Timeline for Can I define the polynomial ring A[x] with an isomorphism f: A ---> A[x]? [closed]
Current License: CC BY-SA 3.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7, 2017 at 21:24 | history | closed |
Stefan Kohl♦ Henry.L Michael Albanese user6976 Yoav Kallus |
Not suitable for this site | |
| Aug 7, 2017 at 16:24 | review | Close votes | |||
| Aug 7, 2017 at 21:24 | |||||
| Sep 11, 2013 at 1:41 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
deleted 4 characters in body; edited tags; edited title
|
| Jul 3, 2010 at 14:29 | vote | accept | Andy Lana | ||
| Dec 12, 2009 at 19:17 | comment | added | Jose Brox | @Scott: I was following the convention that 1 must satisfy 1<>0. | |
| Dec 12, 2009 at 9:36 | comment | added | Andy Lana | (c_0,...,c_n) is in S. Thus, S is a commutative unitary ring with unit 1_s=(1,0,0,0...). Moreover, all the elements (a,0,0,0...) in A are a subring A_1 of S isomorphic to A. We define x as (0,1,0,0...) and x^n as (0,0,0,0...,a,0,0,0...) with n zeroes before a. A polynomial will be an object of the form a_0+a_1 x+...+a_n_0 x^n_0. I consider this to be the "canonical definition of A[x]". I'm sorry if I wasn't clear enough. I surely will be next time. | |
| Dec 12, 2009 at 9:28 | comment | added | Andy Lana | @Dadvi: yes, n is a natural number, and I've never written about a "usual isomorphism", I asked if that could have been an alternate definition to the canonical definition of the polynomial ring over A, A[x]. Let us conisder S the ser of all the infinite sequences (a_0,a_1,...,a_n) of elements of A such that there exists an n_0 in N with the property that for all n > n_0, a_n = 0. S is a subset of A^N. The sum will be thus defined: (a_0,a_1,...a_n)+(b_0,...b_n)=(a_0+b_0,...,a_n+b_n) and the product as (a_0,...,a_n)(b_0,...,b_n)=(c_0,...,c_n), with c= Sum of (a_j *b_j) from j+k=i. | |
| Dec 12, 2009 at 1:32 | comment | added | S. Carnahan♦ | @Jose: This is a minor point, but the zero ring is unital. It just happens to satisfy the equation 0=1. | |
| Dec 12, 2009 at 1:13 | comment | added | David E Speyer | 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. | |
| Dec 12, 2009 at 0:11 | comment | added | Jose Brox | Well, we DO have a trivial example if we consider nonunital rings: we can say that the 0 ring is isomorphic to its ring of polynomials 0[x] (I just wanted to mention this little detail). | |
| Dec 11, 2009 at 22:54 | answer | added | S. Carnahan♦ | timeline score: 5 | |
| Dec 11, 2009 at 20:05 | comment | added | Gabe Cunningham | 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). | |
| Dec 11, 2009 at 19:48 | comment | added | David E Speyer | 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. | |
| Dec 11, 2009 at 19:06 | answer | added | AFK | timeline score: 12 | |
| Dec 11, 2009 at 17:54 | comment | added | Andy Lana | Right, what I really meant was A is a domain, but still, I guess that does not change things. Thanks for the reference | |
| Dec 11, 2009 at 17:51 | comment | added | javier | For most rings, $A$ and $A[x]$ are not isomorphic. I'd be surprised if you can find any ring with that property which is not something extremely huge (like compact operators on some infinite dimensional Hilbert space). For instance, if $A$ is an algebra over a field with nice finiteness properties, the polynomial $A[x]$ will have bigger dimension. | |
| Dec 11, 2009 at 17:49 | comment | added | sdcvvc | If A is a field, then A and A[x] are never isomorphic (since the former one is a field and the latter one isn't). The closest thing you might be searching for is the universal property of the polynomial ring en.wikipedia.org/wiki/…. It doesn't rely on enumeration and basically says "R[x] is the simplest ring that adjoints to R some element x without any special properties" | |
| Dec 11, 2009 at 17:44 | comment | added | Qiaochu Yuan | Whatever that is, it's not an isomorphism. | |
| Dec 11, 2009 at 17:39 | history | asked | Andy Lana | CC BY-SA 2.5 |