Finite Atol, Trace and Norm

Trace and Norm

1. In linear algebra, the norm of a matrix is its determinant, and the trace is the sum of its diagonal elements. Matrix algebra proves that similar matrices have the same trace and norm. By moving to a triangular matrix, we find definition (2).

2. If the matrix is viewed as a linear transformation, the trace is the sum of the eigen values, including multiplicities, and the norm is their product. This is a basis invariant definition, just as the previous algebraic definition was shown to be basis invariant. Trace and norm depend on the transformation, regardless of the representation of the underlying vector space.

3. The trace of an element u in a field extension E/F is the sum of its conjugates, and the norm is the product of those conjugates. You need to consider all the conjugates of u, not just the ones that live inside E. And you need to magnify trace and norm by a dimensionality fudge factor if E is greater than F(u).

4. Represent E as an F vector space, using a basis b₁ through b_n, then express multiplication by u as a matrix in F, according to the selected basis. For instance, u*b₁ is a linear combination of b₁ through b_n, giving the first row of the matrix. Derive trace and norm using this matrix, procedure (1), and obtain the same result as the sum and product of the conjugates. In other words, definitions (3) and (4) agree. The equivalence is demonstrated here. Furthermore, (4) and (1) are equivalent by construction - they use the same matrix algebra.

5. The trace of u, in a field extension E/F, is the sum of all the images of u over all possible embeddings of E into the algebraic closure of F. Since u maps to one of its conjugates when E maps into the closure of F, this is really the same as definition (3). Map F(u) first, which provides a separate embedding for each conjugate of u. Then, if E exceeds F(u), do this over and over as the additional generators of E are mapped to their conjugates in all possible ways. This is the dimensionality fudge factor that I referred to earlier.

6. Let M be a free R module and let e be any R endomorphism from M into itself. This is determined by the image of each basis element, which becomes a linear combination of basis elements. In other words, the endomorphism becomes a matrix, and we can take the trace and norm using definition (1). More on this below.

7. Almost any operator that is applied to a free module is (eventually) generalized to projective modules. The generalization often succeeds, because a projective module is the summand of a free module. Unfortunately this generalization makes the material even more difficult, and it is rarely needed. Throughout number theory, trace and norm are typically applied to free modules. So if you don't understand projective modules, and you don't really want to, you can probably skip those sections with impunity. I'll try to keep them cordoned off.

8. If S/R is a ring extension, one can sometimes take the norm of an ideal in S. If this ideal is principal, generated by u, the norm of the ideal will equal the norm of u. Beyond this, one can take the norm of a fractional ideal in S. This variation of norm is specific to algebraic number theory, and will not be addressed here.

As mentioned above, an endomorphism on a free R module can be expressed as a matrix, whereupon the trace and norm are calculated. The result is a value in the original ring R. Thus trace and norm map endomorphisms back into R.

Let q be an automorphism of M, and write q as an invertible matrix. Essentially, q is a change of basis, as it maps the original basis of M into new linearly independent elements that span. The transformation e, relative to the new basis, is q inverse times e times q. This has the same trace and norm as e, using the proof that was referenced in definition (1). Therefore definition (6) is basis invariant. Trace and norm are determined by the map from M into itself, regardless of the representation of M as a free R module.

When R is a field, definitions (6) and (1) agree. There is nothing to prove; that's the way (6) was constructed.

Let S/R be a ring extension, where S is a free R module. The action of u in S implements an R endomorphism on S, as an R module. Write this as a matrix, and take the norm and trace to obtain norm(u) and trace(u). When S/R is a field extension, this is consistent with definition (4), which is consistent with the other definitions. As long as an element u induces an endomorphism, we can derive trace(u) and norm(u).

Let M be an R module and let E be the ring of endomorphisms of M. Verify that the map trace(E) into R respects addition. The trace of the sum of two endomorphisms is the sum of their traces. Also, for any c in R, c times an endomorphism creates another valid endomorphism, and multiplies the trace by c. Multiplication by c, then by d, multiplies the trace by cd, and c distributes across the sum of two endomorphisms. Therefore, E is an R module, and trace is an R module homomorphism from E into R. (This only works because R is commutative. In the noncommutative world, the ring of endomorphisms need not be an R module at all.)

Let f and g be two endomorphisms in E. Applying f, then g, effectively multiplies the matrices together. This in turn multiplies their determinants. Thus the norm of the product in E is the product of the norms in R.

Trace through a Projective Module

For any module W, there is a natural isomorphism between hom(M,W) and M′×W. Replace W with M and find a natural isomorphism between hom(M,M) and T. The former is the set of endomorphisms of M. Thus an endomorphism maps to a member of T, which is represented as a finite sum of pair generators from M′ cross M. If f is a function in M′ and y is in M, evaluate f(y). Add these up across the pair generators of T that correspond to our endomorphism.

Is this well defined? After all, different sums of pair generators could represent the same member of T. There are two ways to answer this question. In a brute force approach, you might replace f,cy with cf,y. This is essentially the same pair generator, and builds the same entity in T. Either way, this term contributes f(cy) to the trace. This works because f is, itself, an R module homomorphism, and we can pull c in or out, as it acts on the domain or the range.

