Abstract Nonsense

Crushing one theorem at a time

Halmos Chaper One, Section 1: Fields


1.

Problem: Amost all the laws of elementary arithmetic are consequences of the axioms defining a field. Prove, in particular, that if \mathfrak{F} is a field, and if \alpha,\beta and \gamma belong to \mathfrak{F}, then the following relations hold.

a) 0+\alpha=\alpha

b) If \alpha+\beta=\alpha+\gamma then \beta=\gamma

c) \alpha+\left(\beta-\alpha\right)=\beta

d) \alpha0=0\alpha=0

e) (-1)\alpha=-\alpha

f) (-\alpha)(-\beta)=\alpha\beta

g) If \alpha\beta=0 then either \alpha=0 or \beta=0

Proof:

a) By axiom 3 (A3) we know that \alpha+0=\alpha and by the commutativity described in A1 we conclude that 0+\alpha=\alpha+0=\alpha

b) We see that if \alpha+\beta=\alpha+\gamma then \left(\alpha+\beta\right)+-\alpha=\left(\alpha+\gamma\right)+-\alpha which by associativity and commutativity says that \gamma+(\alpha+-\alpha)=\beta+(\alpha+-\alpha) which then implies that \gamma=\gamma+0=\beta+0=\beta.

c) We use associativity and commutativity to rewrite our equations as \beta+(\alpha+-\alpha)=\beta+0=\beta

d) By commutativity of the multiplication it suffices to note that \alpha0=\alpha(0+0)=\alpha0+\alpha0 and thus \alpha0+-\alpha0=\left(\alpha0+\alpha0\right)+-\alpha0 and  by associativity we arrive at 0=\alpha0.

e) We merely note that \alpha+(-1)\alpha=(1+-1)\alpha=0 and thus -\alpha=(-1)\alpha.

f) We use e) to say that (-\alpha)(-\beta)=(-1)\alpha(-1)\beta=(-1)(-1)\alpha\beta. Then, we notice that (-1)(-1)+(-1)=(-1)(-1+1)=0 from where it follows that -(-1)(-1)=-1 and thus (-1)(-1)=1 and the conclusion follows.

g) Suppose that \alpha,\beta\ne0 then since \alpha\ne0 we see that \alpha\beta=0\implies \beta=0\alpha^{-1}=0 which contradicts our choice of b

2.

Problem:

a) Is the set of all positive integers a field?

b) What about the set of all integers?

c) Can the answers to both these question be changed by re-defining addition or multiplication (or both)?

Proof:

a) No, we merely note that there is no additive identity for 1

b) No, there is no multiplicative identity for 2

c) Yes. But first before we justify let us prove a lemma (which is useful),

Lemma: Let \left(\mathfrak{F},+,\cdot\right) be a field with \text{card }\mathfrak{F}=\mathfrak{n}. Then, given any set F with \text{card }F=\mathfrak{n} there are operations \oplus,\odot:F\times F\to F for which \left(F,\oplus,\odot\right) is a field.

Proof: By virtue of their equal cardinalities there exists some bijection \theta:F\to\mathfrak{F}. Then, for \alpha,\beta\in F define

\alpha\oplus\beta=\theta^{-1}\left(\theta(\alpha)+\theta(\beta)\right)

and

\alpha\odot\beta=\theta^{-1}\left(\theta(\alpha)\cdot\theta\left(\beta\right)\right)

We prove that with these operations \left(F,\oplus,\odot\right) is a field. We first note that \oplus,\odot:F\times F\to F and so they are legitimate binary operations. We now begin to show that all the field axioms are satisfied

1) Addition is commutative- This is clear since

\alpha\oplus\beta=\theta^{-1}\left(\theta(\alpha)+\theta(\beta)\right)=\theta^{-1}\left(\theta(\beta)+\theta(\alpha)\right)=\beta+\alpha

