Timeline for Are polynomials with non-($S_n$ or $A_n$) Galois groups discrete?
Current License: CC BY-SA 3.0
14 events
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| Aug 11, 2017 at 2:25 | comment | added | Jacob Bond | @JoeSilverman This may make things more ambiguous, as far as trying to compare the results for $\mathbb{Z}$ and $k$, but in the article by Heintz mentioned above, he uses the Zariski topology on $k^{n+1}$. | |
| Aug 10, 2017 at 22:27 | comment | added | Joe Silverman | ... $\mathbb A_{\mathbb Q}$, but of course, $\mathbb Q$ is discrete in the adeles. This is why results along these lines are often phrased instead in terms of some sort of density, for example the density obtained by ordering the rational numbers by height (as in Dietmann's paper). And as Anthony Quas noted, since you get the same Galois group for $f(x)$ and $f(ax+b)$, and indeed even for $f((ax+b)/(cx+d))$ as long as $ad-bc\ne0$, any topology on $\mathbb Q^n$ that would be useful for this question must be invariant for these transformations. And $\mathbb Q^n/PGL_2(\mathbb Q)$ is a messy set! | |
| Aug 10, 2017 at 22:23 | comment | added | Joe Silverman | I find the question of discreteness in this context to be ambiguous. I gather the you are parameterizing polynomials of degree $n$ with $\mathbb Q$ coefficients by their vector of coefficients in $\mathbb Q^n$. That's fine, but to talk about discreteness, you need a topology, and there really isn't a natural topology on $\mathbb Q$. It appears maybe you're embedding $\mathbb Q$ in $\mathbb R$ and using the induced real topology, but you could as well embed in $\mathbb Q_p$ and use the $p$-adic topology. It actually makes more sense to embed $\mathbb Q$ In the adeles ... | |
| Aug 10, 2017 at 20:09 | comment | added | Jacob Bond | @AnthonyQuas I see, you are setting $f_n(x) = f(x - x/n)$ and considering $\lim f_n(x)$. | |
| Aug 10, 2017 at 18:56 | comment | added | Anthony Quas | No. For a fixed polynomial, the coefficients will converge to their original values since $(1-t)^k\to1$ as $t\to 0$ for any fixed $k$. | |
| Aug 10, 2017 at 18:47 | history | edited | Jacob Bond | CC BY-SA 3.0 |
Fix the original edit
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| Aug 10, 2017 at 18:40 | history | rollback | Jacob Bond |
Rollback to Revision 1
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| Aug 10, 2017 at 18:38 | history | edited | Jacob Bond | CC BY-SA 3.0 |
Rephrase an oversight
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| Aug 10, 2017 at 18:29 | comment | added | Jacob Bond | @AnthonyQuas But under that substitution, the higher power coefficients will tend to 0. | |
| Aug 10, 2017 at 18:20 | comment | added | Anthony Quas | Surely the question makes no sense in $\mathbb Q$: replacing $X$ with $\frac {n-1}nX$ yields a nearby polynomial with the same Galois group. | |
| Aug 10, 2017 at 18:14 | comment | added | MTyson | For a dense subset of $a\in\mathbb{Q}$, $x^n-a$ has Galois group $\mathbb{Z}/n\rtimes(\mathbb{Z}/n)^*$. | |
| Aug 10, 2017 at 17:54 | comment | added | Stanley Yao Xiao | I suspect with existing methods one can prove something like this for degrees 3 and 4, but I am not sure what the answer would be in general. | |
| Aug 10, 2017 at 17:23 | review | First posts | |||
| Aug 10, 2017 at 17:25 | |||||
| Aug 10, 2017 at 17:21 | history | asked | Jacob Bond | CC BY-SA 3.0 |