The Geometry of Numbers

Minkowski’s linear forms theorem can be used to give sharper estimates for a positive real number m for which every ideal I contains an element a such that |N(a)| <= m*N(I). The estimate I used in my first proof is very big, and unwieldy for concrete computations of class groups.

First, a very easy and still pretty bad m is SQRT(|d(K)|). The ideal I has some integral basis x1, x2, …, x(n), whose matrix by definition has determinant d = N(I)*SQRT(|d(K)|). We can let t1, t2, …, t(n) be any numbers we like that satisfy the conditions of the linear forms theorem, say t(i) = d^(1/n) for all i. There exists integers a(j) not all zero such that w(i)(a1x1 + … + a(n)x(n)) <= t(i) for all i, so that the element they define has norm at most t1*t2*…*t(n) = N(I)*SQRT(d(K)).

Recall from the proof of the finiteness of the class group that this shows that every ideal class of K contains a representative with norm <= SQRT(|d(K)|).

For a more sophisticated bound, first write n as r + 2s, where r is the number of real conjugates of K and s is the number of pairs of proper complex conjugates (proper complex ring homomorphisms always come in complex conjugate pairs). Then label the ring homomorphisms w(i) in such a way that the first r homomorphisms are real, and the next 2s are not, with w(r+1) = w(r+s+1), w(r+2) = w(r+s+2), …, w(r+s) = w(r+2s).

Now, let f be a linear transformation from K = Q^n to R^n that sends a to (w1(a), w2(a), …, w(r)(a), Re(w(r+1)(a)), …, Re(w(r+s)(a)), Im(w(r+1)(a)), …, Im(w(r+s)(a))). If a complex number is equal to a + bi where a and b are real, then Re(a + bi) = a and Im(a + bi) = b. Note that the matrix corresponding to this transformation will have determinant with absolute value SQRT(|d(K)|)/2^s since the row operations that take v and v to Re(v) and Im(v) halve the determinant.

In R^n, let the region D consist of all points that in a way correspond to numbers in K with norm less than 1, i.e. points (x1, x2, …, x(n)) such that |x1|*|x2|*…*|x(r)|*|x(r+1)^2 + x(r+s+1)^2|*…*|x(r+s)^2 + x(r+2s)^2| <= 1. D is only convex if r = 1 and s = 0 or r = 0 and s = 1. Otherwise, we need to take some subregions of D.

If S is a convex region contained in D, and the volume of S is k, then we can get a bound of (2^(r+s)/k)*SQRT(|d(K)|). To see why, first let y1, y2, …, y(n) be an integral basis of the ideal I. The matrix corresponding to the basis has determinant of absolute value SQRT(|d(K)|)*N(I); the matrix whose (i, j)th entry is x(i)(y(j)), i.e. the transformation defined by x(i) applied to y(j), then has determinant of absolute value SQRT(|d(K)|)*N(I)/2^s.

Since the determinant is nonzero, the vectors f(y1), f(y2), …, f(y(n)) are linearly independent over R, so they define a lattice of volume SQRT(|d(K)|)*N(I)/2^s. Let u be the nth root of (2^(r+s)/k)*SQRT(|d(K)|)*N(J). Then the region uS has volume (u^n)k = (2^(r+s))*SQRT(|d(K)|)*N(J) = (2^n)*SQRT(|d(K)|)*N(I)/2^s.

This means that for every u‘ > u, there exists some nonzero point of the lattice defined by the f(y(i))’s, call it v, in u‘S. Since for a fixed u‘ there are finitely many such points, we can even find a point of the lattice in uS. That means uS contains a point of the form f(a) where a is in I. All points in S have norm at most 1, so |N(a)| <= u^n = (2^(r+s)/k)*SQRT(|d(K)|)*N(J).

Here I’m going to do a little bit of handwaving. A suitable region S that gives a good bound is the set of all points in R^n such that |x1| + |x2| + … + |x(r)| + 2*SQRT(x(r+1)^2 + x(r+s+1)^2) + … + 2*SQRT(x(r+s)^2 + x(r+2s)^2) <= n. S is fairly self-evidently convex and symmetric, and is contained in D because a theorem that says the geometric average of positive real numbers is at most their arithmetic average (i.e. SQRT(ab) <= (a+b)/2).

It’s possible to prove by induction on r and s that the volume of S is (2^r)*((pi/2)^s)*(n^n)/n!, so that we have a bound m = (4/pi)^s*n!/n^n. This complicated monster is called the Minkowski bound. Although deriving it is hard, it makes proving unique factorization in rings of integers a lot easier.

For example, take Q(SQRT(14)). The original m with respect to the integral basis {1, SQRT(14)} would give (1 + SQRT(14)^2 = 15 + 2*SQRT(14) = 22.48. With respect to the somewhat cleverer choice {1, 4-SQRT(14)}, we get exactly 11. The Minkowski bound gives SQRT(d(K))/2 = SQRT(14) = 3.74. As 2 = (4+SQRT(14))(4-SQRT(14)), the ideal 2 splits into two prime ideals. The ideal (3) is prime since if 3 divides (a + b*SQRT(14))(c + d*SQRT(14)) without dividing any of a, b, c, or d, we get that ac + 14bd and ad + bc are divisible by 3. Modulo 3, 1 and 2 both have the square 1, so ac = bd mod 3 implies abcd = 1 mod 3, and ad + bc = 0 mod 3 implies abcd = -1 mod 3, a contradiction.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: