Timeline for Is it possible to find a $\mathbb{C}$-algebra endomorphism of $\mathbb{C}[x,y]$ that fixes a given $w \in \mathbb{C}[x,y]-\mathbb{C}$?
Current License: CC BY-SA 3.0
11 events
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| Jun 4, 2017 at 13:51 | comment | added | user237522 | Thanks for your comment. (Perhaps I will think a little more about that case myself, before asking again). | |
| Jun 4, 2017 at 10:28 | comment | added | YCor | I would have told you if I had one. Possibly the case when $u$ is algebraic over $\mathbf{C}(w)$ deserves a separate question. | |
| Jun 4, 2017 at 10:03 | comment | added | user237522 | @YCor, please do you have an answer to one of the two cases ($u$ is algebraic over $\mathbb{C}(w)$ or not)? | |
| Jun 4, 2017 at 3:48 | answer | added | Will Sawin | timeline score: 16 | |
| Jun 4, 2017 at 2:26 | history | edited | user237522 | CC BY-SA 3.0 |
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| Jun 4, 2017 at 2:01 | comment | added | user237522 | Actually, I wish to obtain that there is a conjugate $v \in \mathbb{C}(x,y)$ of $u$ which is different from $u$ (conjugate= an element that has the same minimal polynomial as $u$), where it is known that $u$ is integral over $\mathbb{C}[w]$. The existence of such $f$ will guarantee that $\mathbb{C}[x,y] \ni f(u)=:v$ is the desired conjugate. | |
| Jun 4, 2017 at 1:41 | comment | added | user237522 | Good comment. Truly, in what I had in mind $u$ is algebraic over $\mathbb{C}(w)$ (more precisely, $u$ is integral over $\mathbb{C}[w]$). | |
| Jun 4, 2017 at 1:37 | history | edited | user237522 | CC BY-SA 3.0 |
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| Jun 4, 2017 at 1:37 | comment | added | YCor | Are you interested by the special case when $u$ is algebraic over $\mathbb{C}(w)$ or do you want to discard it? | |
| Jun 4, 2017 at 1:23 | history | edited | user237522 | CC BY-SA 3.0 |
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| Jun 4, 2017 at 1:10 | history | asked | user237522 | CC BY-SA 3.0 |