Another approach is to realize that trace is a bilinear map from M′ cross M into R. A bilinear map always factors through the tensor product, so trace is a well defined map from T into R.

The sum of two finite sums in T produces another finite sum. Use this to show trace respects addition in T. The trace of the sum is the sum of the traces. Similarly, let c act on T, apply the trace, and the result is multiplied by c. Therefore trace is an R module homomorphism from T into R.

Take a step back and trace becomes an R homomorphism from E (the endomorphisms of M) into R.

What if M is free of rank n? The map from E into T is a little clearer. Each endomorphism produces a sum of n terms in T: the i^th projection onto R, crossed with the image of the i^th basis element under our endomorphism. This was the i^th row in the matrix in our earlier representation. Apply the projection to the image and extract the i^th entry. Take the sum, and you are adding up all the diagonal elements. The definitions are consistent.

Direct Product

If M is projective then each summand is the summand of a free module, and is projective. Conversely, if each component is the summand of a free module, bring in the other n pieces and show that M itself is the summand of a free module, whence M is projective. Therefore, M is finite flat iff its components are finite flat.

Represent each element of M as a sum of elements across M_i, and represent each function in M′ as a sum of functions from M_i′. The tensor product T, described above, now has, as its pair generators, functions from M_i′ crossed with elements from M_j. However, when i ≠ j, the pair contributes nothing to the trace. We only need consider pairs where the function into R, and the element of M, come from the same component. Group these together by component and find the trace of the endomorphism restricted to each summand. Therefore, the trace of an endomorphism on M is the sum of the traces of the same endomorphism restricted to its submodules.

Return to M a projective module and let V be the other piece, so that M*V is free. Join an endomorphism f on M with the 0 endomorphism on V. The trace of f, with respect to M, is equal to the trace of f with respect to M*V. Do the same for another endomorphism g on M. Invoke the symmetry of trace on the free module M*V, and trace(fg) = trace(gf) on M.

Now we can prove trace is basis invariant, although M doesn't have a basis. Let q be an M automorphism that maps each new generator to a linear combination of the original generators. Apply q, then f, then q inverse to describe f under a new set of generators. The trace of qf/q is the same as the trace of (f/q)*q, which is the trace of f. Once again trace depends on the endomorphism, and not on the generators used to represent M.

Norm through a Projective Module

Applications of the Norm Map

If f is invertible, a unit in the ring of endomorphisms, write fg = 1 and take norms. The norm of the identity map is the determinant of the identity matrix, which is 1, hence |f| is a unit. In other words, the norm of a unit endomorphism is a unit. Conversely, if |f| is a unit, the adjoint over |f| becomes the inverse transformation, and f is a unit. Norm carries units to units, and nonunits to nonunits.

The same holds for x ∈ S. If x is a unit then f = x*S is an invertible transformation, and |x| is a unit. Conversely, let |f| be a unit, so that g is the inverse transformation to f. Let I be the representation of 1 using our basis, written as a column vector. Thus x = fI. Let y be the image of 1 under g. In other words, y = gI. Write I = fgI = f applied to y = xy, and x becomes a unit with inverse y.

Since R is commutative, a left or right inverse implies a two sided inverse, and a unit. What about the matrices over R? Let fg = 1 and take norms. Now |f| is a unit, and that makes f a unit. Let h be the inverse of f and write fh = 1. Subtract, and f*(g-h) = 0. Since units and zero divisors are mutually exclusive, f cannot be a zero divisor, whence g = h. A one sided inverse becomes a two sided inverse, and f becomes a unit.

Since units correspond under norm, f is irreducible whenever |f| is irreducible.

Let c be an element of R and scale the identity matrix by c to find a matrix that realizes multiplication by c. Take the determinant, and |c| = cⁿ. The trace is n*c.

If f is a zero divisor in the ring of endomorphisms, write fg = 0 and take norms, whence |f| is a zero divisor in R. If xy = 0 in S, take norms and |x| is a zero divisor in R.

The converse holds as long as R has a prime ideal, which is assured when R contains 1. Suppose |f| is 0 or a zero divisor, but f is not. If a linear combination of the columns of f yields 0, place the coefficients in a column vector v on the right, and f*v = 0. Replicate this column again and again to build a square matrix g. Now fg = 0, which is a contradiction. Thus the columns of f are linearly independent, spanning a free R module of rank n inside Rⁿ. The same proof, with v on the left, shows the rows of f are linearly independent.

Let P be a minimal prime in R, and localize about P. Free modules remain free after localization, having the same rank. Since localization is flat, a free submodule remains a free submodule, having the same rank. Thus the rows of f still span a free module of rank n. They are still linearly independent. Also, norm and localization commute (we'll prove this later), so that the determinant is still 0, or a zero divisor. The problem has been reduced to a local ring with just one prime ideal P.

A form of gaussian elimination is always valid over such a ring. If the rows are linearly independent, the matrix is converted into an upper triangular matrix, without changing the determinant. The entries down the main diagonal are all units, hence the determinant is a unit. This is not 0 or a zero divisor, hence we have a contradiction. Therefore f is 0 or a zero divisor iff |f| is 0 or a zero divisor.