Timeline for Can a non-surjective polynomial map from an infinite field to itself miss only finitely many points?
Current License: CC BY-SA 2.5
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| Dec 1, 2009 at 5:35 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
Argument for cubic polynomials; deleted 3 characters in body
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| Nov 30, 2009 at 15:16 | comment | added | Greg Kuperberg | Lior: I am throwing in a root of $f(X)-b$ for every $b \in \tilde{k}$ inductively provided that $b \notin A$. So yes, $a$ such that $f(X)-a$ is irreducible might exist, but there are only finitely many of them, contrary to Hilbert's result. | |
| Nov 30, 2009 at 15:10 | comment | added | Greg Kuperberg | (1) My main point is that what everyone else (including me at first) meant merely as a construction is more properly understood as a tangible restatement of the problem. (2) My first point is about topology, with R meant only as an example; but I see that GMS made a similar p-adic argument and I apologize that I missed it. (3) If a popular construction is promoted to an equivalent problem statement, then hopefully that is a useful remark. I would hope that we have room for those if no one solves the problem? | |
| Nov 30, 2009 at 15:06 | comment | added | Lior Bary-Soroker | Greg, how did you understand from what I wrote that $f(X) - T$ is reducible? Clearly it is irreducible. What I'm saying is that your $\tilde k$ is Hilbertian, i.e., there exists $a\in \tilde k$ such that $f(X) - a$ is irreducible. | |
| Nov 30, 2009 at 14:49 | comment | added | Dror Speiser | 1) From comments below you'll see you are not the first or even second to propose trying to build such a construction. 2) Other answers certainly do address this reason for a polynomial not working over the reals. 3) Restating the problem is not an answer. | |
| Nov 30, 2009 at 14:20 | comment | added | Greg Kuperberg | By construction, $f(x)-c$ has a root cofinitely in $\tilde{k}$. So are you telling me that $f(X)-T$ is reducible?? | |
| Nov 30, 2009 at 14:12 | comment | added | Lior Bary-Soroker | I have the feeling that your $\tilde k$ is Hilbertain, if $k$ is (and $A$ is sufficiently large). Here is the rough uncompleted idea: choose a partition of $A$ into two subsets say $A_1$ and $A_2$. Let $M_i$ be the compositum of all splitting fields of $f(X)-a$, $a\in A_i$. I think that if $A$ is sufficiently large, and $A_1$ and $A_2$ are chosen wisely, we could get that $\tilde k$ is not contained in neither $M_1$ nor $M_2$. But clearly it is contained in $M_1M_2$. Thus by Haran's diamond theorem (see link in my post) $\tilde k$ is Hilbertian. | |
| Nov 30, 2009 at 6:35 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |