As the name suggests, algebraic number theory employs modern algebraic techniques to solve problems in number theory. We are therefore interested in integral extensions of the integers Z, and corresponding field extensions of the rationals Q. However, other extensions will prove useful as well.

If you go all the way back to chapter 5, you may recall that we used the gaussian integers to prove there is no integer solution to x4 + y4 = z4. With more complex structures in place, we may be able to answer more questions in number theory.

A global field is a finite separable extension of Q or of Z/p)(t). We are extending the rationals, or the quotients of polynomials in t over the integers mod p.

A number field is a finite extension of Q. Since Q has characteristic 0, extensions of Q are automatically separable. Thus a number field is a global field. In contrast, (Z/p)(t) supports inseparable extensions, such as the pth root of t. We want to avoid those; hence a global field is declared separable.

Within the context of a global field, an algebraic number is algebraic over the base field, either Q or (Z/p)(t). Of course every element in a finite field extension is algebraic, so everything in a global field qualifies as an algebraic number. In contrast, an algebraic integer is integral over the base ring Z or (Z/p)[t]. For example, sqrt(2) is an algebraic integer, while 1/3 is not. The former satisfies the monic integer polynomial x2 - 2, while the latter cannot be integral over Z, since Z is integrally closed.

This section equates the index of a principal ideal in a ring, i.e. the size of the quotient ring, to the index of a sublattice inside a larger lattice, and then to the norm of the element that generates the ideal in its ring. This assumes some familiarity with the concept of a lattice.

Let K be the fraction field of R. Assume the vector space Kn is also a field E. In other words, E/K is a finite field extension.

Let b1 through bn be a basis for a free R module S inside E. Now S is a regular R lattice of E. Furthermore, assume S is closed under multiplication. In other words, S is a subring of E.

Let w be an element of S, which generates a principal ideal H in S. Since H is an R submodule, it becomes a sublattice inside S. Note that H contains wb1 through wbn, linearly independent vectors in E. Therefore H is an n dimensional lattice, and a valid lattice for Kn. In fact wb1 through wbn forms a basis for H, as a free R module and an R lattice of Kn.

What is the index of H in S? A linear transformation carries each bi to wbi, and carries S onto H. Build a matrix M, where the ith row is the representation of wbi, as coefficients on b1 through bn. Thus the entries of M lie in R. The determinant of M gives the index of H in S. In addition, M implements multiplication by w. Its determinant is therefore the norm of w in E/K. The norm of w is the index of the sublattice H in the lattice S. At first this seems counterintuitive, since |w| does not depend on S, but remember that |w| is the index of w*S inside S.

Look again at the quotient ring S/H, this time with R = Z. The elements of S are linear combinations of b1 through bn with integer coefficients. These are lattice points in n space. The sublattice H is spanned by vectors v1 through vn, where each vi is a linear combination of b1 through bn. It is often convenient to think of S as the whole space, and b1 through bn as unit vectors spanning that space, and v1 through vn spanning a sublattice H in that space. Nothing vital is lost under this change in perspective.

Remember that our sublattice H is generated by w. For each i, vi is wbi, represented as coefficients on b1 through bn. We have n linearly independent vectors living in an integer lattice, and defining a sublattice. Start with any point x in S and shift it by a multiple of v1, then by a multiple of v2, and so on through vn, until x lies in the base cell. Each time x is translated by something in H, which does not change the coset of x. Thus everything in S/H is represented by a lattice point in the base cell, i.e. the parallelotope spanned by v1 through vn. Conversely, suppose two points in the base cell represent the same coset of H in R. There difference, a vector strictly inside the base cell, would have to be part of the lattice, and that is a contradiction. Therefore lattice points in the base cell correspond to the elements of S/H.

Since each lattice point represents one unit of volume, the number of lattice points in the base cell equals the volume of that cell, which is the covolume of the sublattice. Here is a more precise proof. Covolume is the determinant of v1 through vn, which is the determinant of wb1 through wbn, which is the norm of w. Put this all together and the norm of w equals the cardinality of S/H, or the index of H in S. As a corollary, every principal ideal in S has a finite index.

Remember that the norm of w is the product of its conjugates, raised to a power according to the dimension of E over K(w). If w generates E, and if E/K is separable, then the index of H in S is the norm of w, or the product of the conjugates of w. We don't require E/K to be a normal extension; just make sure you include all possible conjugates of w, whether they lie in E or not.

Technically the index is the absolute value of the norm. After all, the norm, or the determinant, could come out negative, and the cardinality of S/H is obviously positive. Since the conjugates lie in the complex plane, we are taking the absolute value of a finite product of complex numbers. This in turn is the product of the absolute values of the conjugates of w in the complex plane.

If E goes beyond K(w), and if E/K is separable, the norm of w is the product of the conjugates, raised to the j power, where j is the dimension of E over K(w). Now consider all possible embeddings of S, or E, into the complex plane. These map K(w) into the complex plane, and then E into the complex plane. All the conjugates of w appear in the embeddings of K(w), and they appear again and again, for each embedding of E over K(w). Thus the norm of w is the product of the images of w under all the embeddings of E into the complex plane, and after taking absolute values, the index of H in S is the product of the absolute values of the images of w in the complex plane, under all possible embeddings of E.

Now consider a more exotic ring. Let R = (Z/p)[t] and let K be its fraction field (Z/p)(t). As before, S is a lattice in Kn. Let w generate a principal ideal H in S. The norm of w is the determinant of the matrix that implements multiplication by w. This is a polynomial in t with coefficients in Z/p. Take the degree of this polynomial and raise p to that power. Equivalently, take the index of this polynomial in R. Either way you have a power of p, and thanks to a rather technical theorem in lattices over polynomials, this agrees with the index of H in S. Like number fields, the index of H in S, i.e. the size of S/H, equals the index of the norm of w in R.

