Timeline for $A[x]$ points of an algebraic group
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 20, 2018 at 6:10 | vote | accept | Girish | ||
| Aug 19, 2018 at 22:36 | answer | added | Friedrich Knop | timeline score: 3 | |
| Aug 16, 2018 at 8:10 | comment | added | Girish | @UriyaFirst Thank you for the answer. | |
| Aug 15, 2018 at 15:06 | comment | added | Uriya First | @Girish In that case, as soon as $G$ contains a copy of $\mathbb{G}_a$ as a closed subgroup, the map $G(A)\to G(A[x])$ will not be an isomorphism. This is always the case when $K$ is algebraically closed and $G=\mathrm{O}_n$ with $n>2$ or $\mathrm{Sp}_n$ with $n>1$ (the unipotent radical of a Borel subgroup will contain such a copy). Since you can always take $A$ to an algebraically closed field, $G(A)\to G(A[x])$ will not be an isomorphism for all $A$-s when $G=\mathrm{O}_n$ or $\mathrm{Sp}_n$. | |
| Aug 15, 2018 at 13:28 | comment | added | Girish | @Uriya First I want the natural map to be isomorphism. | |
| Aug 15, 2018 at 12:50 | comment | added | Uriya First | @Girish, is your question about the existence of some group isomorphism $G(A)\cong G(A[x])$ or about the natural map $G(A)\to G(A[x])$ being an isomorphism? | |
| Aug 15, 2018 at 12:50 | comment | added | Uriya First | @abx I agree. The map $G(A)\to G(A[x])$ is rarely an isomorphism. I understood the question as isomorphism of abstract groups. | |
| Aug 15, 2018 at 12:32 | comment | added | abx | @ Uriya First: my understanding is that the OP asks whether the natural map $G(A)\rightarrow G(A[x])$ is an isomorphism. But of course this is not clearly formulated in the post. | |
| Aug 15, 2018 at 11:30 | history | edited | Girish | CC BY-SA 4.0 |
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| Aug 15, 2018 at 11:11 | comment | added | Uriya First | @abx If $K$ is of infinite dimension over it prime subfield, call it $F$, then $\mathbb{G}_a(A)$ and $\mathbb{G}_a(A[x])$ are both $F$-vector spaces of $F$-dimension $\dim_FA=\aleph_0\cdot \dim_F A$ and are therefore isomorphic as abstract groups for any $K$-algebra $A$. | |
| Aug 15, 2018 at 10:56 | comment | added | abx | It is not true for $\operatorname{SL}_2 $, hence not for $\operatorname{Sp}_{2n} $. | |
| Aug 15, 2018 at 10:19 | comment | added | Girish | Yes, For G=$GL_n$ also it is not true. Thats why I asked for $O(n)$ and $Sp_{2n}$. | |
| Aug 15, 2018 at 10:17 | comment | added | abx | Certainly not with $G=\Bbb{G}_a$. | |
| Aug 15, 2018 at 10:07 | history | edited | Girish | CC BY-SA 4.0 |
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| Aug 15, 2018 at 10:03 | comment | added | YCor | 1) "For a ring" means for some ring? for every ring? "a" is not a very clear quantifier. 2) $Sp_{2n}$ is not (at least to me) clearly defined as a scheme over $\mathbf{Z}$... there are several choices of symplectic forms, which do not matter over a field, but over a ring, it's less clear. | |
| Aug 15, 2018 at 9:49 | history | asked | Girish | CC BY-SA 4.0 |