These are the original definitions, but people soon realized that norm and trace can also be defined in terms of eigen values. The norm is the product of the eigen values, including multiplicities, and the trace is the sum. The two definitions are of course equivalent.
This section presents a more general definition of norm and trace, in terms of field extensions. We even allow the extension to be inseparable, which sets us apart from most textbooks. That's why this page is here, under [in]separable fields, instead of somewhere else.
When this definition is applied to vector spaces and matrices, you get the matrix definition back again. (We'll see this below.)
If you haven't had enough of norm and trace, there is an even more general definition involving the exterior product of projective modules, but let's not go there today.
Notice that the definition requires a finite extension and an element u in that extension. Change either u or the extension and the norm will change.
Let's look at a simple example. Consider x3-2 = 0. Let u be the real cube root of 2 and adjoin u to the rationals. This is a field extension of dimension 3. That's certainly finite, so on we go.
There are no "other" cube roots of u in F, so it looks like there is nothing to do; but norm requires us to map u into its normal closure. Adjoin the cube roots of 1, and the complex cube roots of 2 appear. Now map u to any of these three cube roots of 2 and extend the isomorphism to all of F. This produces three isomorphisms, and these are the only isomorphisms. Multiply the three images of u together to get 2.
The trace is the sum of the images of u. In our example the cube roots of 2 are placed symmetrically about a circle in the complex plane, hence their sum is 0. The trace is 0.
If the extension F/K is inseparable, the norm must be raised to the degree of inseparability, and the trace is multiplied by the degree of inseparability. this is a bit of a fudge factor, but it makes things work out properly.
If F/K is inseparable, the degree of inseparability is a power of p, where p is the characteristic of K. When this is multiplied by the trace, the result is 0. The trace is always 0 in an inseparable extension.
Given two elements u and v in an extension F/K, Show that norm(uv) = norm(u)*norm(v), and trace(u+v) = trace(u)+trace(v). This holds even if F/K is partly inseparable.
Let L be the purely inseparable extension of K inside F. The isomorphisms cannot move an inseparable element, hence L is fixed. The K isomorphisms are the same as the L isomorphisms, so count the L isomorphisms instead.
Select an element v in F, but not in L, and build an isomorphism by mapping v onto one of its conjugates. This determines the isomorphism on all of L(v). Next adjoin w, and map w to one of its conjugates. this creates an isomorphism on L(v)(w). Continue this process until you have an isomorphism on F/L. The isomorphism is determined by the image of v, then the image of w, and so on up the line. At each step we select a conjugate, a root of an irreducible polynomial. The number of roots always equals the degree of the irreducible polynomial, which is the dimension of the intermediate extension. By combinatorics, the number of isomorphisms is the product of the number of conjugates at each step, or the product of the dimensions, or the dimension of F/L.
The norm of u is the product of the images of u (which are all u) under all possible isomorphisms of F. The dimension of F/L counts the number of isomorphisms, hence u is raised to this power. finally, bring in the degree of inseparability. This is the dimension of L/K. Put this all together and |u| = un, where n is the dimension of F/K.
Similarly, trace(u) = n×u, where n is the dimension of F/K.
If K(u) is a purely inseparable extension, then the single root u appears n times. There is but one isomorphism, mapping u to u. Raise the single image u to the nth power, the degree of inseparability, and once again the norm is the constant coefficient times (-1)n.
That takes care of u separable and inseparable. The formula holds for any simple extension K(u) whose irreducible polynomial is known.
If u lies in F, the product of its images is symmetric with respect to all L isomorphisms of F. Applying an isomorphism rearranges the images of u, thus multiplying them in a different order. Hence the norm is fixed by all the isomorphisms of F. Since F/L is galois, the norm winds up in L. From there it is raised to the degree of inseparability, and is driven down into K. Once again the norm of u lies in K.
A similar proof shows the trace lies in K.
Similarly, the trace is multiplied by n.
First let's take care of the inseparable fudge factors. Adjoin the purely inseparable elements of F/K, then the purely inseparable elements of E/F. Basis elements are cross-multiplied to give a basis for the purely inseparable extension of K inside E. In other words, the degree of inseparability of E/K is the degree of inseparability of F/K times the degree of inseparability of E/F. (Say that three times fast.) When taking the norm of the norm, or the trace of the trace, degrees of inseparability combine to give the proper result.
Every isomorphism of E into its normal closure implies, by restriction to F, an isomorphism of F into its normal closure. Show that the set of E isomorphisms that fix F form a group. Call this group G.