The definition of a global field requires E/K to be separable. With this in mind, the product of the conjugates of w, raised to a power determined by the dimension of E over K(w), equals the norm of w. We would like to relate this to the images of w under various embeddings of E, as we did with number fields, but we need something analogous to the complex plane. Let C be the algebraic closure of K, or, if you prefer, let C be the algebraic closure of the completion of K, using the valuation induced by R. Now C has a valuation that is consistent with R. The sum of the valuations gives the valuation of the product, and so, the index of H in S is p raised to the sum of the valuations of the images of w in C, for all possible embeddings of E in C. If you didn't understand this part that's ok; the earlier result on number fields embedded in the complex plane is the primary result.

A global field is a finite separable extension, and the base ring, either Z or (Z/p)[t], is a pid. Apply a general theorem describing the ring of integers over a dedekind domain, and the algebraic integers become dedekind. In addition, the algebraic integers form a free R module of rank n, where R is the base ring. We can describe the integral ring as the span of a basis b1 through bn over R.

Show every global field has a finite class group in 3 steps.

  1. Every ideal has finite index. This is true for Z and (Z/p)[t]. In the integers, there is an ideal for each positive integer, which acts as generator for that principal ideal. If an ideal is generated by g then the quotient ring has index g, represented by the integers 0 through g-1, also known as mod g. Turning to (Z/p)[t], which is also a pid, there is an ideal for each monic polynomial. The index is p raised to the degree of that polynomial, which is finite. This finite index property propagates upward to the integral ring.

  2. There are finitely many ideals with index below c, for any integer c. In the integers, there are c-1 generators g below c. Similarly, there are finitely many monic polynomials having degree less then log base p of c. This property propagates upward to the integral ring.

  3. Every ideal H in the integral ring has an x such that the index of x in H is no larger than a fixed integer l. This becomes a sufficient condition for a finite class group. We'll prove this for Z, and then for (Z/p)[t].

Let a number field E/Q have dimension n, and let S be the algebraic integers therein. Since Z is a pid, S is a free Z module. Give S a basis b1 through bn.

Let H be an ideal in S. Choose m so that mn ≤ |S/H| < (m+1)n. This can be done because H has finite index in S, as shown in step 1.

Consider the set of elements in S spanned by b1 through bn with coefficients between 0 and m inclusive. Its cardinality exceeds the number of cosets of H, so two elements lie in the same coset. Their difference x is a nonzero element in H, with coefficients ≤ m in absolute value.

As shown in the previous section, the index of x in S equals its norm, and the norm of x is the product of the absolute values of the images of x under all possible embeddings. Let c() be an embedding of S in the complex plane. In fact there are n embeddings, c1(S) through cn(S). Let c be any one of these embeddings.

The absolute value of a sum of vectors in the complex plane is at most the sum of the absolute values of the individual vectors. Thus the absolute value of c(x) is at most m times the sum of abs(c(bj)).

Taking the product over all embeddings pulls out n factors of m, giving mn. What remains is a product over all embeddings, of the sum of the absolute values of the images of b1 through bn. Call this expression l. Now |x| ≤ mn×l. Put this all together.

|H| ≥ mn { that's how m was selected }
|H| × l ≥ mn × l
|x| ≤ mn × l
|x| ≤ |H| × l
|x| / |H| ≤ l
index of x in H ≤ l

Let's look at a fun example, adjoining the square root of -5. (We'll prove this is dedekind in the next chapter.) Use 1 and sqrt(5)i as a basis. The embeddings are the identity map and complex conjugation. Apply the first embedding and find the sum of the absolute values, which is |1|+|sqrt(5)i|, or 1+sqrt(5). The other embedding produces the same sum. Take the square of 1+sqrt(5) and get approximately 10.472, which is our value of l.

It may seem strange to think of the index of x in H as a decimal, like 10.472. After all, there are an integer number of cosets of x in H. However, the value of l, derived from images of the basis elements in the complex plane, acts as an upper bound. There are 10 or fewer cosets of x in H.

When looking for the generators of this finite class group, (which is nontrivial since the ring is not a ufd), one can restrict attention to primes with index bounded by l, primes with index < 11. Such a prime Q in S, with index < 11, contracts to a prime P in Z, and S/Q is at least as large as Z/P. Thus we can restrict attention to the primes in S lying over 2, 3, 5, and 7.

When b1 and b2 map to two vectors in the complex plane, as in the above example, l can be improved. These vectors, I'll call them u and v, span a lattice in the plane. Let the coefficients range from -m to m and build a parallelogram in the plane. Its farthest corner is m times the distance to u+v, or u-v if the angle uv exceeds 90 degrees. In our example the angle is precisely 90 degrees for both embeddings, the shape is a rectangle, and the corner is m times sqrt(6) from the origin. Thus l = sqrt(6) × sqrt(6), or 6. We only need look at the primes over 2 3 and 5. Let's give that a whirl.

Let w be the square root of -5. There are two primes over 3, Q1 generated by 3 and 2+w, and Q2 generated by 3 and 2-w. These are nonprincipal, thus nontrivial in the class group. The product Q1Q2 is generated by four different multiples of 3, and spans 3 via (2+w)×(2-w) - 3×(2+w) - 3×(2-w) = 3, thus giving the principal ideal generated by 3. Q1 and Q2 are inverses in the class group. On the other hand, multiply Q1 by the fractional ideal generated by (2-w)/3, giving Q2. Thus Q1 and Q2 are equal, and represent a nontrivial involution in the group.

5 is ramified, with w generating the prime over {5}, but that is principal, so we can move on to 2. The prime over 2 is Qr, generated by 2 and 1+w. This too is ramified, with Qr2 = {2}. Multiply Q1 by Qr and get the generators 6, 3±3w, and 4+2w. The product ideal has index 6, which is within our bound of 6, so it could represent a new member of the class group; we need to check. The product includes 1-w, which generates all four generators. The result is principal, hence Qr = Q1. We have characterized our first nontrivial class group: Q adjoin the square root of -5 yields Z/2.

One wonders if there is a number field having a class group G for any given finite abelian group G.

A finite separable extension of (Z/p)(t) also admits a finite class group, but that is not addressed here.

Let S be the integral ring in a number field E/Q. Remember that S can be expressed as a free Z module of rank n. Build a basis for S consisting of b1 through bn. The embeddings of E into the algebraic closure of Q define, and are defined by, the images of the basis. Such an embedding can be followed by conjugation in the complex plane, giving another embedding. These embeddings coincide iff the image of each bi is real. Thus the embeddings can be partitioned into r1 real embeddings and r2 pairs of conjugate embeddings in the complex plane.

