ac.commutative algebra - Do there exist non-PIDs in which every countably generated ideal is principal?

The question is fully settled by Hugh Thomas' anwer, but let me mention this related interesting fact.

Theorem. There is a ring R and ideal I on R, such that every countable subset
of I is contained in a principal subideal of I, but I is
not principal.

Proof. Let I be the ideal of nonstationary subsets of
ω₁, in the power set P(ω₁), which is a Boolean
algebra and hence a Boolean ring. That is, I consists of those subsets of ω₁ that
are disjoint from a closed unbounded subset of ω₁. It is an elementary set-theoretic fact that the intersection of any countably many closed unbounded subsets of ω₁ is still closed and unbounded, and thus the union of countably many
non-stationary sets remains non-stationary. Thus, every
countable subset of I is contained in a principal subideal
of I. But I is not principal, since the complement of any
singleton is stationary. QED

In the previous example, the ideal I is not maximal. If one assumes the existence of a measurable cardinal (a large cardinal notion), however, then the example can be made with I maximal.

Theorem. If there is a measurable cardinal, then there is a
ring R with a maximal ideal I, such that every countable
subset of I is contained in a principal sub-ideal of I, but
I is not principal.

Proof. Let κ be a measurable cardinal, which means that
there is a nonprincipal κ-complete ultrafilter U on the
power set P(κ), which is a Boolean algebra and thus a
Boolean ring. The ideal I dual to U is also κ-complete,
which means that I is closed under unions of size less than
κ. In particular, since kappa is uncountable, this
means that the union of any countably many elements of I
remains in I, and this union set generates a principal
subideal of I containing the given countable set. The ideal I is maximal since U was an ultrafilter. QED

I'm not sure at the moment whether the situation of this last theorem requires a measurable cardinal or not, but I'll think about it.

rna sequencing - RNA seq and using of Poly(A) or non-Poly(A) based amplification of RNA

I'm studying "Deep sequencing the circadian and diurnal transcriptome of Drosophila brain" Hughes et al., 2012. I've got some problems with the materials and methods.

Before RNAseq, the authors amplify RNA. They use two kits: Poly(A) based and non-poly(A) based. For the samples Poly(A) based amplified they do a sequencing to generate 100 bp paired-end and for the samples Non poly(A) based they do a sequencing to generate 75 single-ends.

I read in the strategy:

Total RNA was amplified, and ribosomal RNA was depleted, using a non-poly(A)-based amplification kit, which significantly diminishes the 3' bias in downstream libraries used for RNA-seq (see Methods).

1) I don't understand why they use both kits. Actually, they say that they want to asses the bias introduced by an alternative amplification methods BUT the following sequencing is not the same in both cases so there is more than one parameter which change between the both protocols so I don't understand how they can compare.

2) For the poly(A) based I understand that the rRNA are not polyadenylated so they remove rRNA from the amplified sample. However, don't they remove non-coding RNA too (long or short)?

3) For the non poly(A) based, I don't understand how they remove the rRNA but I understand that since they don't select the RNA according to their poly(A) tails they keep the rest of the RNAs which are not kept in the poly(A) based amplification.

4) When I look to the RUM statistics, I see that there is less total aligned read in non-poly(A) based: 70-80% (non polyA) and 80-90% (polyA). So it seems like we loose some information with non-poly(A) based.

To resume, why the authors use Poly(A) and non-poly(A)? Why do they show all the rest of the results from the non-poly(A) amplification since they have less aligned read? How do they compare since the following sequencing is not the same? How the rRNA are removed in both techniques and especially in non-poly(A)?

I didn't find the protocols kit with exactly the same name but here are the links to the protocols which are closer (I think).

http://www.mscience.com.au/upload/pages/nugen/nugen_ov_rna_seq-brochure.pdf

http://excilone.com/client/document/ug-arcturusE%E2%80%9E%C2%A2-riboamp%C3%82-hs-plus-amplification-kit-user-guide_38.pdf
Thank you very much,

zeta functions - Is there an analogue of the Lefschetz fixed point theorem for discrete dynamical systems?

