Timeline for Extending an automorphism from a sub-algebra to the algebra
Current License: CC BY-SA 3.0
34 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Mar 26, 2017 at 1:57 | comment | added | Lubin | Comparatively preliminary. I’d rather not put a URL out here. If you’re interested, e-mail me. I have to say that in the comment above, I overstated my thesis. The notes that I have concentrate on a situation where the automorphism group of the downstairs field is a topological group, and the extendable elements of it form a subgroup of finite index. | |
| Mar 25, 2017 at 22:47 | comment | added | user237522 | @Lubin, thank you, sounds interesting. Please do you have any notes/references of your recent talk? | |
| Mar 25, 2017 at 22:04 | vote | accept | user237522 | ||
| Mar 25, 2017 at 21:55 | comment | added | R. van Dobben de Bruyn | Clearly there are no examples where both $A$ and $B$ are finite field extensions of $k$ (since $k$ is algebraically closed). However, in my answer I give many examples where $A$ and $B$ are transcendence degree $1$ field extensions of an algebraically closed field $k$. | |
| Mar 25, 2017 at 21:52 | answer | added | R. van Dobben de Bruyn | timeline score: 4 | |
| Mar 25, 2017 at 20:25 | comment | added | user237522 | @R.vanDobbendeBruyn, but what if we further assume that the field $k$ is also algebraically closed (in addition of being of characteristic zero)? Can you still find a counter-example? (I am curious to know if in Lubin's talk the fields are algebraically closed). Thanks. | |
| Mar 24, 2017 at 18:18 | comment | added | Lubin | Indeed, @R.vanDobbendeBruyn, I just gave a relatively elementary talk taking off from the same example you give, arguing that for a finite separable extension of fields, you can hardly ever extend an automorphism of the smaller field to the larger. | |
| Mar 24, 2017 at 14:56 | comment | added | user237522 | @R.vanDobbendeBruyn, thanks for your counter-example. | |
| Mar 24, 2017 at 14:54 | comment | added | user237522 | Exactly, and this is why I have said that this case seems true to me (because I "believe" in a positive answer to the two-dimensional Jacobian Conjecture). However, I do not know yet how to prove that an involution on $T:=k[p,q]$ extends to an involution on $k[x,y]$, where $k[p,q] \subseteq k[x,y]$ and $\{p,q\}$ have an invertible Jacobian. I only think I know how to prove that this implies the two-dimensional Jacobian Conjecture (and it is enough to consider involutions on $T$ with Jacobian $-1$). | |
| Mar 24, 2017 at 7:34 | comment | added | YCor | If two elements of $k[x,y]$ have invertible Jacobian, unless I miss something, the Jacobian conjecture tells you that the pair generates the polynomial ring. So this is unlikely to give any example. | |
| Mar 24, 2017 at 3:22 | comment | added | R. van Dobben de Bruyn | Even for finite field extensions of $k = \mathbb Q$ there are counterexamples! For example, the conjugation on $\mathbb Q(\sqrt{2})$ cannot be extended to an automorphism of $\mathbb Q(\sqrt[4]{2})$, since $\sqrt{2}$ is a square in the latter but $-\sqrt{2}$ isn't (e.g. use a real embedding $\mathbb Q(\sqrt[4]{2}) \to \mathbb R$ to see this). | |
| Mar 24, 2017 at 2:03 | comment | added | user237522 | Such $A \subseteq B$ is a separable ring extension. Perhaps this should help somehow? | |
| Mar 24, 2017 at 1:51 | comment | added | user237522 | In your counter-example for (3), the Jacobian of $\{t,u^2\}$ is $2u$ which is not invertible in $k[t,u^2]$. What if $A$ is generated by two elements of $B$ that have an invertible Jacobian (namely, in $k^*$) ?. This case I had in mind when I have said that it seems true to me, but I do not have a proof yet. | |
| Mar 24, 2017 at 1:48 | comment | added | user237522 | Nice counter-example! | |
| Mar 24, 2017 at 1:41 | comment | added | YCor | When $A$ is a polynomial ring I already gave you examples. With $A$ and $B$ both polynomial rings, take $B=k[t,u]$ and $A=k[t,u^2]\subset B$. Then the flip $t\leftrightarrow u^2$ of $A$ does not extends to an automorphism of $B$, since $t$ is not a square in $B$ but $u^2$ is. Variant with one variable: $B=k[t]$, $A=k[t^2]\subset B$. Then writing $u=t^2$, the automorphism of $A$ mapping $u$ to $u+1$ does not extend, since $u+1$ is not a square in $B$ (assuming $k$ not of char 2) while $u$ is. | |
