field | Teaching blog

305 -Fields (5)

February 27, 2009

At the end of last lecture we stated a theorem giving an easy characterization of subfields of a given field ${\mathbb F}.$ We begin by proving this result.

Theorem 18. Suppose ${\mathbb F}$ is a field and $S\subseteq{\mathbb F}.$ If $S$ satisfies the following 5 conditions, then $S$ s a subfield of ${\mathbb F}:$

$S$ is closed under addition.

$S$ is closed under multiplication.

$-a\in S$ whenever $a\in S.$

$a^{-1}\in S$ whenever $a\in S$ and $a\ne0.$

$S$ has at least two elements.

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2 Comments | 305: Abstract Algebra I | Tagged: arithmetic mean-geometric mean inequality, Ferdinand von Lindemann, field, pi, transcendental | Permalink
Posted by andrescaicedo

Suppose that ${\mathbb F}$ is a field and that $S\subset{\mathbb F}.$ It may be that $S$ is also a field, using the same operations of ${\mathbb F}.$ For example, if ${\mathbb F}={\mathbb R},$ then we could have $S={\mathbb Q}.$

Definition 15. If ${\mathbb F}$ is a field and $S\subset{\mathbb F},$ we say that $S$ is a subfield of ${\mathbb F}$ if $S$ is a field with the operations of ${\mathbb F}.$

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1 Comment | 305: Abstract Algebra I | Tagged: field | Permalink
Posted by andrescaicedo

305 -Fields (3)

February 18, 2009

At the end of last lecture we arrived at the question of whether every finite field is a ${\mathbb Z}_p$ for some prime $p.$

In this lecture we show that this is not the case, by exhibiting a field of 4 elements. We also find some general properties of finite fields. Finite fields have many interesting applications (in cryptography, for example), but we will not deal much with them as our focus through the course is on number fields, that we will begin discussing next lecture.

We begin by proving the following result:

Lemma 13. Suppose that ${\mathbb F}$ is a finite field. Then there is some natural number $n>0$ such that the sum of $n$ ones vanishes, $1+\dots+1=0.$ The least such $n$ is a prime that divides the size of the field.

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2 Comments | 305: Abstract Algebra I | Tagged: field, finite field | Permalink
Posted by andrescaicedo

305 -Fields (2)

February 13, 2009

We continue from last lecture. Examination of a few particular cases finally allows us to complete Example 6. The answer to whether ${\mathbb Z}_n$ (with the operations and constants defined last time) is a field splits into three parts. The first is straightforward.

Lemma 10. For all positive integers $n>1,$ ${\mathbb Z}_n$ satisfies all the properties of fields except possible the existence of multiplicative inverses. $\Box$ .

This reduces the question of whether ${\mathbb Z}_n$ is a field to the problem of finding multiplicative inverses, which turns out to be related to properties of $n.$
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1 Comment | 305: Abstract Algebra I | Tagged: field, zero divisor | Permalink
Posted by andrescaicedo

305 -4. Fields.

February 11, 2009

Definition 1. Let ${\mathbb F}$ be a set. We say that the quintuple $({\mathbb F},+,\times,0,1)$ is a field iff the following conditions hold:

$+:{\mathbb F}\times {\mathbb F}\to {\mathbb F}.$
$\times:{\mathbb F}\times {\mathbb F}\to {\mathbb F.}$ (We say that ${\mathbb F}$ is closed under addition and multiplication.)
$0,1\in{\mathbb F}.$
$0\ne1.$
Properties 1–9 of the Theorem from last lecture hold with elements of ${\mathbb F}$ in the place of complex numbers, ${}0$ in the place of $\hat0,$ and ${}1$ in the place of $\hat1.$

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4 Comments | 305: Abstract Algebra I | Tagged: equivalence relation, field | Permalink
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Answer by Andres Caicedo for If ZFC has a transitive model, does it have one of arbitrary size? May 5, 2013

