I refer to the textbook for the basic notion and properties of *vector spaces*. A *field* is a triple satisfying the axioms listed in pages 2, 3 of the textbook as properties of see also https://andrescaicedo.wordpress.com/2009/02/11/305-4-fields/ and surrounding lectures, in this blog. I am writing this note mostly to record the exercise, the question, and the statement of Steinitz lemma, so I am not recording the proofs we discussed in lecture.

The following example, that I want to leave as a (voluntary) exercise, is due to John Conway.

Exercise 1DefineNim-additionandNim-multiplicationon as follows:

is the result of adding without carrying the binary expansions of and . For example,is computed by applying the following rules:

is commutative.distributes overLetting we have and for

Show that is a field.

In lecture, the basic properties of vector spaces were presented. Recall the following notions:

Definition 1Let be a vector space over a field and let

Thespanof is the collection of all finite linear combinations of elements of The empty combination is understood as the vector We say that spans iff and that spans (or that is spanning) iff spansThe set isindependentiff for all vectors we have thatThe space isfinite dimensionaliff there is a finite spanning set Otherwise, is infinite dimensional.is abasisfor iff it is independent and spans.

The following is very useful; in many algebraic context, it is taken as a basic property that any decent *independence relation* must satisfy:

Lemma 2 (Steinitz exchange lemma)For any subset of a vector space and any vectors if

then

Corollary 3If a subset of a vector space is independent, and is such that then is independent.

Theorem 4Any vector space admits a basis.

Remark 1The result is true for all vector spaces, whether they are finite dimensional or not, but the argument in the infinite dimensional case involves Zorn’s lemma. In lecture, we only proved the finite dimensional case. The argument shows that any spanning set contains a basis. The proof we gave in lecture does not seem to adapt to the infinite dimensional case. In particular, in principle we could need to deal with the following pathological situation:

Question 1Can we have a vector space and a set of vectors such that the are pairwise distinct and yet

for all ?

(I encourage you to think about this question, although just as with the exercise, I am not requesting that you turn it in.)

Steinitz exchange lemma can be used to prove the following fundamental result, with the proof of which we will begin the next lecture.

Lemma 5If is a finite dimensional vector space, and are bases for then

Remark 2If is infinite dimensional, and are bases for then it is still the case that in the sense that there is a bijection between and

Definition 6In light of Lemma 5 and the subsequent remark, we can define thedimensionof a vector space as the size of any basis for

*Typeset using LaTeX2WP. Here is a printable version of this post.*

In the exercise, when we define the Fermat numbers, can n be 0, or is it strictly greater than 0.

Thanks,

Nick

Hi Nick.

can be 0 there, so for example.

I was also curious about one property about the nim-multiplication. Is defined by the above definition?

I guess we need to add that is associative.

Then and 🙂

I had erroneously claimed above that we obtain an algebraically closed field this way. This is incorrect. The definition is due to John Conway, who in fact defined , for all ordinals. He showed that the ordinal is the algebraic closure of the field of two elements. (And is a subfield of this, of course.)