Timeline for non-isomorphic stably isomorphic fields
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 15, 2012 at 15:33 | vote | accept | Hugo Chapdelaine | ||
| May 14, 2012 at 8:55 | answer | added | Laurent Moret-Bailly | timeline score: 8 | |
| May 13, 2012 at 2:24 | comment | added | Kevin Ventullo | @Hugo: This is false. The multiplicative group of a number field is isomorphic to the roots of unity times a free group of countable rank. | |
| May 12, 2012 at 21:21 | comment | added | Hugo Chapdelaine | @Martin, yes you are right but forget for the moment the isomorphism between $L[x]$ and $K[x]$. Assume that you only know that $K$ and $L$ are fields and that you have a group isomorphism between $K^{\times}$ and $L^{\times}$ does this imply that $K$ is isomorphic to $L$ as fields? It could be but I never thought about that. | |
| May 12, 2012 at 20:43 | comment | added | Ralph | I don't know if it's of practical help, but $K(x) \cong L(x)$ implies $K \cong L$ iff there is an isomorphism $K(x) \xrightarrow[]{\sim} L(x)$ that maps a transcendence base of $K|F$ onto a transcendence base of $L|F$ (where $F$ denotes the prime field). | |
| May 12, 2012 at 19:47 | comment | added | Martin Brandenburg | @Hugo: Let $L,K$ be fields. When $K[x] \to L[x]$ is an isomorphism of rings, it maps $K$ onto $L$ (since it maps $K^*$ onto $L^*$) and therefore gives $K \cong L$ as rings. Where's the problem? | |
| May 12, 2012 at 18:13 | comment | added | Hugo Chapdelaine | @Martin, I'm not sure I understand your comment. If $R$ is a domain then you always have $R[x]^{\times}=R^{\times}$ and therefore $R^{\times}\simeq S^{\times}$. But this is just the multiplicative structure. So in a field, for non-zero elements $a,b$ we have $a+b=b(ab^{-1}+1)$ and therefore you still need to know what what it means to add $1$ to a field element... Si if $\phi:K^{\times}\rightarrow L^{\times}$ is a group isomorphism then we get $\phi(a+b)=\phi(a)\phi(1+ba^{-1})$, but this is not clear to me that this is equal to $\phi(a)+\phi(b)$... | |
| May 12, 2012 at 17:57 | comment | added | Hugo Chapdelaine | Thanks Ralph for the answer, yes indeed, an isomorphism takes algebraic elements over the prime field to algebraic elements over the prime field. | |
| May 12, 2012 at 17:50 | comment | added | Hugo Chapdelaine | Here is one possible way of constructing such examples: Let $\iota_1:G\rightarrow S_{n}$ and $\iota_2:G\rightarrow S_{m}$ be two embeddings of a finite group $G$ where $S_k$ denotes the symmetric group of degree $k$. Let $K_n$ be the field of rational functions over $\mathbf{Q}$ in $n$ variables then it is easy to see that $K_n^{\iota_1(G)}$ and $K_{m}^{\iota_2(G)}$ are stable isomorphic, but in general I don't see any reason why they should be isomorphic. Of course one needs to choose the group $G$ carefully since for "many" $G$'s $K_n^{\iota_1(G)}$ will always be purely transcendental. | |
| May 12, 2012 at 16:26 | comment | added | Ralph | Concerning Q2: A sufficient condition is that $K,L$ are algebraic extensions of the prime field. | |
| May 12, 2012 at 16:02 | answer | added | Angelo | timeline score: 14 | |
| May 12, 2012 at 15:15 | comment | added | Martin Brandenburg | Just a comment: The question $R[x] \cong S[x] \Rightarrow R \cong S$ has been studied in the literature since the 70s (google for "isomorphic polynomial rings" or see math.stackexchange.com/questions/13504); Hochster has constructed counterexamples. On the other hand, this is true for fields (consider units). So in Q1, we cannot expect $L[x] \cong K[x]$ to hold. By the way: 1+, since I don't know an example for Q1 at all. | |
| May 12, 2012 at 15:01 | history | asked | Hugo Chapdelaine | CC BY-SA 3.0 |