Follow a primitive element in E/Q, and the number of embeddings equals the dimension of E over Q, hence r1+2r2 = n.

The number of embeddings of E into real space does not depend on the basis. Thus both r1 and r2 are basis invariant.

Let M be the matrix that runs basis elements through their images in the complex plane. Thus Mi,j = ci(bj) for the embeddings ci(S). Multiply M by a column vector of integers on the right to get a lattice. The first coordinate takes on the values of S while the remaining coordinates track the images of S under the various embeddings.

The matrix M implements a transformation from the rectilinear lattice in Rn onto a lattice that lives in n dimensional complex space. Of course the first r1 dimensions may as well be real, since the first r1 rows of M are real. But the remaining 2r2 rows are complex.

Let u be the determinant of this matrix. This gives the covolume of the lattice, or the volume of the base cell. Of course the lattice is an n dimensional Z module, living in a space that has n + 2r2 real dimensions. This because each complex dimension is two real dimensions put together. Let's see if we can pull this back down to Rn.

Build a new matrix N, with real entries, as follows. For each pair of complex rows in M, add the second row to the first, then subtract half the first from the second. This separates the real and imaginary components of the two rows, and does not change the determinant. Divide the purely imaginary rows by i, and cut the real rows (from the complex pairs) in half. This gives the real and imaginary components of the complex embeddings as separate real vectors in Rn. M is the "standard" embedding of S into complex space, and N is the "component" embedding of S into real space.

Let v be det(N) in real space, the determinant of the component matrix. Ignoring the factors of i, v is u divided by 2 to the r2.

Remember that the field homomorphisms from E into the algebraic closure of Q are linearly independent. M is nonsingular, with a nonzero determinant u. Divide by a power of 2 and N has a nonzero determinant v, and is nonsingular as well. The revised matrix N that maps the rectilinear lattice into real space produces an n dimensional lattice in Rn.

An ideal H is a sublattice of S, and maps (via N) to a sublattice in real space having covolume v×|H|.

If H is principal, generated by x, x*b1 through x*bn form a linearly independent set in S. If they didn't, we could cancel x in the integral domain, and b1 through bn would not be independent. Thus H is an n dimensional sublattice of S, and so is its image through N. If H is not principal it contains a principal ideal, and the same result holds. Each ideal is an n dimensional sublattice of S, mapping to an n dimensional sublattice in the range of N.

Let H be a fractional ideal of S, so that d*H lies in S. Build a superlattice around the integer lattice, using the multiples of 1/d as coefficients on the basis. Apply our transformation and find a superlattice in Rn containing the image of S. One can give this a covolume of v/d. Let H have an index of j inside our superlattice. Relative to the original lattice S, H has an index of j/d, and its image has a covolume of vj/d. The covolume is a rational number, but that just means we need a finer lattice to contain both H and S.

A subring of S is also a lattice, with a certain covolume. If the subring contains n independent vectors, it is an n dimensional sublattice with a nonzero covolume, and the same holds after passing through N.

This theorem, which is the foundation for much of algebraic number theory, is due to Minkowski. It places certain constraints on discriminants and class groups of number fields. Needless to say, it's clever!

Let S be the ring of integers in a number field E/Q, and let H be an ideal in S. Embed S into complex space, and real space, using the matrices M and N described in the previous section. Represent x using the basis of S, and write it as a column vector on the right of M, i.e. M*x. The result is a column vector, the n images of x. The norm of x is the product of these images, which is related to the sum of these images. We can place a bound on the sum by keeping M*x inside a certain shape in complex space. I'll describe that shape first, then use it to constrain x in H, and build a bound l, as we did in the previous section. That in turn constrains the class group.

Build a generalized octahedron W in real and complex space as follows. The space that contains W has r1 real dimensions and r2 complex dimensions. Only one of each pair of conjugate rows of M is used. After all, the distance to a point p in the complex plane is equal to the distance to p. We'll just count the distance to p twice. Thus the sum of the absolute values of the real coordinates, plus twice the sum of the absolute values of the complex coordinates, must be less than t. We are interested in the volume of such a shape. This is an exercise in multiple integration.

To start, volume scales as tn, so we can set t = 1, and multiply by tn later.

Concentrate on 1 section, with all coordinates positive, and multiply by 2n later.

Run through the real coordinates first. The coordinates are bounded below by 0, and above by the plane wherein the coordinates sum to 1. The shape is a simplex: a segment in 1 dimension, a right triangle in 2 dimensions, a right tetrahedron in 3 dimensions, and so on. In 1 dimension the volume is 1. In 2 dimensions the right triangle runs from 0 to 1 along the x and y axes, with a hypotenuse of x + y = 1, and has an area of 1/2. In 3 dimensions, integrate as z runs from 1 to 0. With u set to 1-z, the cross section is a right triangle with legs of length u. Area is u2/2, and as you integrate from 0 to 1, the volume is 1/6. Moving to 4 dimensions, the integrand is u3/6, and the resulting volume is 1/24. By induction, the volume of a unit right simplex in k dimensions is 1/k!. Scale the shape by u and the volume is uk/k!.

Move to the first complex coordinate and apply a similar inductive argument to the quarter circle in the complex plane with radius ½. In polar coordinates, the integrand becomes r × (1-2r)k/k!. Integrate by parts, wherein (1-2r)k/k! is the derivative of -½(1-2r)k+1/(k+1)!. Let r run from 0 to ½, and let θ run from 0 to π/2, and get π/8 × 1/(k+2)!.

Do this for each of the r2 complex coordinates and the volume of our region is (π/8)r2/n!. Multiply this by (2t)n to get the volume of W.

View the same space as n dimensional real space, and notice that the matrix is now equal to N. The lattice N*S and the octahedron are superimposed.

Let v be the determinant of N, as it was in the previous section. This is the covolume of the image of S. Let |H| be the index of H in S. Thus the covolume of the image of H is v×|H|.