Let M be a map from F onto a field isomorphic to F, inside the normal closure of F. Premultiply M by any isomorphism in the group G and find another isomorphism that carries F onto the same image. In other words, the E isomorphisms that restrict to M are a coset of the subgroup G inside the larger group of E isomorphisms.
These groups are finite, and the cosets of G all have the same size. For any map M, the isomorphisms in the class M, i.e. the isomorphisms that look like M when restricted to F, correspond 1-1 with the elements of G.
Let u be any element in E and take its norm. We can multiply the images of u in any order we like. consider the isomorphisms in G first. When multiplied together, these images give the norm of u with respect to E/F.
Next consider the isomorphisms in the coset determined by some map M. The product of the images of u, under this coset, is the image of the norm of u under M. This holds for every map M, i.e. every isomorphism of F. When these are multiplied together, the result is the product of the images of the norm of u, for every isommorphism of F, which is the norm of the norm of u. Since every E isomorphism has been considered, the norm of the norm is the norm.
You can show that the trace of the trace is the trace. The proof is essentially the same.
Multiplication by u implements a linear transformation on the K vector space. It maps 1 → u, u → u2, u2 → u3, and so on up to un-1 → un, which becomes a linear combination of the lower powers of u.
Build a matrix that implements multiplication by u. If M is the matrix, we want x*M (where x is a linear combination of powers of u) to equal ux. The first row of M is the image of 1, which is u. Thus the first row is [0,1,0,0,0,…0]. The second row is the image of u, or [0,0,1,0,0,…0]. This pattern continues, with ones just above the main diagonal. The last row is un, expressed as lower powers of u.
What is the determinant of M? Apply definition 2. Only one permutation produces a nonzero product. Multiply the ones above the main diagonal by the lower left entry. Call this lower left entry c, hence the determinant is c times the parity of the permutation. The parity is even if n is odd, odd if n is even.
If u has irreducible polynomial p(x), then p has constant term -c. We already showed that the norm of u, with irreducible polynomial p, is the constant coefficient of p times (-1)n. Bring in another -1, because the constant term of p is -c. Therefore our new definition of norm, calculated through p(x), agrees with the matrix definition.
Now return to the matrix M and consider its trace. This is the lower right entry, which is minus the second coefficient in the irreducible polynomial p, which is the trace. Once again the two definitions agree.
What if we select a different basis B for F/K? Multiplication by u is implemented by some matrix relative to this new basis B. It may look nothing like our matrix M above.
Let Q be the nonsingular matrix that describes the new basis B in terms of the original basis (i.e. the powers of u). Thus Q is our change of basis matrix.
Take any element x and represent x as a linear combination of elements from B. Thus x is a vector of coefficients drawn from K, and applied to the basis elements B. We will multiply x by u in an indirect fashion. First multiply by Q to represent x as a linear combination of powers of u. Then multiply by M, which effectively multiplies the element by u. Then multiply by Q inverse to return to the basis B. The new matrix QM/Q is similar to M, and implements multiplication by u. Since QM/Q is similar to M, it has the same trace and norm.
Trace and norm are basis invariant. To find the norm of u in K(u), select any basis and write the matrix that implements multiplication by u relative to that basis. The determinant is the norm and the sum of the diagonal elements is the trace.
Once the norm of u is computed with respect to K(u), we can easily find the norm of u with respect to E, where E properly contains K(u). Raise the norm to the jth power, where j is the dimension of E over K(u). This was discussed earlier.
How does this look when u acts on all of E? Does the norm still agree with the determinant of the larger matrix? Again, any basis will do - let's select a convenient one. Start with K(u), and let the powers of u build a basis for K(u)/u. Then adjoin other elements, building a basis v1 through vj for E/K(u). The cross product of these basis elements, the powers of u times v1 through vj, builds a basis for E/K. Consider multiplication by u. This acts on K(u), and modifies the coefficients on v1 through vj. By unique representation, this must be the action of u on E. In other words, u does not move v1 to v2, or anything inconvenient like that. The matrix that implements multiplication by u is block diagonal, with j blocks for v1 through vj. The determinant of each block is the norm of u in K(u)/K, as described above. There are j blocks, so the determinant of the entire matrix is the norm raised to the j, which is the norm of u in E/K. similarly, the trace is multiplied by j.
For any element u in a finite extension, the algebraic definitions of trace and norm agree with the matrix definitions, using any basis for E as a K vector space.