# non-isomorphic stably isomorphic fields

Q1: What is the simplest example of two non-isomorphic fields $L$ and $K$ of characteristic $0$ such that $L(x)\simeq K(x)$ (here $x$ is an indeterminate)?

Q2: Do we have a sufficient criterion for a general field $K$ of characteristic $0$ which guarantees that if $K(x_1,\ldots,x_n)\simeq L(x_1,\ldots, x_n)$ (here $L$ is a field and the $x_i$'s are indeterminates) then $K\simeq L$?

• 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. Commented May 12, 2012 at 15:15
• Concerning Q2: A sufficient condition is that $K,L$ are algebraic extensions of the prime field. Commented May 12, 2012 at 16:26
• 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. Commented May 12, 2012 at 17:50
• Thanks Ralph for the answer, yes indeed, an isomorphism takes algebraic elements over the prime field to algebraic elements over the prime field. Commented May 12, 2012 at 17:57
• 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). Commented May 12, 2012 at 20:43

I don't think that there are any really easy examples. In the famous paper of Beauville, Colliot-Thélène, Sansuc and Swinnerton-Dyer "Variétés stablement rationnelles non rationnelles" they construct surfaces $S$ over $\mathbb Q$ that are not rational, but such that the products $S \times \mathbb P^3$ are rational. You get an example by taking $K$ to be a purely transcendental extension of the function field of $S$ of transcendence degree $d$, and a purely transcendental extension of $\mathbb Q$ of transcendence degree $d+2$, for some $d$ between $0$ and $3$ (I don't know the correct value of $d$).
• Silly question: what is wrong with the example $K=\mathbb{Q}$ and $L=\mathbb{Q}(x)$? Commented May 14, 2012 at 4:20
• @Mahdi: Well, $K(t) \not\cong L(t)$. Commented May 14, 2012 at 14:53
• @Martin: I see. What I had in mind was to say $K(x)\cong L(x)$ (same $x$ as in $L=\mathbb{Q}(x)$), but I see now, $x$ is not an indeterminate over $\mathbb{Q}(x)$. Thanks! Commented May 14, 2012 at 15:38
An answer to Q2, generalizing Ralph's comment: "$K$ is algebraically closed" is a sufficient condition. Indeed, you can characterize $K$ inside $K(x_1,\dots,x_n)$ as the set of elements having $m$-th roots for infinitely many integers $m$. More generally, it is enough to assume that for some $m>1$, the $m$-th power map on $K$ is onto. Examples: $K$ perfect of positive characteristic, or $K=\mathbb{R}$.