Set t to the nth root of n!×(8/π)r2×v×|H|. W has a volume of 2n×v×|H|.

Since W is closed, convex, and symmetric, apply the point lattice theorem. Some nonzero x in H has its image in W. The sum of the absolute values of the conjugates of x is bounded by t. Their average is bounded by t/n.

The arithmetic mean bounds the geometric mean, hence the product of these norms, which is the norm of x, is bounded by the arithmetic mean raised to the n, or (t/n)n. Substitute for t and get this.

|x| ≤ n!/nn × (8/π)r2 × v × |H|

Everything on the right hand side, other than |H|, is going to play the role of l, which places a bound on the class group.

l = n!/nn × (8/π)r2 × v

Remember the relationship between the matrices M and N, and their determinants u and v. v is u divided by 2r2.

l = n!/nn × (4/π)r2 × u

M*S is a lattice in complex space, with covolume u, and discriminant u2. Let d be the discriminant, u2. d is also the determinant of M transpose times M. At the same time, MT*M is a matrix whose i,j entry is the trace of bibj. Each trace is a real number, MT*M is a real matrix, and d is real. Since discriminant is well defined up to the square of a unit, we may as well call d positive.

l = n!/nn × (4/π)r2 × sqrt(d)

In fact d is a positive integer. A typical entry y = bibj lies in S and is integral over Z. With Z integrally closed, the minimum polynomial of y has integer coefficients, and proves y is integral. All the conjugates of y satisfy this polynomial and are integral. The trace of y is integral over Z, and lies in the base field Q, and since Z is integrally closed, the trace is an integer. The matrix consists of integers, and the discriminant is an integer.

You might be thinking, "Gee, maybe if I pick a different basis I could find a better value of l." Not so - d is basis invariant, and strictly a function of S. Let a1 through an be an alternate basis for S as a Z module. Let Q be the matrix that performs the linear transformation. In other words, the row vector b1 through bn times Q gives the row vector a1 through an. Since an inverse matrix performs the reverse transformation, and both matrices have integer entries, the determinant of Q is ±1. Let ci be one of the automorphisms of E into its normal closure. The ith row of M is ci applied to the basis b. Since Ci fixes Q, this row can be multiplied by Q, or you can multiply b by Q and then apply ci; the result is the same. Thus M*Q is a matrix wherein the ith row is ci applied to the basis a1 through an. Multiply the transpose of MQ by MQ and take the determinant to find the discriminant relative to the new basis. Since the determinant of Q is ±1, the result is the same as det(MTM), which is our original discriminant. The discriminant is basis invariant, and only depends on S. Since S is a function of E, each number field has a well defined discriminant.

Return to the example of Z adjoin sqrt(-5). In an earlier section we found l = 10.472, and then wrangled it down to 6. Let's see what happens here. Use the standard basis of 1 and sqrt(-5). Build the 2×2 matrix of trace of bibj. The antidiagonal products are pure imaginary, and their trace is 0. The diagonal products are 1 and -5, and these are doubled for the trace. The determinant, or discriminant, is -20. In this extension n = 2 and r2 = 1. Plug and chug, and l = 2.847. We know that this extension is not a ufd. It has a nontrivial class group. A generator for this group must have index 2, and must lie over 2. We saw earlier that there is one ramified prime over 2, and its square is a principal ideal. Therefore the class group is Z/2.

Consider Z adjoin sqrt(2). The discriminant is 8. With n = 2 and r2 = 0, l = 1.414. A proper ideal cannot have index less than 2, hence this extension is a pid.

Adjoin sqrt(3) and get l = 1.732, another pid.

Adjoin (1+sqrt(5))/2 and get l = 1.118, another pid.

Let the integral ring S act as its own ideal. In other words, H = S. Select any nonzero x in S and its norm is an integer. The conjugates of x are all nonzero, hence |x| is at least 1. If the discriminant d is too low, there is some x with |x| < 1, and that is impossible. Set |x| = 1 and solve for d, to see how small d can be.

sqrt(d) ≥ nn/n! × (π/4)r2

If n = 1 the extension is trivial. The basis is 1, the trace is 1, and the discriminant is 1. Assume n > 1.

Pair up i with n-i in n!. The product of these two factors is the square of their geometric mean. This only increases if we use the square of the arithmetic mean, giving (n/2)2. Rais this to the (n-1)/2 or (n-2)/2 for the number of pairs. Then there's another n at the end, and if n is even, there's an n/2 in the middle. The denominator n! is at most nn/2n-1.

sqrt(d) ≥ 2n-1 × (π/4)r2

sqrt(d) ≥ 2r1+2r2 × (π/4)r2 / 2

sqrt(d) ≥ 2r1 × πr2 / 2

Since π is less than 4, this is smallest when the embeddings are all complex. Thus d is at least (π/2)2, or 2.467. Since d is an integer, the discriminant is at least 3.

By adjoining (1+sqrt(-3))/2, the discriminant can indeed be 3.

If T is an integral domain, and a free Z module of rank n, (thus integral over Z), let E be its fraction field, and extend T up to S, the integral ring. Since T is a subring of S it creates a sublattice, with an even larger covolume. This translates to a larger discriminant. Thus the discriminant of T is at least 3, and at least 12 if T is a proper subring of S, i.e. if T is not integrally closed. For example, adjoin sqrt(5) to Z. This is a subring of the pid that was described earlier. Its discriminant is 20, whereas the discriminant of the pid, produced by adjoining (1+sqrt(5))/2, is 5.

To review, the previous theorem gives us the following inequality.

sqrt(d) ≥ 2r1 × πr2 / 2

This holds for rings of integers, and for subrings thereof, which have even larger discriminants.

As n steps through 1 2 3 etc, the right side is at least 1, 2, π, 2π, π2, 2π2, etc. For a given d, there is a bound on n, hence a bound on the dimension of E/Q. Fix the values of n, r1, and r2, and ask how many number fields satisfy these constraints. If the answer is finite, then we are halfway there.

Let S/Z be an integrally closed extension with discriminant d. Map S into real and complex space as usual. Build the octahedron W that was described in the previous section. W depends only on d, and not on S.

