## 305 -Homework set 2

February 9, 2009

This set is due February 20 at the beginning of lecture. Consult the syllabus for details on the homework policy. I do not think this set is particularly difficult, but it is on the longish side of things, so make sure you leave yourself enough time to work on it.

1. Gauß’ fundamental theorem of algebra states that any equation $p(x)=0,$ where $p$ is a polynomial with complex coefficients, has at least one complex root $z.$ This means that $z$ is a complex number and $p(z)=0.$ Show that $p$ has at most $n$ roots, where $n$ is its degree, and that if we count roots up to multiplicity, then it has exactly $n$ roots. Since the multiplicity of a root $z$ is by definition the largest $m$ such that $(x-z)^m$ is a factor of $p(x),$ you may want to verify that $p(z)=0$ iff $(x-z)$ is a factor of $p.$

2. Let $p(x)$ be a polynomial with real coefficients, and let $z$ be a complex root of $p.$ Show that $p(\bar z)=0$ as well. Conclude that if the degree $n$ of $p$ is odd and the coefficients of $p$ are real, then $p$ has at least one real root. (You may use the fundamental theorem of algebra, if needed.) Conclude also that if $p$ is of degree four and has real coefficients, then $p$ can be factored as the product of two quadratic polynomials with real coefficients. (Does this follow “directly” from the argument described in lecture?)

3. Solve exercises 54-56 from Chapter 3 of the book.

4. Show directly that if $a,b,c$ are real numbers, then at least one of the solutions of $x^3+ax^2+bx+c=0$ is a real number. What I mean is that, rather than appealing to problem 2, you want to look at the solutions obtained by Cardano’s method as described in lecture, and argue directly from the formulas so obtained that at least one of the solutions must be real. Be careful, since your argument should not give you that all three roots are real, since this is not true in general.

5. Show directly that a quartic with complex coefficients admits only 4 roots. What I mean is that, rather than appealing to problem 1, you want to look at the solutions obtained by Ferrari’s method as described in lecture, and argue directly that they only produce 4 roots, even though, in principle, they produce 24 (since they involve solving a cubic and then taking a square root to obtain parameters from which four solutions are then found).

## 305 -3. Complex numbers.

February 9, 2009

Mathematicians first approached complex numbers cautiously. Although it was clear that they were useful in solving certain problems at least formally (for example, they are needed to even make sense of the formulas we found in the previous lectures) what was not clear was that they made sense. Perhaps indiscriminate use of them would lead to contradictions.

Gauß solved this problem by realizing that one can define ${\mathbb C}$ and its operations in terms of ${\mathbb R}$ and its operations. As long as we are willing to accept that ${\mathbb R}$ makes sense, then no contradictions will come up from the use of complex numbers.

## 580 -Cardinal arithmetic (3)

February 9, 2009

It is easy to solve negatively the question immediately following Homework problem 5 on lecture II.1. I asked whether if $X$ is Dedekind-finite but ${\mathcal P}(X)$ is Dedekind-infinite, then it followed that there is an infinite Dedekind-finite set $Y$ such that ${\mathcal P}(Y)\preceq X$.

To exhibit a counterexample, it is enough to know that it is consistent to have an infinite Dedekind finite set $X$ that is the countable union of finite sets (in fact, sets of size 2). Notice that $\omega$ is a surjective image of $X,$ so ${\mathcal P}(X)$ is Dedekind-infinite. Suppose that ${\mathcal P}(Y)\preceq X.$ Then certainly $Y\preceq X,$ so $Y$ is a countable union of finite sets $Y_n.$ If $Y$ is infinite then $Y_n\ne\emptyset$ for infinitely many values of $n.$ But then $\omega$ is also a surjective image of $Y$, so $\omega$ (and in fact $P(\omega)$) injects into ${\mathcal P}(Y)$ and therefore into $X,$ contradiction.

At the end of last lecture we showed Theorem 10, a general result that allows us to compute products $\kappa^\lambda$ for infinite cardinals $\kappa,\lambda,$ namely:

Let $\kappa$ and $\lambda$ be infinite cardinals. Let $\tau=\sup_{\rho<\kappa}|\rho|^\lambda.$ Then

$\displaystyle \kappa^\lambda=\left\{\begin{array}{cl} 2^\lambda & \mbox{if }\kappa\le 2^\lambda,\\ \kappa\cdot\tau & \mbox{if }\lambda<{\rm cf}(\kappa),\\ \tau & \begin{array}{l}\mbox{if }{\rm cf}(\kappa)\le\lambda,2^\lambda<\kappa,\mbox{ and }\\ \rho\mapsto|\rho|^\lambda\mbox{ is eventually constant below }\kappa,\end{array}\\ \kappa^{{\rm cf}(\kappa)} & \mbox{otherwise.}\end{array}\right.$

## 580 -Cardinal arithmetic (2)

February 7, 2009

At the end of last lecture, we showed Theorem 7, König’s lemma, stating that if $\kappa_i<\lambda_i$ for all $i\in I,$ then $\sum_{i\in I}\kappa_i<\prod_i\lambda_i.$ We begin by looking at some  corollaries:

Corollary 8.