| Mar 24, 2017 at 1:38 | comment | added | user237522 | I guess one can find a counter-example for (1) and for (2)? But (3) seems to me more difficult (it seems true to me, but I do not have a proof yet). | |
| Mar 24, 2017 at 1:34 | comment | added | user237522 | I am not sure, perhaps each of the three options is interesting for me: (1) A is a polynomial ring. (2) B is a polynomial ring. (3) Both $A$ and $B$ are polynomial rings (with the same Krull dimension). | |
| Mar 24, 2017 at 1:30 | comment | added | YCor | You need to specify what is supposed to be polynomial ring: the small or the large ring? | |
| Mar 24, 2017 at 1:28 | comment | added | YCor | Simpler (but not domains): $k\times k\subset k\times k[t]$: the flip (of order 2) extends to an endomorphism, but not to an automorphism. | |
| Mar 24, 2017 at 1:28 | comment | added | user237522 | Very nice! What about polynomial rings? Do you think you can prove or give a counter-example in that case? | |
| Mar 24, 2017 at 1:23 | comment | added | YCor | Here's one localization where $f$ extends to an endomorphism but not an automorphism: assume that $R$ is a domain of characteristic zero, and consider the inclusion $A=R[t]\subset B=S^{-1}R[t]$, where $S=\{t+n:n\ge 0\}$. Then $f:t\mapsto t+1$ extends from $A$ to $B$, but the unique extension is not surjective, since $t^{-1}$ is not in the image. | |
| Mar 24, 2017 at 1:20 | comment | added | YCor | Whether it extends to an endomorphism is another question, but in my example it doesn't, by the same argument. On the other hand, for localization of domains, the possible extension is unique and hence if $f^n$ and the extension $g$ exists, then $g^n=1$. | |
| Mar 24, 2017 at 1:16 | comment | added | user237522 | @YCor, thanks for you comments! Concerning your first comment, I would elaborate (1) whether $f$ extends to an endomorphism or to an automorphism (is this what you meant?). Your second and third comments are interesting (you can write them as an answer if you like, though I had in mind different algebras than your counter-example). | |
| Mar 24, 2017 at 1:06 | comment | added | YCor | Easier and more explicit: if $A\subset B=A[1/u]$, then an automorphism of $A$ extending to $B$ implies geometrically that it corresponds to an automorphism of variety stabilizing the zero locus of $f$. This yields to an obvious example: the automorphism $t\mapsto 1-t$ (of order 2) of $R[t]$ ($R$ any nonzero commutative ground ring, where we work in $R$-algebras) does not extend to $R[t,t^{-1}]$ (direct argument: because in the large ring $t$ is invertible but not $1-t$). | |
| Mar 24, 2017 at 1:03 | comment | added | YCor | Curves of genus $\ge 2$ have finite automorphism group. Hence there are many 1-dimensional domains (f.g. over field $k$) $A$ with an embedding of $k[t]$ (or $k[t,t^{-1}]$), and very few of the automorphism of these subfield will extend. | |
| Mar 24, 2017 at 1:00 | comment | added | YCor | When $f$ is, say an involution, there are 2 distinct questions (1) whether $f$ extends (2) whether $f$ extends to an involution. | |
| S Mar 24, 2017 at 0:23 | history | suggested | Konstantinos Kanakoglou |
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| Mar 24, 2017 at 0:19 | comment | added | user237522 | Thank you very much for your comment. Indeed, an answer to the question you cited should help to answer my question also. | |
| Mar 24, 2017 at 0:11 | review | Suggested edits | |||
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| Mar 24, 2017 at 0:10 | comment | added | Konstantinos Kanakoglou | I do not know the answer but i think your question is intimately connected to mathoverflow.net/q/111881/85967. | |
| Mar 23, 2017 at 23:57 | history | edited | user237522 | CC BY-SA 3.0 |
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| Mar 23, 2017 at 21:57 | history | asked | user237522 | CC BY-SA 3.0 |