Well, I seem to have got myself in a right old tangle here; I hope one of you guys can help untangle me. I was brought up to believe it is necessary to make a distinction between polynomial forms and polynomial functions, since (depending on the field over which they are defined) different poly. forms can "refer" to the same poly. function. Then, as I recall, it is customary, for the polynomial form \(p(x)=\sum \nolimits^n_{n=0}a_nx^n\), to regard the \(\{x^i\},\,\,\,\,i=0,1,2,.....,n\) as indeterminates, rather than as variables. Wolfram seems to believe that a polynomial form is homogeneous. But this seems to make no sense, even if, as the Wiki asserts, that the \(\{x^i\}\) are "merely placeholders for the coefficients", since the usual rules for multiplying "genuine" exponents apply to the ordering indices, if that's what they are? Are they? Assuming so, multiplication, like \((a_i x^i)( b_j x^j)= a_ib_j x^{i+j}\) still makes perfect sense, even though the exponents are not powers in the usual sense. Does this imply that a polynomial form is homogeneous? Can't see it. And then again, I have a reputable text here which asserts that, for a poly. form, the notation \(x^n\) refers in the usual way to \(xxxx...x\) n times. I cannot bring these into register for poly. forms as defined above, assuming I have understood correctly (unlikely!). Grr
I believe this is just a terminological issue, so once you sort out which book/reference means what by what you are all set. The term "form" is interpreted by Wolfram as in "quadratic form" which means it must be homogeneous, so that quadratic forms become a special case of polynomial forms.
temur, I thank you, you have been a good friend to me on this forum. I lead on to the following; see if you, or anyone else for that matter, thinks it a valid argument (personally I have my self-doubts). Suppose that \(K\) is a field, and that I denote by \(K[x]\) the ring of polynomial forms over \(K\). It is easily shown that this is in fact a vector space, so that the polynomial form \(p(x) \in K[x]\) is a vector. Now by definition a polynomial form may have only finitely many terms in the summand, so that only for finitely many \(\alpha_kx^k\) in the summand may I have that \(\alpha_k\) is non-zero. But since for arbitrary \(p(x) \in K[x]\) the zero and non-zero coefficients may "fall" anywhere, I am obliged to use as a basis for this vector space the countably infinite set \( \{x^0,\,x^1,\,x^2,\,....\}\) and any polynomial can be expressed as the sum of scalar operations on a subset of this basis (these are linearly independent since the existence of the zero vector \(p(x) =0 \Rightarrow \alpha_i =0\) for all \(i\) by the above.) In other words, as a vector space \(K[x]\) is infinite-dimensional. I now note that any field \(K\) has all the properties of a vector space over itself, so to speak. I also suppose this field/vector space is such that a sensible basis might be \(\{1\}\). I now make the vector space \(K \oplus K \oplus K \oplus....\), countably. A possible basis for this infinite dimensional vector space might be \(\{(1,\,0,\,0,\,...), (0,\,1,\,0,\,0,\,...), (0,\,0,\,1,\,0, .....),....\}\). I rashly claim a vector space isomorphism \(K[x] \simeq K\oplus K \oplus K \oplus....\). I am a little uneasy about this. Both these vector spaces are infinite dimensional: for finite dimensional spaces, equal dimensionality is sufficient to establish isomorphism, in the infinite dimensional case this need not be so. However, regarding the \(x^i\) as placeholders, there is a clear one-to-one correspondence \(\{x^0,\,x^1,\,x^2,....\} \Left\Rightarrow\{(1,\,0,\,0,\,...), (0,\,1,\,0,\,0,\,...), (0,\,0,\,1,\,0.....),....\}\), may I say my claim is valid? This would imply that no vector in \(K\oplus K \oplus K \oplus....\) may have no more than finitely many non-zero coefficients. Or in other words, this monster has only finitely many non-zero elements, a result I would like to be true. Have I shown it?
Your argument shows that \(K[x]\) and \(K \oplus K \oplus K \oplus\ldots\) are isomorphic, but this does not show that vectors in \(K \oplus K \oplus K \oplus \ldots\) cannot have infinitely many non-zero coefficients. The latter is true simply by definition of direct sum, and you have used this definition in your argument to show the isomorphism, because this definition fixes what you mean by a basis of the two objects under consideration. To contrast, if you allow infinite linear combinations, the polynomials \(K[x]\) and the direct sum \(K \oplus K \oplus K \oplus\ldots\) must be replaced respectively by the formal power series \(K[[x]]\) and the direct product \(K \times K \times K \times\ldots\). They pairwise have exactly the same bases \(\{x^0,\,x^1,\,x^2,....\}\) and \(\{(1,\,0,\,0,\,...), (0,\,1,\,0,\,0,\,...), (0,\,0,\,1,\,0, .....),....\}\), but the meaning in which these are bases is different in each case, so you have to be aware of this meaning when you want to derive a general property just by looking at bases. To add: What you are referring to as polynomial forms seem to be just polynomials. I am wondering what distinction you have in mind.
