Weekly 7: David Hilbert's 9th Problem

For my final weekly post, we are going to have some fun. I will be trying to interpret the meaning of David Hilbert's 9th problem the best I can. But, before I state the actual problem, I would like to build up the terminology and appropriate knowledge required to understand it. When I type in a different color, that just signifies my thoughts and opinions on the statement at hand. So first, observe the following table I took from Wikipedia. Here, f(n) = n^2 - 5 and the column after f(n) gives the prime factorizations of f(n).

http://en.wikipedia.org/wiki/Quadratic_reciprocity#Other_statements

Whats interesting about this table is that for the prime factorizations, other than the numbers 2,5, the prime numbers that appear as factors end in either a 1 or 9. There is a cool way to state this: the primes q for which there exists an n such that n^2 = 5 (mod q) are precisely 2, 5, and those primes q such that q = 1 (mod 5) (so q = 11, 31, 41 ...) or q = 4 (mod 5) (q = 19, 29, 89, ...). The law of quadratic reciprocity gives something similar like the example above of prime divisors of f(n) = n^2 − c for any integer c. So how about we state that law, with my thoughts in blue. Law of quadratic reciprocity:

Let p, q be distinct odd primes. Then:
(i) If p = 1 (mod 4) (p = 5, 13, 17, 29,...) or q ≡ 1 (mod 4) (q = 5, 13, 17, 29,...),

p is a square mod q (so p = n^2 (mod q), which is equivalent to saying n^2 = p (mod q)) if and only if q is a square mod p (so q = n^2 (mod p), or n^2 = q (mod p)).

(ii) If p = q = 3 (mod 4) (so p = q = 3, 7, 11, 19, 23, ...), p is a square mod q  if and only if  q is not a square mod p.

The statement of the theorem is almost directly from my first reference at the bottom. This theorem seems pretty confusing. So how about we explain it by using an example; naturally, we will use the example I stated at the beginning of the blog. In that example, p = 5, which means p = 1 (mod 4) so we use the (i) statement of the reciprocity law. Also, it was given in the example that n^2 = 5 (mod q) so we know p (which is 5)  is a square mod q. Then by the law of quadratic reciprocity part (i), we can see that q is a square mod 5, which means q = n^2 (mod 5). Now what does this mean? It means that q is a prime where q = 1^2 (mod 5), or q = 2^2 (mod 5). This means q = 1 (mod 5) (so q = 11, 31, 41, ...) or q = 4 (mod 5) (so q = 19, 29, 89, ...) and there you go, we used the reciprocity law to get the same results as in our example.

     But, your thinking, HOLD ON KYLE, what about q = 0 (mod 5), q = 9 (mod 5), q = 16 (mod 5), q = 25 (mod 5), and so on. Well first, if q = 0 (mod 5) then q = 0,5,10,15, ...; where none of those numbers will ever be prime because they are multiples of 5, so we can throw that case out. But, for q = 9 (mod 5), we have q = 9 + 5c = 1 + 5 + 5c = 1 + 5(1+c) so q = 1 (mod 5), but that would mean q = 9 (mod 5) is redundant. I could go on with other cases, but how about not. Recall in MTH 310 when we had congruence classes. I won't go to far in depth, because every math major has to take the class anyway, but it's known that if a = b (mod 5), then a is congruent to either: 0 (mod 5), 1 (mod 5), 2 (mod 5), 3 (mod 5), or 4 (mod 5), there is no need to go on because each case will just revert back to the 5 cases stated above, ie. redundancy. The 5 cases above are known as the congruence classes of Z (mod 5). But, now you are saying: Kyle, what if we hit a case in which q = n^2 (mod 5) where it reverts back to cases 2 (mod 5), or 3 (mod 5). That won't happen though. If you want to know why, refer to my daily 13. Now back to my explanation of the Law of quadratic reciprocity.

So we went through statement (i), so then what about (ii)? Well if you understood my explanation of (i), then surely you can understand (ii). So there you go, you have a very rough understanding of the law of reciprocity. There are actually many ways to state this law: Euler, Lagrange, and Gauss had some I found them more confusing than the way stated above. Gauss is the one who actually proved the law. Now let's switch gears and talk about something a bit different.

So from Math 310, we all remember rings: sets with 2 binary operations that satisfy basic properties, like the set of rational numbers with addition and multiplication. Then a field was a ring with a few more properties: multiplication was commutative, had an identity, and each element had an inverse. Again, the rational numbers are an example of a field. Now let A be a field and let B be a field that contains A and has the same operations as A. Then B would be a field extension of A. So then with the field Q (set of rationals), the set L = {a + b*sqrt(5), where a,b are rationals} would be a field extension of Q. Also, the term algebraic number field is any finite field extension of Q. 

So gathered with all this information, here is David Hilbert's 9th problem:

Find the most general law of the reciprocity theorem in any algebraic number field.

So this problem is like the law of reciprocity, except that we are working in a more general sense: not just rational numbers, but any field extension of the raitionals. Sadly, this problem has been partially. Luckily the guy supposedly only solved the law of reciprocity for abelian extensions (another confusing concept) of the rationals. So the non-abelian case is up for grabs. Now that you understand the problem you can go solve it, Good Luck!

References

PETE L. CLARK. QUADRATIC RECIPROCITY I, http://math.uga.edu/~pete/4400qrlaw.pdf

http://en.wikipedia.org/wiki/Hilbert's_problems

http://en.wikipedia.org/wiki/Quadratic_reciprocity