## 305 -Homework set 5

March 4, 2009

This set is due March 11 at the beginning of lecture. Details of the homework policy can be found on the syllabus and here.

Solve exercises 7, 35, 40, 43, 44, 45 from Chapter 6 of the book.

## 305 -5. Extensions by radicals

March 4, 2009

(This post was typeset using Luca Trevisan‘s LaTeX2WP program.)

Last lecture we characterized subfields and used the characterization to provide many new examples of fields. Now we start to explore systematically which subfields of the complex numbers are suitable to study the question of which polynomial equations can be solved. Read the rest of this entry »

## 580 -Cardinal arithmetic (7)

March 4, 2009

[This document was typeset using Luca Trevisan‘s LaTeX2WP. I will refer to result $n$ (or definition $n$) from last lecture as $3.n.$]

A. The Galvin-Hajnal rank and an improvement of Theorem 3.1

Last lecture, I covered the first theorem of the Galvin-Hajnal paper and several corollaries. Recall that the result, Theorem 3.1, states that if ${\kappa}$ and ${\lambda}$ are uncountable regular cardinals, and ${\lambda}$ is ${\kappa}$-inaccessible, then ${\prod_{\alpha<\kappa}\kappa_\alpha<\aleph_\lambda}$ for any sequence ${(\kappa_\alpha:\alpha<\lambda)}$ of cardinals such that ${\prod_{\alpha<\beta}\kappa_\alpha<\aleph_\lambda}$ for all ${\beta<\kappa.}$

In particular (see, for example, Corollary 3.7), if ${{\rm cf}(\xi)>\omega}$ and ${\aleph_\xi}$ is strong limit, then ${2^{\aleph_\xi}<\aleph_{(|\xi|^{{\rm cf}(\xi)})^+}.}$

The argument relied in the notion of an almost disjoint transversal. Assume that ${\kappa}$ is regular and uncountable, and recall that if ${{\mathcal A}=(A_\alpha:\alpha<\kappa)}$ is a sequence of sets, then ${T({\mathcal A})=\sup\{|{\mathcal F}|:{\mathcal F}}$ is an a.d.t. for ${{\mathcal A}\}.}$ Here, ${{\mathcal F}}$ is an a.d.t. for ${{\mathcal A}}$ iff ${{\mathcal F}\subseteq\prod{\mathcal A}:=\prod_{\alpha<\kappa}A_\alpha}$ and whenever ${f\ne g\in{\mathcal F},}$ then ${\{\alpha<\kappa:f(\alpha)=g(\alpha)\}}$ is bounded.

With ${\kappa,\lambda}$ as above, Theorem 3.1 was proved by showing that there is an a.d.t. for ${(\prod_{\beta<\alpha}\kappa_\beta:\alpha<\kappa)}$ of size ${\prod_{\alpha<\kappa}\kappa_\alpha,}$ and then proving that, provided that ${|A_\alpha|<\aleph_\lambda}$ for all ${\alpha<\kappa,}$ then ${T({\mathcal A})<\aleph_\lambda.}$

In fact, the argument showed a bit more. Recall that if ${f:\kappa\rightarrow{\sf ORD},}$ then ${\aleph_f=(\aleph_{f(\alpha)}:\alpha<\kappa).}$ Then, for any ${f:\kappa\rightarrow\lambda}$, ${T(\aleph_f)<\aleph_\lambda.}$

The proof of this result was inductive, taking advantage of the well-foundedness of the partial order ${<_{b,\kappa}}$ defined on ${{}^\kappa{\sf ORD}}$ by ${f<_{b,\kappa}g}$ iff ${\{\alpha:f(\alpha)\ge g(\alpha)\}}$ is bounded in ${\kappa.}$ That ${<_{b,\kappa}}$ is well-founded allows us to define a rank ${|f|_b}$ for each ${f:\kappa\rightarrow{\sf ORD},}$ and we can argue by considering a counterexample of least possible rank to the statement from the previous paragraph.

In fact, more precise results are possible. Galvin and Hajnal observed that replacing the ideal of bounded sets with the nonstationary ideal (or, really, any normal ideal), results in a quantitative improvement of Theorem 3.1. Read the rest of this entry »