I'm sorry if this isn't an appropriate question for MO. I've been reading here for a while, but I still haven't got a good grasp of what's a *good* question.  
  
Given a field A and the polynomial ring A[x], we order the elements of A in any sequence and we define the isomorphism $f\colon A\to A[x]$ such that every element an$\mapsto$an xn, an $\in$ A, xn $\in$ A[x].


Can this be considered an alternate definition for A[x], is it just wrong, or is it the same as the canonical one?


Andy



Can I define the polynomial ring A[x] with an isomorphism f: A ---> A[x]?
The question seems to involve a construction of a set-theoretic map, and the indexing (natural numbers?) suggests that A is assumed to have a countable underlying set. That map doesn't even yield a surjection of sets.


I would like to reinterpret the question in the following way: How much structure do we need to forget in order for there to exist an isomorphism $A \to A[x]$? YBL pointed out that there is never an A-algebra isomorphism (if A is nonzero) and that there can be a ring-theoretic isomorphism if A is big enough. If A has an infinite underlying set, then there exist isomorphisms on the underlying sets. It is potentially interesting to ask when we get isomorphisms on the underlying additive groups: it is sufficient for A to have a polynomial ring structure, but that is far from necessary: e.g., A could be any field of infinite dimension over its prime field.


Regarding your last question, you can define a polynomial ring using a sequence of embeddings $f\_n: a \mapsto ax^n$ together with a specified multiplication law. This is a special case of the monoid ring construction. I'm not sure if this was the construction you initially had in mind, but it doesn't yield an isomorphism, since it isn't a single map. 


