Algebraic closure of the field of two elements

I think I learned about this from a comment on MathOverflow.

Recall that the field of two elements \mathbf F_2 is the ring \mathbf Z/2\mathbf Z of integers modulo 2. In other words, it consists of the elements 0 and 1 with addition 1+1 = 0 and the obvious multiplication. Clearly every nonzero element is invertible, so \mathbf F_2 is a field.

Lemma. The field \mathbf F_2 is algebraically closed.

Proof. We need to show that every non-constant polynomial f \in \mathbf F_2[x] has a root. Suppose f does not have a root, so that f(0) \neq 0 and f(1) \neq 0. Then f(0) = f(1) = 1, so f is the constant polynomial 1. This contradicts the assumption that f is non-constant. \qedsymbol

Number theory is heavy metal

As some of you may be aware, I am a musician as well as a mathematician. I often like to compare my experiences between the two. For example, I found that my approach to the creative process is not dissimilar (in both, I work on the more technical side, with a particular interest in the larger structure), and I face the same problems of excessive perfectionism (never able to finish anything).

This post is about some observations about the material itself, as opposed to my interaction with it.

Universal donor.

A look at the arXiv listing for algebraic geometry shows the breadth of the subject. Ideas from algebraic geometry find their way into many different areas of mathematics; from representation theory and abstract algebra to combinatorics and from number theory to mathematical physics. But when I say algebraic geometry is a universal donor (of ideas, techniques, etc.), I also mean that its applications to other fields far outnumber the applications of other fields to algebraic geometry, and that the field of algebraic geometry is largely self-contained.

Much the same role is played by classical music inside musical composition. Common practise theory is used throughout Western music, whether you’re listening to hip hop, trance, blues, ambient electronic, bluegrass, or hard rock. Conversely, the influence of popular genres music on [contemporary] classical is comparatively little. One could therefore argue that classical music is a universal donor in the field of musical composition.

Universal receptor.

The opposite is the case for number theory. Another vast area, number theory often uses ingenious arguments combining ideas from algebra, combinatorics, analysis, geometry, and many other areas. In practical terms, the amount of material that a number theorist needs to master is immense: whatever solves your particular problem.

So is there any musical genre that plays a similar role? I claim that metal fits the bill. With a vast list of subgenres including thrash metal, black metal, doom metal, progressive metal, death metal, symphonic metal, nu metal, grindcore, hair metal, power metal, and deathcore, metal writing often contains creative combinations of other genres, from classical music to ambient electronic, and from free jazz to hip hop. As Steven Wilson of Porcupine Tree said about his rediscovery of metal:

I said to myself, this is where all the interesting musicians are working! Because for a long time I couldn’t find where all these creative musicians were going… You know in the 70’s they had a lot of creative musicians like Carlos Santana, Jimmy Page, Frank Zappa, Neil Young, I was thinking “where are all these people now?” and I found them, they were working in extreme metal.

I sometimes think of metal as a small microcosmos reflecting the full range of Western (and some non-Western) music, tied together by distorted guitars and fast, technical drumming.

In summary, algebraic geometry is classical music, and number theory is heavy metal.

Furstenberg’s proof of the infinitude of the primes

Another very classical proof that’s just too beautiful to ignore. This time, the theorem is millennia old, but it’s really the proof that I’m interested in.

Lemma. There are infinitely many primes.

Proof. Define the sets

    \[ U_{a,n} = a + n\Z = \{x \in \Z\ |\ x \equiv a \pmod{n}\}, \]

for a, n \in \Z with n \neq 0. Note that if U_{a,n} \cap U_{b,m} \neq \varnothing; say it contains some c \in \Z, then

    \[ U_{a,n} \cap U_{b,m} = U_{c, \lcm (n,m)}. \]

Hence, the intersection of two sets of this form is again of this form (or empty). Thus, they form the basis for a topology \scr{T} on \Z. Notice that the sets

    \[ U_{a,n}\comp = \bigcup_{b \not \equiv a} U_{b,n} \]

are also open.

Now suppose that there were finitely many primes. Then the intersection

    \[ V = \bigcap_{p \text{ prime}} U_{0,p}\comp \]

is a finite intersection of opens, hence open. On the other hand, it equals the set of integers divisible by no prime number, which is \{1, -1\}. But all basic open sets are infinite, so this set can never be open. \qedsymbol

Remark. This proof is in essence the usual proof, where we consider p_1 \cdots p_n + 1 and conclude that some prime has to divide it and this prime can be none of the p_i. Indeed, when we showed that the intersection of basic opens is open, we went to a set whose period is \lcm(n,m) (which for V gives period p_1 \cdots p_n). Since 1 is in V, so should p_1 \cdots p_n + 1 be.

That said, it doesn’t seem possible to adopt this proof to reprove other theorems like ‘there are infinitely many primes congruent to 3 modulo 4’. The problem is that the above proof seems to rely on a global analysis of the set V: it has to be an infinite set. So the method is too crude to prove theorems with congruence restrictions.