2) Addition is associative- This is also clear since

\alpha\oplus\left(\beta\oplus\gamma\right)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\beta\oplus\gamma\right)\right)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\theta^{-1}\left(\theta(\beta)+\theta(\gamma)\right)\right)\right)

which is equal to

\theta^{-1}\left(\theta(\alpha)+\left(\theta(\beta)+\theta(\gamma)\right)\right)=\theta^{-1}\left(\left(\theta(\alpha)+\theta(\beta)\right)+\theta(\gamma)\right)

which finally is equal to

\theta^{-1}\left(\theta^{-1}\left(\theta\left(\theta(\alpha)+\theta(\beta)\right)\right)+\theta(\gamma)\right)=\theta^{-1}\left(\theta\left(\alpha\oplus\beta\right)+\theta(\gamma)\right)=\left(\alpha\oplus\beta\right)\oplus\gamma

3) There exists a zero element- Let 0 be the zero element of \mathfrak{F} then \theta^{-1}(0) is clearly the zero element of F. To see this we note that

\alpha\oplus\theta^{-1}(0)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\theta^{-1}\left(0\right)\right)\right)=\theta^{-1}\left(\theta(\alpha)+0\right)=\theta^{-1}\left(\theta\left(\alpha\right)\right)=\alpha

for every \alpha\in F.

4) Existence of inverse element- If \alpha\in F we note that

\alpha\oplus\theta^{-1}\left(-\theta(\alpha)\right)=\theta^{-1}\left(\theta(\alpha)+\theta\left(\theta^{-1}\left(-\theta(\alpha)\right)\right)\right)

which equals

\theta^{-1}\left(\theta(\alpha)+-\theta(\alpha)\right)=\theta^{-1}(0)

which is the identity element of F

5-8 are the analogous axioms for multiplication, which are (for the most part) the exact same as the above.

9) Distributivity- We note that

\alpha\odot\left(\beta\oplus\gamma\right)=\theta^{-1}\left(\theta(\alpha)\cdot\theta\left(\beta\oplus\gamma\right)\right)

which equals

\theta^{-1}\left(\theta(\alpha)\cdot\theta\left(\theta^{-1}\left(\theta(\beta)+\theta(\gamma)\right)\right)\right)=\theta^{-1}\left(\theta(\alpha)\cdot\left(\theta(\beta)+\theta(\gamma)\right)\right)

from where the rest is obvious.

This completes the lemma \blacksquare

Now, we may answer the question. Since \mathbb{Q} is a field and \mathbb{Q}\cong\mathbb{N}\cong\mathbb{Z} the above lemma implies there exists addition and multiplications on \mathbb{N} and \mathbb{Z} which make them into fields.

3.

Problem: Let m\in\mathbb{N}-\{1\} and let \mathbb{Z}_m denote the integers \text{mod }m.

a) Prove this is a field precisely when m is prime

b) What is -1 in \mathbb{Z}_5?

c) What is \tfrac{1}{3} in \mathbb{Z}_7?

Proof:

a) We appeal to the well-known fact that ax=1\text{ mod }m is solvable precisely when (a,m)=1. From there we may immediately  disqualify non-primes since the number of multiplicatively invertible elements of \mathbb{Z}_m is \varphi(m) and \varphi(m)<m-1 when m is not a prime. When m is a prime the only thing worth noting is that every non-zero element of \mathbb{Z}_m has a multiplicative inverse. The actual work of showing the axioms hold is busy work, and I’ve done it before.

b) It’s clearly 4. Since 1+4=5=0

c) It’s 5. To see this we note that 5\cdot3=15=1

4.

Problem : Let \mathfrak{F} be a field and define

c:\mathbb{N}\to\mathfrak{F}:n\mapsto\underbrace{1+\cdots+1}_{n\text{ times}}

show that either there is no n such that c(n)=0 or that if there is, the smallest such n is prime

Proof: Assume that c^{-1}\left(\{0\}\right)\ne\varnothing and p=\min c^{-1}\left(\{0\}\right). Now, suppose that p=ab where 1<a,b<p. We see then that

c(a)c(b)=(\underbrace{1+\cdots+1}_{a\text{ times}})c(b)=\underbrace{c(b)+\cdots+c(b)}_{a\text{ times}}

which upon expansion equals

\underbrace{(\underbrace{1+\cdots+1}_{b\text{ times}})+\cdots+(\underbrace{1+\cdots1}_{b\text{ times}}}_{a\text{ times}})

which by associativity and grouping is equal to

\underbrace{1+\cdots+1}_{ab\text{ times}}=\underbrace{1+\cdots+1}_{p\text{ times}}=0

which by concatenation of the equations yields

c(a)c(b)=0

but since \mathfrak{F} is a field it follows that c(a)=0 or c(b)=0, either way the minimality of p is violated.

