Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?


share|improve this question
Whatever that is, it's not an isomorphism. –  Qiaochu Yuan Dec 11 '09 at 17:44
I hope this doesn't come off too harshly: I don't understand this question. You say you will define a map from A to A[x]. The proposed map is supposed to be described by the formula a_n \to a_n x^n. This formula doesn't make sense, because I don't know what n is. For example, if R is the real numbers, where do I send 17? Does it go to 17, to 17 x, to 17 x^{17}, or to someplace completely different? Where does \pi go? I don't know whether the question you are thinking of is a good question, but the question you have written is not, because is not clear what you are thinking of. –  David Speyer Dec 11 '09 at 19:48
The question says: "we order the elements of A in any sequence", so my understanding is that we arbitrarily order A as a_1, a_2, .... Of course, given that we want the indices to be natural numbers, we can't do this for any uncountable field (or ring, for that matter). –  Gabe Cunningham Dec 11 '09 at 20:05
OK, I think I understand what you are thinking, thanks to Gabe's comment. But f isn't much of a map. It is not a map of rings, as f(a+b) is not f(a)+f(b), and f(ab) is not f(a)f(b). It is very far from covering A[x] - it doesn't hit any polynomial which has more than one term in it. So this is not an isomorphism; it has basically none of the properties an isomorphism should have. I also don't know what you could have meant by "the usual isomorphism". A guideline to what makes a good question might be that you understand what all the words in it mean. –  David Speyer Dec 12 '09 at 1:13
@Jose: This is a minor point, but the zero ring is unital. It just happens to satisfy the equation 0=1. –  S. Carnahan Dec 12 '09 at 1:32
show 7 more comments

2 Answers

up vote 4 down vote accepted

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.

share|improve this answer
The construction I had in mind was something like that, thanks. Still, I can't really remember why I wanted it to be an isomorphism. Thanks a lot for the answer –  Andy Lana Dec 12 '09 at 9:37
add comment

$A$ is never isomorphic to $A[x]$ as an $A$-algebra. Because, for any $A$-algebra $B$, the set of $A$-algebras homomorphisms

  • $Hom_A(A,B)$ has only one element: the unit map $a\mapsto a\cdot 1$.
  • $Hom_A(A[x],B)$ is canonically identified with the set $B$: $b$ corresponds to $\sum a_nx^n\mapsto \sum a_n b^n$.

But it is possible for $A$ and $A[x]$ to be isomorphic as rings. Take $A = k[x_1,\ldots,x_n,\ldots]$ a polynomial ring in a infinity of variables, then $A \to A[x_0]$ that sends $x_{i+1}$ to $x_i$ is an isomorphism of $k$-algebras (not of $A$-algebras).

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.