# Problem on space of polynomials in two variables, is there something wrong ?

1. Sep 9, 2009

### winter85

1. The problem statement, all variables and given/known data

Let P(n,m) be the space of all polynomials z with complex coefficients, in two
variables s and t, such that either z = 0 or else the degree of z(s, t) is <= m - 1
for each fixed s and <= n - 1 for each fixed t.

Prove that there exists an isomorphism between Pn (x) Pm (tensor product of Pn and Pm) and P(n,m) such that the element z of P(n,m) that corresponds to a (x) b (tensor product of vectors, a in Pn, b in Pm) is given by z(s,t) = a(s)b(t).

3. The attempt at a solution

Never mind the definition of tensor product, it seems to be that the conclusion can't be correct? because if there is such an isomorphism then every element z in P(n,m) can be written as z(s,t) = a(s)b(t) where a and b are polynomials, however z(s,t) = s + t clearly is an element P(n,m) but cannot be written in this form?

There's probably something that i'm not getting so I would appreciate it if anyone can point it out for me. Thanks.

2. Sep 9, 2009

### HallsofIvy

Staff Emeritus
I think you had better "mind the definition of tensor product"! Such an isomorphism would mean that every element in P(n,m) can be written as a sum of terms of the form a(s)b(t)- and s+ t is exactly of that form.

3. Sep 9, 2009

### winter85

oh i see, Thanks HallsofIvy. I know now where my confusion comes from.

The definition in the book i'm using (Halmos) says:
the tensor product of two spaces U and V is the dual space of the space of all bilinear forms on UxV.
the tensor product of vectors u in U and v in V is z in U(x)V defined by z(w) = w(u,v) for every bilinear form w on UxV.

I read about tensor products in other places and it seems that's not the standard definition (it says in the book that this definition only works for finite dimensional spaces). It seems that U(x)V is the vector space generated by all tensor products u (x) v of vectors u in U and v in V. Is this correct?

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook