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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

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closed as off-topic by Stefan Kohl, Henry.L, Michael Albanese, Mark Sapir, Yoav Kallus Aug 7 '17 at 21:24

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question does not appear to be about research level mathematics within the scope defined in the help center." – Stefan Kohl, Henry.L, Michael Albanese, Mark Sapir, Yoav Kallus
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Whatever that is, it's not an isomorphism. $\endgroup$ – Qiaochu Yuan Dec 11 '09 at 17:44
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    $\begingroup$ 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. $\endgroup$ – David E Speyer Dec 11 '09 at 19:48
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    $\begingroup$ 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). $\endgroup$ – Gabe Cunningham Dec 11 '09 at 20:05
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    $\begingroup$ 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. $\endgroup$ – David E Speyer Dec 12 '09 at 1:13
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    $\begingroup$ @Jose: This is a minor point, but the zero ring is unital. It just happens to satisfy the equation 0=1. $\endgroup$ – S. Carnahan Dec 12 '09 at 1:32
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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.

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  • $\begingroup$ 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 $\endgroup$ – Andy Lana Dec 12 '09 at 9:37
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$A$ 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).

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