Assume there is at least one real coordinate, i.e. one real embedding of S. Stretch W along this axis, while shrinking it in all other dimensions, so that the volume remains constant. The other dimensions will drop from t down to ½. The preferred axis expands from t to (2t)n/2.

The shape still contains a nonzero lattice point, which is the image of some nonzero x in S.

Remember that the norm of x is at least 1. The coordinates of the image of x track the conjugates of x, and all of them are smaller than 1, in absolute value, except for one conjugate. Thus one conjugate has a norm that is strictly larger than all the others.

Suppose x generates an intermediate field F inside E/Q. There are then E/F copies of each conjugate of x in the image of x. In other words, the embeddings of x are repeated as necessary, as the rest of E is mapped into the algebraic closure of Q. This contradicts the fact that precisely one coordinate is larger than the others. Therefore E is Q adjoin x. This is a geometric version of the primitive element theorem.

What if E/Q has no real embeddings? Expand the imaginary component of a complex pair, at the expense of the other coordinates. Two conjugates of x have absolute value larger than all the others, but they are complex conjugates, and cannot be equal. In fact they are at opposite ends of the long octahedron. Reason as above to show x generates E.

Let p be the minimum polynomial of x. This is irreducible over Q, of degree n. The absolute values of the roots of p, which are the conjugates of x, are all bounded. This means their product is bounded, as is their sum. These are the final and second coefficients on p, respectively. The other coefficients are also bounded. Since coefficients are integers, there are finitely many polynomials to choose from, and finitely many field extensions E/Q.

Since integral rings and field extensions correspond, there are finitely many integral rings for each discriminant.

What about subrings thereof? Given an integral extension T/Z that is a finite Z module and an integral domain, take the fraction field to get E/Q. Then take the integral closure of T, so that T lives inside S, which is the integral ring of E. T is now a submodule of a free Z module of rank n, hence T is free of rank ≤ n. Assume T contains a primitive element for E/Q, or by some other means T contains n linearly independent elements. Now T is free of rank n. Let T have discriminant d. Since T is a subring of S, it creates a sublattice of index k. The covolume of T in S is K, and the discriminant of T is k2 times the discriminant of S. Therefore S has a discriminant of d/k2. Since d has finitely many factors, and each of these leads to finitely many integral rings, there are a finite number of choices for S. Select one, and remember that T corresponds to a particular sublattice of index k. There are finitely many sublattices of index k, thus finitely many subrings of S. That completes the proof.

Let S be the integral ring of E/Q/Z. Map S onto the standard lattice in real/complex space with r1 real dimensions and 2r2 complex dimensions. The Dirichlet unit theorem uses this lattice to constrain the units of S.

Let L be euclidean space, i.e. real space, with r1+r2 dimensions. Apply a continuous map from complex space into L by taking the log of the absolute value of each component. Conjugate pairs have the same absolute value, so we just add these absolute values together to make one real value. This map is not defined on the hyperplanes where any of the components are equal to 0. Remember that a conjugate of x is 0 iff all conjugates are 0, iff x = 0, so this restriction only excludes the point 0 from S. Everything else maps into complex space, and then into L. I will call this the "log" embedding of S into euclidean space. It should not be confused with the standard embedding, or the component embedding, as described earlier.

The log of the absolute value is a continuous function, hence the map into L is continuous per component, and continuous wherever it is defined.

Let U be the group of units in S. Group action is multiplication in S as usual. Multiplication commutes with each embedding, hence the product in S becomes the product in each coordinate of complex space. Multiplication also commutes with absolute value. After taking logs, multiplication becomes addition in each coordinate of L. Thus we have a group homomorphism from U into L, where multiplication in U becomes addition in L.

Let K be the kernel, the units that map to 0. These units have absolute values of 1 over all embeddings. Define a bounded shape in complex space, with components bounded by 1. There are finitely many lattice points in this set, which pull back to finitely many points of S. Therefore K is finite.

A multiplicative finite group in a field is cyclic, hence K is cyclic, generated by some rth root of 1. Conversely, every root of 1 in E is integral over Z, and lies in S. Therefore K consists of the roots of 1 in E, a finite cyclic group.

Note that K contains all torsion elements of U. This because a torsion element in a multiplicative group is always a root of 1. Therefore the quotient U/K, which is the image of U, is torsion free.

The image of U is closed under addition in L. If 0 is not a cluster point, we have a lattice. Place a small closed ball about 0 in L, and pull back, by continuity, to a closed bounded region in complex space. This contains finitely many lattice points. Thus there are finitely many points of S, and of U, near 0 in L, and U becomes a lattice in L.

This is a Z lattice, or a free Z module, whose rank is bounded by r1+r2, the dimension of L. In fact, the rank is one less. The norm of a unit times the norm of its inverse equals the norm of 1, which is 1. Therefore the norm of any unit in S is a unit in Z, which is ±1. The norm is the product of the conjugates. Thus the product of the conjugates in complex space becomes ±1. Map this into L by taking logs, and the sum of the logs equals log(1), which is 0. The image of U lies in an r1+r2-1 dimensional hyperplane passing through L. Specifically, the coordinates of L must sum to 0.

In summary, U is finitely generated, with a cyclic torsion subgroup consisting of the roots of 1 in E, and a free subgroup with rank ≤ r1+r2-1.

Adjoin the square root of a negative number, and r1+r2-1 = 0, and the units are just the roots of 1 in E. These are ±1, ±i for the gaussian integers, and the sixth roots of 1 for the eisenstein integers.

Let T be a subring between Z and S, where T is a free Z module of rank n, with fraction field E. Let x be an element of T. The conjugates of x depend on E/Q, and do not depend on T or S. Thus the map from T into complex space, and into L, is compatible with the map on S. Since T* is a subgroup of S*, it maps to a sublattice in the hyperplane passing through L.

Let S be the ring of integers in a number field. Remember that S is dedekind. A divisor is a function nM that assigns an integer to each maximal ideal M, such that most of these integers are zero. In other words, finitely many are nonzero. Add two divisors together by adding the two values of nM for each maximal ideal M. The set of divisors is an abelian group, with generators that set one maximal ideal to 1 and the others to 0. This free abelian group is canonically isomorphic to the multiplicative group of fractional ideals of S. Factor any fractional ideal into prime ideals, and use the exponents as values nM to build the corresponding divisor.