Background/Motivation

Let $(X, f)$ be a discrete dynamical system. For now, $X$ is just a set and $f$ is just a function $f : X to X$. Suppose that $f^n$ has a finite number of fixed points for every $n$. Then the dynamical (Artin-Mazur) zeta function $zeta_f$ is given by

$displaystyle zeta_f(t) = exp left( sum_{n ge 1} frac{text{Fix } f^n}{n} t^n right)$.

The coefficients of $zeta_f(t)$ have a nice combinatorial interpretation that seems to have homological significance. A particularly famous case of this construction is that $X$ is the set of points of a variety over $overline{ mathbb{F}_p }$ and $f$ is the Frobenius map; then $zeta_f$ is a local zeta function, since $text{Fix } f^n$ is precisely the number of points of the variety over $mathbb{F}_{p^n}$.

Now give $X$ the additional structure of a compact triangulable space and let $f$ be continuous. Again suppose that $f^n$ has a finite number of fixed points for every $n$ and let $i(f, x)$ denote the index of a fixed point $x$, and let $L(f)$ be the sum of the indices $i(f, x)$ over all fixed points of $x$. Thus $L(f)$ generalizes the number $text{Fix } f$ to the case that the indices are not all equal to $1$. Similarly one defines the Lefschetz zeta function by

$displaystyle zeta_f(t) = exp left( sum_{n ge 1} frac{L(f^n)}{n} t^n right)$.

The Lefschetz fixed point theorem is then equivalent to the statement that $zeta_f$ is equal to the alternating product of the characteristic polynomial of the induced action of $f$ on the singular homology groups $H_k(X, mathbb{Q})$; in particular, $zeta_f$ is rational because there are finitely many such groups. Weil famously suggested that if one could define an analogue of singular homology for varieties over finite fields, an analogue of the Lefschetz fixed point theorem would prove the Weil conjectures. This was eventually done, and is known as etale cohomology.

However, I'm interested in a simpler dynamical system than a variety over a finite field. Let $G$ be a finite (directed, possibly with loops) graph, let $X(G)$ be the set of aperiodic closed walks on $G$ with a distinguished vertex, and let $f : X(G) to X(G)$ be the function which sends the distinguished vertex of an aperiodic closed walk to the next vertex in the walk. (An aperiodic closed walk is analogous to a point together with all of its Galois conjugates, and $f$ is conjugation.) Then $text{Fix } f^n$ is precisely the number of closed walks of length $n$ on $G$. A basic result in algebraic combinatorics then tells us that $text{Fix } f^n = text{tr } mathbf{A}^n$, where $mathbf{A}$ is the adjacency matrix of $G$, and this is equivalent to the statement that

$displaystyle zeta_f(t) = frac{1}{det(mathbf{I} - mathbf{A}t)}$.

What this suggests to me is that there is an analogue of the Lefschetz zeta function at work and that it is telling me that $X(G)$ has one nontrivial homology group on which $f$ acts as $mathbf{A}$, but I don't know if this is a reasonable interpretation. Hence my questions!

Edit, 1/8/10: Let me give an example where I can introduce another "homology group." Let $H$ be a proper subgraph of $G$, and let $X(G, H)$ denote the set of aperiodic closed walks on $G$ with a distinguished vertex and with the property that at least one edge or vertex of the closed walk is not in $H$; $f$ is the same as above. If $mathbf{B}$ denotes the adjacency matrix of $H$, it then follows that $text{Fix } f^n = text{tr } left( mathbf{A}^n - mathbf{B}^n right)$, hence

$displaystyle zeta_f(t) = frac{det(mathbf{I} - mathbf{B}t)}{det(mathbf{I} - mathbf{A}t)}$.

Questions

• What is a sensible definition of the (say, integral) homology of a discrete dynamical system without any further structure? What conditions on $X$ are necessary to ensure that there are only finitely many homology groups, and do they hold for $X(G)$ and/or $X(G, H)$?




