-6
$\begingroup$

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

Improve this question
$\endgroup$
12
  • 5
    $\begingroup$ Whatever that is, it's not an isomorphism. $\endgroup$
    – Qiaochu Yuan
    Commented Dec 11, 2009 at 17:44
  • 4
    $\begingroup$ 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
    Commented Dec 11, 2009 at 19:48
  • 1
    $\begingroup$ 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
    Commented Dec 11, 2009 at 20:05
  • 1
    $\begingroup$ 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
    Commented Dec 12, 2009 at 1:13
  • 2
    $\begingroup$ @Jose: This is a minor point, but the zero ring is unital. It just happens to satisfy the equation 0=1. $\endgroup$
    – S. Carnahan ♦
    Commented Dec 12, 2009 at 1:32

2 Answers 2

Reset to default
5
$\begingroup$

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.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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
    Commented Dec 12, 2009 at 9:37
12
$\begingroup$

$A$ 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).

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged .