Necessity of the Cauchy product?

Question

I was trying to justify the necessity of defining the multiplication between two polynomials by means of the Cauchy product. Is there any other "suitable" option giving rise to a ring anyhow?

Answer 1

Okay, so given the usual $R$-module structure on $R[X]$, I think it's most natural to think of the polynomial ring as a "universal" space of functions, ergo the point of a polynomial as a general object is that it can be evaluated anywhere it makes sense. Hence, the natural ring structure on $R[X]$ should satisfy the following property.

Let $A$ be an $R$-algebra and let $a\in A$. Then $e_a: R[X]\to A$ given by $e_a(f)=f(a)$ is clearly a homomorphism of $R$-modules, but it is natural to want this to be a homomorphism of $R$-algebras.

Now, if $f\neq 0$, we claim that there exists $A$ and $a$ as above such that $e_a(f)\neq 0$. Indeed, this happens when $A$ is $R[X]$ with the usual Cauchy Product and $a=x$.

So fix $f=x$ and $g_n=x^n$ (the symbol) and note that, under this assumption, $e_a(f^n)= e_a(f)^n=a^n=e_a(g_n)$ Hence, $e_a(f^n-g_n)=0$ for any evaluation map and by the above, $f^n=g_n$. This implies that $R[X]$ is equipped with the Cauchy product.

Conversely, it's clear that the usual Cauchy product does make every $e_a$ an algebra-homomorphism.

Answer 2

Given a polynomial $\,p(x) := a_0 + a_1 x + a_2 x^2 + \dots \,$ and a polynomial $\,q(x) := b_0 + b_1 x + b_2 x^2 + \dots \,$ then define the Hadamard product polynomial $\, p(x)\odot q(x) = a_0b_0 + a_1b_1 x + a_2b_2 x^2 + \dots.\,$ . This is easily shown to give rise to a ring multiplication operation.

This Hadamard product is very different than the Cauchy product because all of the $\,x^n\,$ are pairwise "orthogonal" since their product is zero and hence are algebraically independent. In the usual Cauchy product, the monomial $\,x\,$ and its powers form a linear basis for all polynomials.