In a comment to François's answer, I point out that the least $\alpha$ such that $L_\alpha$ is a model of $\mathsf{ZFC}$ is countable. In what follows, "model" means "transitive model of $\mathsf{ZFC}$." If $M$ is transitive, then $L^M=L_\beta\models\mathsf{ZFC}$ for $\beta=\mathsf{ORD}\cap M$, so the least height of a model is count […]

Andres Caicedo

Answer by Andres Caicedo for How additive is Lebesgue measure in ZF+AD ? December 7, 2010

Ricky: I think I see how to answer the problem under a stronger assumption. Rather than $\mathsf{AD}$, work in $$ {\sf AD}^+ + V=L({\mathcal P}({\mathbb R})). $$ This is a bit unsatisfying, since it is very possible the question can be answered assuming only $\mathsf{AD}$. In any case, $\mathsf{AD}^+$ is potentially harmless, since it may be that $\mathsf{AD […]

Andres Caicedo

Answer by Andres Caicedo for Basis theorem (due to Solovay?) May 1, 2013

The result is indeed Solovay's Basis Theorem. It is a consequence of Moschovakis's Coding Lemma, and sometimes it is referred to as (a version of) the reflection theorem (for example, in section 8 of Steel's Handbook article). Perhaps the optimal reference (it is self-contained, and easily accessible) is Section 2.4 of Peter Koellner, and W. H […]

Andres Caicedo

Answer by Andres Caicedo for Counterintuitive consequences of the Axiom of Determinacy? April 28, 2013

Let's see: $\mathsf{AD}$ implies that all sets of reals are Lebesgue measurable, have the Baire property, and the perfect set property (so, a version of the continuum hypothesis holds). It is conjectured that it also implies that all sets of reals are Ramsey. All these statements fail (badly) in the presence of choice. (There is a technical strengthenin […]

Andres Caicedo

Answer by Andres Caicedo for Models of Determinacy March 14, 2013

Certainly, $L(\mathbb R)$ is not a mouse (rather, a weasel) over a countable set, and the only way I see of thinking of it as a mouse and still capturing all the reals is making it a mouse over $\mathbb R$, in which case the answer is trivial. There is interesting work on $\mathbb R$-premice, but I do not think this is what you had in mind here. More relevan […]

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Math.StackExchange

Answer by Andres Caicedo for When does equality holds in $A\subseteq P(\cup A)$ May 17, 2013

If equality holds, then $A=\mathcal P(X)$ for some set $X$, namely $X=\bigcup A$. Conversely, note that if $A=\mathcal P(X)$, then $\bigcup A=X$. That is: $A=\mathcal P(\bigcup A)$ iff $A$ is the power set of some set $X$, in which case $X=\bigcup A$.

Andres Caicedo

Answer by Andres Caicedo for The relationship of ${\frak m+m=m}$ to AC May 16, 2013

Suppose that $A$ is infinite and Dedekind finite. Then $\mathfrak m=|A\cup\mathbb N|$ satisfies that $|A|

Andres Caicedo

Answer by Andres Caicedo for Growth-rate vs totality May 11, 2013

There is a hierarchy of fast-growing functions $f_\alpha:\mathbb N\to\mathbb N$ indexed by ordinals $\alpha

Andres Caicedo

Answer by Andres Caicedo for Model theory question with finiteness May 13, 2013

[Edit: I address first the original version of the question, where the requirement that $\Psi$ is transitive was not stated. The last paragraph addresses this case.] As stated, this is false, and it requires too many wrinkles to fix appropriately: (Assume that) $\mathsf{ZF}$ is consistent, and that $\Psi$ is a model of $\mathsf{ZF}+\lnot\mathrm{Con}(\mathsf{ […]

Andres Caicedo

Answer by Andres Caicedo for What are some good open problems about countable ordinals? May 3, 2013

There is still much to explore in the area of partition calculus. The notation $$\alpha\to(\beta,\gamma)^2$$ means that if one colors the $2$-element subsets of $\alpha$ with two colors, red and blue, then there is an $H\subseteq\alpha$, all of whose $2$-sized subsets have the same color and, either the color is red and $H$ has order type $\beta$, or the col […]

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