Polynomials are formal sums. So in particular, is infinite-dimensional over , even if e.g. , the field with two elements. This is true even though e.g. is the zero function on , as you can check by substituting and for .
Polynomial functions are functions e.g. on the field itself (or, more generally, on some object that can be locally parameterized by the field). But let’s simply consider , the set of polynomial functions from the field to itself. is a ring with addition and multiplication given point-wise on function values. Any polynomial can be considered a polynomial function in the obvious manner, and this defines a surjection of rings:
In the case where is a finite field, is a proper quotient of i.e. . To see this, consider the multiplicative group . Then , where . Therefore for any , and so for any . We’ve shown therefore that the non-zero polynomial is in , since it is maps to the zero function on . This is just the generalization of our observation for , above.
In fact, we’ve found a generator for the kernel, i.e. . One way to see this is to check that the polynomial functions defined by first monomials are linearly independent as functions on , which can be done using the Vandermonde matrix and the distinct non-zero field elements at your disposal. Another way to see this, since we are working over a finite field, is to simply count the elements of the quotient and of . There are clearly elements in the quotient, but what about ? It turns out that consists of all functions . To see this, given any function on , just use Lagrange interpolation to build yourself a polynomial of degree that matches the function at all points. There are functions , and so that’s the size of .
Thus the quotient has the same number of elements as . But we have a chain of surjections
so .
On the other hand, if the field is infinite, then is zero, i.e. polynomials over and polynomial functions on are isomorphic. To see this just show that the set of all monomials is linearly independent, using the Vandermonde matrix and the endless supply of distinct non-zero elements of .
The multivariate case is entirely analogous
All functions are polynomial functions. To see this, just mimic the Lagrange interpolation of the univariate case (it’s sufficient to convince yourself that, given any point in , you can write down a polynomial function that takes the value 1 there, and the value 0 everywhere else). Thus, since there are functions , this is the size of , the ring of polynomial functions on .
Now write for the obvious surjection of rings. Then it follows immediately from the univariate case that we have an inclusion of ideals Now notice that the quotient has elements. This is the same number as above, and so we conclude that we’ve already found the full kernel, i.e.
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