1. If $\beta$ is a limit ordinal and $(\kappa_i:i<\beta)$ is a strictly increasing sequence of nonzero cardinals, then $\sum_{\alpha<\beta}\kappa_\alpha<\prod_{\alpha<\beta}\kappa_\alpha.$
2. If $(\kappa_i:i\in I)$ is an $I$-indexed sequence of nonzero cardinals and $\kappa_i<\sum_{j\in I}\kappa_j$ for all $i\in I,$ then $\sum_i\kappa_i<\left(\sum_i\kappa_i\right)^{|I|}.$
3. (Cantor) $\kappa<2^\kappa.$
4. For any infinite $\kappa$, one has $\kappa<\kappa^{{\rm cf}(\kappa)}.$
5. ${\rm cf}(2^\kappa)>\kappa.$

## 305 -Solving cubic and quartic polynomials (3)

February 5, 2009

As briefly mentioned in the book, for polynomial equations of degree five or higher we cannot find solutions the same way that we found solutions to the equations of degree at most four.

Implicit in this statement is the notion of solution’ that we are interested in. From the point of view of Abstract Algebra, the expressions we are looking for are what one usually calls solutions by radicals,’ although we have not made this too precise yet. Informally, we look for a formula in which we are allowed to use:

• The coefficients of the given polynomial, and
• complex and real numbers,

and in which we can make use of

• the elementary operations $+,-,\times,\div$, and
• radicals—i.e., we can take $n$-th roots of any of the expressions we can obtain, for any positive integer $n.$

The amazing result that motivates our work through this course is that, indeed, no such formulas exist for polynomials of degree five or higher.

However, it is a bit misleading to read this as saying that no `formulas’ exist at all. The situation is similar to what mathematicians had to face before having the notion of complex numbers. Then, some equations could not be solved, since their solutions would involve the extraction of square roots of negative numbers. Nowadays, we understand that we can solve those equations, as long as the use of complex numbers is allowed, and we cannot otherwise.

Indeed, although no solutions by radicals are possible for the roots of quintic polynomials, if we allow a larger class of operations to be used, then solutions exist. For example, more general hypergeometric series than $n$-th roots allow us to find the roots of the quintic. Although not nearly as popular now as they once were, hypergeometric series (a particular kind of power series) are still fairly used, for example in partition theory.

There is a well known poster from Wolfram Research explaining how Mathematica can be used to solve the quintic with the help of these functions; their webpage actually is very interesting.

Besides the usefulness of hypergeometric functions, Felix Klein found an approach using the symmetries of the Icosahedron. His book, Lectures on the Icosahedron and the Solution of Equations of the Fifth Degree, is available online for download from Cornell.

These approaches essentially propagate to higher degree equations, with some new obstacles appearing along the way, but in this sense, one can solve any polynomial equation. The solutions are less satisfactory in that general hypergeometric functions are less well understood and much less intuitive than $n$-th roots. For the purposes of the course, we will concentrate on solutions by radicals, since also these are the ones that lend themselves naturally to algebraic (rather than analytic) study, and the arguments work in much more generality than just over the complex numbers.

## 580 -II. Cardinal arithmetic

February 5, 2009

Homework problem 5. (${\sf ZF}$). Show that if $\omega\preceq{\mathcal P}(X)$ then in fact ${\mathcal P}(\omega)\preceq{\mathcal P}(X)$.

Question. If $X$ is Dedekind-finite but ${\mathcal P}(X)$ is Dedekind-infinite, does it follow that there is an infinite Dedekind-finite set $Y$ such that ${\mathcal P}(Y)\preceq X$?

From now on, until we cover determinacy, we assume the axiom of choice unless stated otherwise.

## Steel VIG (January 30-February 1, 2009)

February 3, 2009

John Steel was one of my advisors at UC Berkeley. The fifteenth Very Informal Gathering of Logicians at UCLA (VIG) was in honor of John Steel on his 60th birthday.

The meeting was excellent, with some very interesting talks and nice shared memories. Plus, Benjamin Miller worked miracles and secured the photo above without John finding out.

## 305 -Solving cubic and quartic polynomials (2)

February 3, 2009

Let’s return to the problem of solving quartic polynomials. In the first lecture on this topic, we reduced the problem of solving an equation like

$x^4+ax^3+bx^2+cx+d=0$

to solving the similar problem

$y^4+py^2+qy+r=0,$

in which the coefficient of $y^3$ is zero. This is achieved by a simple translation, simply set $x=y-a/4$. This was motivated and explained in that lecture. Let us now see how we can approach this problem.

## 580 -Some choiceless results (5)

February 2, 2009

[This lecture was covered by Marion Scheepers. Many thanks! The notes below also cover lecture 6.]

We want to prove the following result and a few related facts.

Theorem. (Specker). ${\rm CH}({\mathfrak m})$ and ${\rm CH}(2^{\mathfrak m})$ imply $2^{\mathfrak m}=\aleph({\mathfrak m})$.

It follows immediately from the theorem that ${\sf GCH}$ implies ${\sf AC}$, since the result gives that any (infinite) ${\mathfrak m}$ embeds into $\aleph({\mathfrak m})$.