5.

Problem: Let \mathbb{Q}(\sqrt{2})=\left\{a+b\sqrt{2}:a,b\in\mathbb{Q}\right\}

a) Is \mathbb{Q}(\sqrt{2}) a field?

b) What if \alpha,\beta are required to be integers?

Proof:

a) This is a classic yet tedious exercise, I will not do it here.

b) No. For example, consider 1+\sqrt{2}. Then, we have that

\displaystyle \left(1+3\sqrt{2}\right)^{-1}=\frac{1-3\sqrt{2}}{1-18}=\frac{-1}{17}+\frac{3}{17}\sqrt{2}\notin\mathbb{Z}(\sqrt{2})

6.

Problem:

a) Doest the set of all polynomials with integer coefficients (\mathbb{Z}[x]) form a field?

b) What about \mathbb{R}[x]?

Proof:

a) No.

b) No. I’ll let you figure these out (it’s really easy)

7.

Problem:

Let \mathfrak{F} be the set of all ordered pairs (a,b) of real numbers

a) Is \mathfrak{F} a field if addition and multiplication are done coordinate wise?

b) If addition and multiplication are done as one multiplies complex numbers?

Proof:

a) No. Consider that (0,1) is a not the additive identity but it has no multiplicative inverse.

b) Yes, this is just field isomorphic to \mathbb{C}

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September 21, 2010 - Posted by | Fun Problems, Halmos, Munkres, Topology | , , , , , ,

4 Comments »

  1. OMG! I’ve been dying to read this book for so long but
    never allowed myself due to the lack of solutions! It’s
    not exactly an easy text, as your solutions show, but
    wow I can tell you these are appreciated! Once I
    finish Axler & Shilov I think I will give this
    book my best! 😀 Thank you! Will you be adding the
    answers for parts 2,3 & 4?

    Comment by Gerard | October 19, 2010 | Reply

    • Gerard,

      If you mean sections 2,3, and 4, I did them but apparently didn’t tag them as Halmos solutions, if you didn’t see them in the “Halmos Category”. I just did though, so if you click again you shall surely see them.

      That said…I’m glad you enjoy them!! I agree, this isn’t an easy book. I am concurrently studying out of A Second Course in Linear Algebra by Brown, and that’s what I call a hard book.

      A word to the wise about my solutions.

      a) Sometimes I omit some which are messy but simple, and in doing so renumber the problems.

      b) Change the notation to fit my own needs/preferences/habits

      c) Do things in a global perspective, possibly appealing to theorems which aren’t strictly in the text.

      That said, I feel as though my solutions can be helpful to you if you do choose to read the book. 🙂

      If you have any questions whatsoever, feel free to contact me on here! Also, in case you weren’t aware I am also in the process of doing Munkres, Rudin, and a book on Category Theory..if you are at all interested in any of those.

      Comment by drexel28 | October 19, 2010 | Reply

  2. Well Rudin is of course on my list coming up in a few months,
    I must admit having your solutions handy would be really
    helpful for me to compare my own against when that time comes, albeit a few weeks after they’re first written by me 😉
    Of course nobody like to have the need for a solutions manual
    but when you’re beginning abstract mathematics without any
    teacher to guide you they are simply a necessity so I’m just
    going to thank you for offering this to me, particularly
    with Halmos as I’ve read a few chapters of his set theory
    book & found it extremely clear & Interesting.

    Comment by Gerard | October 19, 2010 | Reply

  3. You’re entirely welcome, and I understand. No one, I don’t care who you are doesn’t get stumped by a problem every now and then. That said, the problems which stump me may seem easy to your and vice versa. Thus, it makes sense to some kind of extra-personal reference.

    Comment by drexel28 | October 19, 2010 | Reply


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