• Under what conditions does an analogue of the Lefschetz fixed point theorem hold for this homology theory, and can it be made to correctly reproduce the $X(G)$ and $X(G, H)$ computations above?

noncommutative geometry - Kontsevich, and Geometric, Quantization and the Podles sphere

There exist a large family of noncommutative spaces that arise from the quantum matrices. These algebraic objects $q$-deform the coordinate rings of certain varieties. For example, take quantum $SU(2)$, this is the algebra $< a,b,c,d >$ quotiented by the ideal generated by
$$
ab−qba, ~~ ac−qca, ~~ bc−cb, ~~ bd−qdb, ~~ cd−qdc, ~~ ad−da−(q−q^{−1})bc,
$$
and the "q-det" relation
$$
ad−qbc−1
$$
where $q$ is some complex number. Clearly, when $q=1$ we get back the coordinate ring of $SU(2)$. In the classical case $S^2 = SU(2)/U(1)$ (the famous Hopf fibration). This generalises to the q-case: the $U(1)$-action generalises to a $U(1)$-coaction with an invariant subalgebra that q-deforms the coordinate algebra of $S^2$ - the famous Podles sphere. There exist such q-matrix deformations of all flag manifolds.

Since all such manifolds are Kahler, we can also apply Kontsevich deformation to them to obtain a q-defomation. My question is: What is the relationship between these two approaches?

Alternatively, we can apply Kostant-Souriau geometric quantization to a flag manifold. How does alegbra relate to its q-matrix deformation?

bacteriology - Are there grass or fiber eating birds?

My understanding, that may be wrong, is that cellulose/fibre has little nutritional value to many animals because it's hard to break down thus making consumption inefficient. However, Ruminating mammals, other mammals such as black bears, and many insects such as grasshoppers digest grasses.

Are there any birds with the gut flora to break down fibre? Or even some that eat grass but digest it in another way?

lo.logic - What does it mean to 'discharge assumptions or premises'?

Apollo is correct. A slightly more technical way of putting it is that "discharging" is an application of a theorem of metalogic called the deduction theorem:

T,P|-Q iff T|-P->Q

The single turnstile symbol "|-" stands for the syntactic consequences relation. The deduction theorem basically says "Q is derivable from T and P iff if P then Q is derivable from T alone". T may, of course, be an empty class of statements, in which case P->Q is tautologous.

Many systems of natural deduction introduce conditional proof as a primitive rule, but there are simpler systems that are just as powerful in which the deduction theorem is proved and conditional proof is a derived rule supported by the deduction theorem. The deduction theorem is important because it shows you don't need conditional proof as a primitive rule, and this makes the proof of other theorems in metalogic a whole lot simpler. Basically, if you have as few rules as possible it gives you fewer cases to check. For practical purposes, however, it's a whole lot easier to teach and use a system that introduces lots and lots of primitive rules as opposed to one that uses as few rules as possible.

Mathematicians use conditional proof all the time, by the way. For example, in a proof of Q by cases you get conditionals P1->Q, P2->Q, etc. by for each case supposing the antecedents, deriving Q from the supposition, then "discharging" the supposition. Then you show the disjunction of the antecedents is exhaustive.

ac.commutative algebra - For which fields K is every subring of K...?

Let me put together the previous two answers (plus epsilon) to give an answer to all three questions.

Step 1: By Gilmer's theorem, a field $K$ has all its subrings Noetherian iff:

(i) It is a finite extension of $mathbb{Q}$, or
(ii) It is an algebraic extension of $mathbb{F}_p$ or a finite extension of $mathbb{F}_p(t)$.

Step 2: Suppose $K$ is a number field which is not $mathbb{Q}$. We may write $K = mathbb{Q}[alpha]$ for some algebraic integer $alpha$. Then $R = mathbb{Z}[2alpha]$ is a non-integrally closed subring of $K$ so is not a Dedekind domain. So the only field of characteristic $0$ which has every subring a Dedekind domain is $mathbb{Q}$, in which case (by the previous question) every subring is a PID.

Step 3: Suppose $K$ has characteristic $p > 0$. If $K$ is algebraic over $mathbb{F}_p$, then every subring is a field, hence also Dedekind and a PID. If $K$ is a finite extension of $mathbb{F}_p(t)$ then it admits a subring of the form $mathbb{F}_p[t^2,t^3]$, which is not integrally closed.

So the fields for which every subring is a Dedekind ring are $mathbb{Q}$ and the algebraic extensions of $mathbb{F}_p$. For all such fields, every subring is in fact a PID.