Give this group the discrete topology, and it becomes a topological group. Call this space D1.

Every nonzero element x in E has some multiplicity for each prime M; namely, the number of factors of M in the ideal {x}. If X is in S then the multiplicities reproduce the exponents on the factorization of x*S as a product of prime ideals in S. If x is not in S, the multiplicities determine the factorization of x*S as a fractional ideal inside E. These multiplicities build a particular divisor in D1. Such a divisor is called a principal divisor, corresponding to a principal fractional ideal. Other divisors may exist, that are not principal, if the product of the maximal ideals to their respective powers does not yield a principal fractional ideal. If S is a pid then every fractional ideal, and every divisor, is principal.

An effective divisor has nM ≥ 0 for each M. This corresponds to a nonzero ideal in S.

A group homomorphism takes the nonzero elements of E onto the principal divisors. Multiplication in E becomes addition across nM. The units of S map to the 0 divisor, having no prime factors across the board. D1 mod the image of E is the class group of S.

The log embedding, described in the previous section, maps E* into euclidean space. Negate each coordinate of this log embedding, so that -log(|x|) is the sum of the coordinates of the image of x in euclidean space. This is the neglog embedding.

Euclidean space under addition is a topological group. Call this space D2, and let G be the direct product of D1 and D2. This too is a topological group. The elements of G are called Arakelov divisors.

Two different points in G, different in D1 or D2 or both, can be placed in disjoint open sets, hence G is hausdorff. It follows that every point of G, every arakelov divisor, is closed.

A map from E* into G determines, and is determined by, the component maps into D1 and D2. This is a group homomorphism, because it is a group homomorphism per component. The image is the set of principal divisors in G. The kernel is the roots of unity contained in E, as these are the only elements that map to 0 in D2. (We'll prove conjugation permutes the roots of 1 in a later chapter.) The cokernel is, by definition, the arakelov picard group, denoted pic. This can be viewed as a topological group, by applying the quotient space topology from G.

The units of S become 0 in D1, and a lattice along a hyperplane in D2. This lattice is the image of the units in G. Given x in E, all the associates of x map to a given divisor in D1, and a shifted version of the unit lattice in D2. The image of E is then all the principal divisors of D1, each crossed with a shifted lattice in D2. Cosets of this structure become the cokernel, the elements of pic.

Open sets in pic are not obvious, so let's build one. Cross the divisor a in D1 with any open set O in D2. This is open in G, but is it open in pic? Add in every principal divisor in G. Such a divisor changes a to b, and shifts the open set O to another open set in D2. The result is a union of open sets in G, hence open in G, hence open in pic.

A principal divisor is effective in G if it is effective in D1, and its neglog coefficients are all nonnegative. In other words, x is in S, and x and all its conjugates lie in or on the unit circle in the complex plane. Are any units effective? A unit u is 0 in D1, belonging to no maximal ideals, so look at D2. We already mentioned that the roots of 1 remain on the unit circle through all conjugations, hence they are 0 in D2, and are effective. If u lies outside the unit circle in the complex plane it still has norm 1, hence one of its conjugates lies inside the unit circle. u produces conjugate logs that are both positive and negative, hence u is not effective.

A degree homomorphism takes G into the reals by evaluating the sum, over all M in S, of nM × log(|M|); then add in the coordinates of D2. If you are curious, you can read about the norm of an ideal here; or you can just accept the definition of |M| as p raised to the residue degree of M over p. Verify that addition in G becomes addition in the reals, giving a group homomorphism. Set everything to 0 except for one of the coordinates in D2, and degree maps onto all of R.

Evaluate this expression on a principal divisor x. The image in D2 is -log(|x|). Let's see what happens in D1. Start with the norm of x, which is the same as the norm of the ideal {x}. Norm commutes with product, so replace {x} with the product of |M| raised to the nM, for each prime factor M in {x}. Take the log, and find the aforementioned sum in D1. The degree is 0. The image of E, through G, and into R, is 0.

At this point I must foist more notation upon you. Let G be the topological group of arakelov divisors as described above, and let G+ be the monoid of effective divisors. Let pic be the arakelov picard group, G mod the principal divisors. Since a principal divisor always has degree 0, degree is a well defined homomorphism from pic onto R. Let Gr, G+r, and picr be the elements of said spaces having degree r.

Is the quotient space pic hausdorff? Take two divisors a and b that do not differ by a principal divisor, thus representing different cosets in the cokernel, and different elements of pic. Let a1 and a2 be the projections of a into D1 and D2, and similarly for b. We can build the two open sets a1*D2 and b1*D2, but these two open sets need not be disjoint. Maybe two points in D2 cause a1 and b1 to differ by a principal divisor. These points become the same in pic, whence the open sets overlap. Let w in E generate such a principal divisor. Assume w′ is another element of E that connects points in our open sets. Now w/w′ creates a divisor that is 0 throughout D1. This is a unit in S. Conversely, w times any unit in S will connect two points in our open sets. As mentioned earlier, the units of S form a lattice H in the hyperplane of D2. Points are isolated, and H is a closed set. Shift H by the image of w and we still have a closed set. Shift H again by a2, and b2 is not one of the lattice points. This because a and b are distinct in pic. So there is a euclidean distance δ between b2 and the closest lattice point. Reduce δ further, so that it is no larger that the shortest distance between the lattice points of H. Then cut δ in half. Now place a2 and b2 in open balls of radius δ. Shift a2 by the image of w as we did before, and it seeds a shifted version of H, wherein each lattice point is the center of an open ball, and each ball contains only that lattice point. These balls do not intersect the ball about b. Therefore pic is hausdorff.

Since picr is a subspace of pic, each picr is also hausdorff.

