following diagram on the left. Note that the blue pair are glued together with a twist, while the red pair are glued straight’.

Gluing the edges of a square to form a Klein Bottle

Well, the answer is yes’. Let $m,n\in\mathbb{N}$ be such that $m$ is even and $n$ is odd. Let $C$ be a  rectangle which is divided into an $m\times n$ chessboard, coloured as usual so that white squares and black squares alternate. I think of the bottom’ or horizontal’ edges as being $m$-squares long, and the side’ or vertical’ edge as being $n$-squares long. Below on the right is a picture for the case $m=5,n=6$.

even-by-odd chessboard

Gluing the edges of this chessboard together in the way described above results in a Klein bottle which is tiled in the usual alternating black and white’ pattern that Sonic sees all around him. This shows that $\mathcal{S}$ could be a Klein bottle.

Note that we’ve only taken into account topological information. In later posts I want to tackle the geometric’ question.

# Sylvy’s weekly puzzle #4a

Cross-reference to puzzle page on my website

It seems that my first-year students (who form a large part of the target audience for my puzzles) are unfamiliar with the countable’ and uncountable’ terminology that I used in puzzle #4. …oops!

Instead of introducing it now, I’ll give a completely different puzzle which was originally shown to me by a friend (more on this later).

Sylvy’s weekly puzzle #4a

Show by means of a geometrical drawing’ that

$\arctan(1)+\arctan(2)+\arctan(3)=\pi$.

deadline: 11am Monday 9th of November

how to hand-in: send typed solution by email or put hand-written solution into the folder on my office door

prize: I’m very pleased to announce that the winner of this puzzle will get a book of mathematical puzzles!!

Have fun!

# Sylvy’s weekly puzzle #4

Cross-reference to puzzle page on my website

This is an old one, but good fun.

Sylvy’s weekly puzzle #4

Let $\mathcal{P}(\mathbb{N})$ denote the powerset of the natural numbers. For clarity, we will consider 0 to be a natural number. Then the pair

$(\mathcal{P}(\mathbb{N}),\subseteq)$

is a partial order: the (weak) ordering is reflexive, antisymmetric, and transitive. But it is very definitely not a total order: given $X,Y\in\mathcal{P}(\mathbb{N})$ we may have neither $X\subseteq Y$ nor $Y\subseteq X$. A chain is a subset $\mathcal{C}$ of $\mathcal{P}(\mathbb{N})$ on which $\subseteq$ is a total order. For example:

$\mathcal{C}_{0}=\big\{\emptyset,\{0\},\{0,1,\},\{0,1,2\},...\big\}$

is a chain, but note that $\mathcal{C}_{0}$ is countable.

This week’s problem is to find an uncountable chain, that is a subset $\mathcal{C}$ of $\mathcal{P}(\mathbb{N})$ which is totally ordered by $\subseteq$ and uncountable.

Deadline: 10am Thursday 5th of November

(you can either send your solution by email or put a hand-written solution into the folder on my office door)

Prize: I’m very please to anounce that the winner will get a book of mathematical puzzles!!

Solution class: there will be a class at 10am on Thursday 5th of November in my office to go over the solution to this puzzle.

The webpage for these puzzles is http://anscombe.sdf.org/puzzle.html

Have fun!

# Two old-ish Terry Tao posts about model theory

I can’t get enough of Terry Tao’s blog What’s New, he writes so much and so brilliantly. Anyway, I found two really nice old-ish posts about model theory:

• this which talks about completeness, compactness, zeroth-order logic, and Skolemisation;
• and this which talks about nonstandard analysis, notions of elementary convergence’ and `elementary completion’, countable saturation, compactness and saturation re-written from an analytical point-of-view, and the Szemerédi regularity lemma (something I always want to know more about).

I don’t have time to write anything more now, but later I will get back to it.