Dear Peter!

Many thanks for your answer and response!

Let z=(z1,z2,...,zn) and let C(z) denote the field of all rational functions with n variables z1,z2,...,zn, that is, ratios of polynomials with the variables above, with complex coefficients.

I'm interested in the question of whether any mxm matrix with entries in the field C(z) has a Jordan Normal Form or has the Schur form, that is, it can be brought to an upper triangular form (or lower triangular form) by some unitary matrix. Obviously, for the definition of unitary matrices, one needs to define an inner product over C(z)^m, and this is another question.

The question that you have raised is another question that should be asked: whether the field C(z) is closed?

These questions have led me to the isomorphism between C the field of complex numbers and the field of C(z) for rational functions as above. But since the proof of the isomorphism uses Zorn's Lemma and therefore does not supply any constructive way, I had to find another way to deal with the problem.

I will be very happy if you will have any clue for answering these questions and if you could give some references for authors dealing with these problems.

All The Best!

Dear Peter!

Thank you very much for sharing your great intelligent and elegant answers and for sacrificing your precious time!

Thank you for bringing to my attention that C(z) is not isomorphic to C; therefore, the field C(z) is incomplete. I thought I saw proof of isomorphism between C(z) and C, but I probably was wrong.

So, regarding the Schur decomposition of any mxm matrix over a field F, we must need that F would be complete? or that the decomposition must live over the closure of F? Is there any way out of this!?

Regarding the definition of an inner product over C(z) (say, for a single variable z), it must be defined for any two elements of C(z). And if we would define it as

$=\int_{0}^{2\pi}g(e^{it})^{*}f(e^{it})dt$, as you have proposed (if I understood you well) then it would not be defined for f or g having poles on the unit circle. So, how can this problem be solved? If we would join the value of infinity, the structure would not be norm-complete, since small perturbations in the coefficients of the rational functions could cause a jump to infinity in the norm and we would lose continuity! Moreover, in completing the norm, we would have to include all the holomorphic functions over the open unit disk, with a finite 2-norm. This would end up with the Hardy space $H_{2}({\mathbb D})$, of all holomorphic functions over the unit disk with finite 2-norm, which is not C(z) (but every function there can be approximated by a sequence of functions from C(z), for any given approximation error). So, how can we define an inner product on C(z), for which C(z) itself is norm-complete, without adding to it any element?

Many Thanks Again!