Let p be a point in an open set in pic. This is an open set in G, covered by base open sets, Thus p is represented by a divisor of G that lives in a base open set in D1 cross a base open set in D2. This is a particular divisor of D1 cross an open ball in D2. We saw this before. This cross product implies an open set in pic. Thus every open set in pic is covered by open sets of this form. Indeed, these are the base open sets of pic.

pic is a group, and a topological space; show that pic is a topological group. In other words, addition is continuous. Write a + b = c, where c lives in a particular base open set. The topology of D1 is discrete, so a1 and b1 can stand alone. In D2, place a2 and b2 in open balls so that their sum lives in the open ball about c2. If a or b is adjusted by a principal divisor, the result lives in the open ball about c, adjusted by that same principal divisor. The open sets of pic surrounding a and b map into the open set of pic that surrounds c, and addition is continuous.

As with any topological group, a subgroup inherits the same topology, and every coset of that subgroup is homeomorphic to that subgroup by translation. Therefore pic0 is homeomorphic to picr for every real number r. We will prove pic0 is compact by showing picr is compact for a sufficiently large value of r.

Note that G is not compact, since its continuous image under the degree homomorphism is the real line, which is not compact. The same holds for pic, since that too maps onto all of R. But pic0 stands a chance, as it maps to 0, which is certainly compact in the real line.

Consider a divisor a in G, having components a1 and a2. Let l(a) be the elements of E that drive a1 into an effective divisor in D1. l(a) is a fractional ideal in E, specifically, l(a) = the product of M to the -nM. This is a consequence of to contain is to divide. If x belongs to this fractional ideal then x*S is contained in this fractional ideal, and when that is factored the exponent on M is at least -nM. Multiply a by x, thus adding these exponents to a1, and the result is everywhere nonnegative, giving an effective divisor.

As a fractional ideal, l(a) forms a lattice in real space via the component embedding. In an earlier section we said S has covolume v in this space. Equivalently, the covolume of s in complex space is v times 2 to the r2, also designated u. This in turn is the square root of the discriminant. We'll get back to this later. For now, the lattice defined by l(a), in component space, has covolume v times the norm of l(a), or v times the product of |M| to the -nM.

Next consider a2, the second half of the arakelov divisor a. These coordinates are often interpreted as logs, so in that spirit I'm going to reverse that by taking the exponential of each coordinate. This builds a box of sorts, box(a). The first r1 coordinates build a rectangular box in euclidean space. If y is the exponential of the coordinate, the box runs from -y to y. The remaining r2 coordinates each create a circle in the complex plane, with radius sqrt(y). Put this all together and the volume of the box is 2r1 × πr2 × the product of the exponential of each coordinate in a2. The last factor is the exponential of the sum of the coordinates in a2.

For a divisor a in G, raise e to the degree of a. The result is e to the sum of the coordinates of a2, as described above, times the norm of the ideal represented by a1. Multiply this by the covolume of the lattice l(a), and the norm of the ideal is in the numerator and denominator, and cancels. You are left with v times e to the sum of the coordinates of a2. This is a couple of constants away from the volume of the box.

volume = edegree(a) × covolume(l(a)) / v × 2r1 × πr2

v is a property of the number field and does not depend on a. The ratio of the volume of the box determined by the divisor a, to the covolume of the lattice l(a), depends only on the degree of a. Choose the degree large enough so that the ratio is at least 2n. It's not vital to know what this degree is, but let's have a look anyways.

Take logs, and move everything away from degree(d). Start with n×log(2) - r1×log(2), giving 2r2×log(2), or r2×log(4). Subtract r2×log(π) and get r2×log(4/π). Finally add in log(v). If z is the discriminant of S, replace v with sqrt(z)/2r2. Therefore the degree is at least r2×log(2/π) + log(z)/2. In any case, we have a degree r that makes the box much bigger than the unit cell in the component lattice.

Superimpose the box on the component lattice, placing each complex dimension on top of two real dimensions. Apply the point lattice theorem, and some point x in l(a) lives in the box.

Multiply a by x, and the result is effective in D1. This because x is a point in l(a).

Next consider the ith conjugate of x. If this embedding is real, the conjugate is trapped between -y and y, since x lies in the box. If the conjugate is complex, it lies in a circle of radius sqrt(y). Multiply by the other complex conjugate in the pair and the product lies in a circle of radius y. Take logs, and all the conjugates have logs bounded by the corresponding components of a2.

E maps into D2 via the neglog embedding. Multiplying by x subtracts these logs from a2. The result is nonnegative in each coordinate, hence an effective divisor in G.

Multiplication by the principal divisor x does not change the degree. Thus, when you mod out by the principal divisors, then restrict to the elements of degree r, each member of the quotient group can be represented by an effective arakelov divisor. Mod out Gr, or G+r, by principal divisors, and the result is the same, namely picr.

With this in mind, show G+r is compact. Each effective divisor implies an ideal in S, whose norm cannot exceed er. Yet the number of ideals in S with bounded norm is finite. For each such ideal b1, b2 takes up the slack to give degree r. This is a section of a hyperplane whose coordinates sum to a given value, and remain nonnegative. This is a compact set in euclidean space. G+r is a finite union of compact sets, and is compact. The quotient map takes this compact set onto picr, hence picr is compact, and by translation, pic0 is compact.

The dirichlet unit theorem says the lattice of units has dimension at most r1 + r2 - 1. Call the lattice L, living in a hyperplane P whose coordinates sum to 1. This is the second part of that theorem, and it says the dimension of L is exactly r1 + r2 - 1. In other words, L spans P.

Let G be the arakelov divisors of the integral ring S. Let Y be a finite subset of the prime ideals in S, with Z as complement. Let H be the subgroup of G that vanishes on Z. H includes 0 cross D2. The quotient group J = G/H is generated by the primes of Z. This is a direct product; G = H*J.

Write two exact sequences, one above the other, as follows.


Vertical arrows turn nonzero elements of E into principal divisors. Apply the serpent lemma, (described in a later chapter on homology), to get an exact sequence of kernels and cokernels of the vertical homomorphisms. Start by assuming Y is empty. Divisors that vanish on Z are 0 in D1. This means H = D2, and J = D1. The first kernel, as 0 maps into H, is of course 0. The nonzero elements of E that map to 0 in G are the roots of 1. The nonzero elements of E that map to 0 in D1 are the units of S. Now move to the cokernels. The cokernel of 0 into H is H, or Rr1+r2. The next cokernel is pic, and the next cokernel is D1 mod the principal divisors, or the class group of S. Write this as an exact sequence.

