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.

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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).

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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

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


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!!

webpage: http://anscombe.sdf.org/puzzle.html

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


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:


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.