0 → roots of 1 in E → units of S → Rr1+r2 → pic → class group → 0

Change the bottom exact sequence so that the middle group is G0. Principal divisors have degree 0, so the vertical arrow already maps into G0, and there is no trouble. Take as subgroup H the divisors of degree 0 that vanish on the primes in Z. Again, Y is empty and Z is everything. H is the hyperplane whose coordinates sum to 0, which I denoted P. A coset in J is represented by a fractional ideal in E, and then a real number, along a designated axis, that makes the degree 0. The real number that indicates this coset is - the log of the norm of the fractional ideal, and it carries no new information. H can be added to this without changing the degree, thus building the coset of J. Again, G0 is a direct product of these two groups.

Run the serpent lemma again. The kernels of E are still the roots of 1 and the units of S. These map to 0, just as they did before. The first cokernel is the hyperplane P. Next we get pic0, and then the class group with the degree adjustment going along for the ride.

0 → roots of 1 in E → units of S → P → pic0 → class group → 0

Represent an element of pic0 as a member of the class group, along with its degree adjustment, plus a point in P.

The arrows are not just group homomorphisms; they are continuous maps on topological spaces. For instance, J is a quotient space of G, and the same holds if we restrict to the subspace G0. Its topology comes from the fractional ideals in D1 with a real number going along for the ride, hence discrete. The cokernel is a second application of a quotient map, and the quotient space of a discrete space is discrete, hence the topology is discrete. Since class groups are finite, we are not really surprised by this result.

P mod L injects into pic0, and forms the kernel of pic0. The kernel is the preimage of 0 in J, and is a closed set. Since pic0 is compact and the kernel is closed, the kernel is compact. Therefore P mod L is compact. If L has fewer dimensions, the quotient space is unbounded in the remaining dimensions, and cannot be compact. Therefore L spans all of P, and P mod L is a nice compact parallelotope. The units of S form a lattice with rank r1 + r2 - 1.

Units that form a basis for the lattice L in the hyperplane P are called the fundamental units. Let v be the covolume of this lattice within the space P. This is basis invariant.

Draw a vector perpendicular to P, with all coordinates set to 1. This vector has length z = sqrt(r1+r2). The volume spanned by the fundamental units and this vector is vz.

Project P onto any of the standard hyperplanes perpendicular to an axis. I'll illustrate by squashing the last coordinate down to 0. Keep the first r1 + r2 - 1 coordinates the same, and add their values into the last coordinate. Since coordinates sum to 0, the last component becomes 0, and we're on the floor. The transforming matrix has ones down the main diagonal, and ones down the right most column, and the determinant is 1. Volume does not change. The transformed parallelotope still has volume vz.

The new vector that was perpendicular to P now has r1 + r2 in the last column. Its height off the floor is z2. The volume of the projected basis of fundamental units is v/z. The regulator of S is v/z, or the covolume of the base cell projected into any one of the standard hyperplanes. To calculate the regulator, throw out any coordinate and take the determinant.

The regulator is 0 iff the number field is derived from the square root of a negative number. r2 = 1, and there are no units off the unit circle. The regulator of a positive quadratic extension is the log of its fundamental unit.

If you go all the way back to chapter 5, there is a section on the eighth roots of 1. This is a 4 dimensional extension via the irreducible polynomial y4 + 1. I presented a new unit, 1+y-y3, not on the unit circle, having inverse -1+y-y3. This is the fundamental unit for this number field. Since r1 = 0 and r2 = 2, there are no other units. Every unit is some power of the fundamental unit, positive or negative, times a power of y from 0 to 7.

Let S be a simple extension R[u], where u is a root of a monic polynomial p(x) of degree n. Thus S is a free R module with powers of u acting as a basis. Also, S is an integral domain. Let E and F be the fraction fields of R and S, where E is separable over F. Suppose p(x) factors into q(x) * r(x) over F. Multiply both factors by a common denominator, and two polynomials over R have a product that is a scale multiple of p. Substitute u for x, and something nonzero times something nonzero = p(u) which is 0. This contradicts S an integral domain, hence p is the minimum polynomial for u. In other words, E = F(u).

Let the powers of u (1, u, u2, etc) be the first row of a matrix W. The second row of W is the powers of u run through another embedding of E into the algebraic closure of F. These are the powers of one of the conjugates of u. This continues all the way down the rows of W. Each row is the powers of a different conjugate of u.

Notice that W is a vandermonde matrix. Its determinant is the product of the pairwise differences of the conjugates of u. Since E/F is separable, the roots are all distinct, each difference is nonzero, and the product is nonzero.

Let M be W transpose times W, in that order. Mi,j is the trace of the ith basis element times the jth basis element. The determinant of M is, by definition, the discriminant of the extension, but it is also det(W)2. And det(W) is the product of the pairwise differences of the conjugates of u. Put this all together and the discriminant is the square of the product of the pairwise differences of the conjugates of u.

Oddly enough, the same result holds if u is purely inseparable. The trace of anything in such an extension is 0, the matrix M is 0, and the discriminant is 0. At the same time, the roots of p(x) are equal, their differences are 0, and the discriminant is 0.

The discriminant is a mathematical function of u, and does not depend on the base ring. Push R all the way up to F, for example. If a linear combination of the powers of u is 0 then multiply through by a common denominator and find the same over R. This is a contradiction, hence the powers of u are linearly independent over F. The discriminant is the same over R or over F, or over any fraction ring in between. In other words, the discriminant is unchanged by localization.

Let R be integrally closed and let S/R be a ring extension that is an integral domain. Let S be a free R module of rank n. Give S a basis of b1 through bn.

Let a group of automorphisms G act on S, such that G fixes R. Pass to the fraction fields E and F, and G fixes F. In fact E/F is galois, with galois group G.

Let the discriminant of S = d, where d is a unit in R. Trap the integral closure of S between two free R modules, generated by b and b/d. Since d is a unit, these free modules are the same. Therefore S equals its integral closure, and S is integrally closed.