Timeline for Algebraic dependency over $\mathbb{F}_{2}$
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2014 at 22:33 | history | bounty ended | eig | ||
| Dec 2, 2014 at 1:00 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 404 characters in body
|
| Dec 1, 2014 at 19:09 | comment | added | David E Speyer | @PeterMueller Thanks! Okay, now we need to deal with $n>2$. | |
| Dec 1, 2014 at 17:24 | comment | added | Peter Mueller | @David: If $h(g(x))$ is a polynomial, where $g$ is a polynomial and $h$ is a rational function, then $h$ is actually a polynomial. (Write $h(z)=p(z)/q(z)$, then $1=u(z)p(z)+v(z)q(z)$ by Bezout, so $1=u(g(x))p(g(x))+v(g(x))q(g(x))$, thus the numerator and denominator of $p(g(x))/q(g(x))$ are relatively prime, so $q$ is a constant.) | |
| Dec 1, 2014 at 15:57 | comment | added | Gorav Jindal | @DavidSpeyer, even something weaker would suffice. If $g_i$'s are well defined on all of $\mathbb{F}_{2}^{n}$ then also we should be done. | |
| Dec 1, 2014 at 15:50 | comment | added | David E Speyer | @GoravJindal Good point! I also checked the Schinzel reference, it does give that $g$ is a polynonial (in the $n=2$ case) but it doesn't say in an obvious place that the $h_j$ are polynomials. So, I agree, if unirational implied rational and if we knew the $g_j$ were polynomials, we'd be done. I'm still skeptical, though. | |
| Dec 1, 2014 at 15:42 | comment | added | Gorav Jindal | @PeterMueller, I think it is okay even if we have $f_{i}=\frac{h_{i}(g)}{k_{i}(g)}$. I can assume $k_{i}(g)$ to be non zero since $f_{i}$ is defined on all of $\mathbb{F}_{2}^{2}$, then his argument of at most 2 different values still works. | |
| Dec 1, 2014 at 15:28 | comment | added | David E Speyer | @GoravJindal Besides needing the $g_i$ to be polynomials, don't you need the $h_j$ to be polynomials, where $f_j = h_j(g_1, \ldots, g_{n-1})$? | |
| Dec 1, 2014 at 15:27 | comment | added | David E Speyer | The second is that, if $f_i = h_i(g_1, \ldots, g_{n-1})$ for $h_i$ a rational function and $g_j$ some rational functions, then $f_i(a)$ need not be determined by $(g_1(a), \ldots, g_{n-1}(a))$. Now that I think about it, I'm a little confused about how to write PeterMueller's argument to address this in the $n=2$ case as well. I think about it in terms of abstract algebraic curves, but I run into the same problem once I get to surfaces. | |
| Dec 1, 2014 at 15:22 | comment | added | David E Speyer | @eig There are two flaws with this argument. The first is that unirational (what you can deduce from being a subfield of $\mathbb{F}_2(x_1, \ldots, x_n)$) is weaker than rational (isomorphic to $\mathbb{F}_2(g_1, \ldots, g_{n-1})$) once $n>2$. See en.wikipedia.org/wiki/Zariski_surface . | |
| Dec 1, 2014 at 15:21 | comment | added | Gorav Jindal | @PeterMueller, Let's try to extend your proof for larger $n$. Assume that if $\mathbb{F}_{2}[f_{1},f_{2},\ldots,f_{n}]$ has transcendence degree $k<n$ over $\mathbb{F}_{2}$ then $\mathbb{F}_{2}(f_{1},f_{2},\ldots,f_{n})=\mathbb{F}_{2}(g_{1},g_{2},\ldots,g_{k})$ for $g_{i}\in\mathbb{F}_{2}(x_{1},x_{2},\ldots,x_{n})$, is the only issue is that how to make sure that these $g_{i}$ can be chosen from $\mathbb{F}_{2}[x_{1},x_{2},\ldots,x_{n}]$ instead of $\mathbb{F}_{2}(x_{1},x_{2},\ldots,x_{n})$? In particular, we just need that all $g_{i}'s$ are defined on all of $\mathbb{F}_{2}^{n}$? | |
| Dec 1, 2014 at 15:19 | comment | added | eig | @PeterMueller if $f_1,\dots,f_n$ are algebraically dependent, can we say that there are polynomials $g_1,\dots,g_{n-1} \in \mathbb{F}_2[x_1,\dots,x_n]$ where $\mathbb{F}_2(f_1,\dots,f_n)=\mathbb{F}_2(g_1,\dots,g_{n-1})$ ? If this is the case, then we can generalized the arguement to any $n$ right ? | |
| Dec 1, 2014 at 15:14 | comment | added | David E Speyer | I forgot to add the keyword in my previous comment: "smooth RATIONAL cubic surface". (It contains a pair of skew lines defined over $\mathbb{F}_2$, namely $X=W=0$ and $Y=Z=0$.) | |
| Dec 1, 2014 at 14:33 | comment | added | David E Speyer | @DavidLampert I think Zariski surfaces en.wikipedia.org/wiki/Zariski_surface should give a counter-example to unirational --> rational in this setting. But the bigger problem is that rational surfaces are just much wilder than rational curves. For example, $X^2 Y + X Y^2 + W^2 Z + W Z^2 =0$ is a smooth cubic surface over $\mathbb{F}^2$ with $15$ $\mathbb{F}_2$ points. | |
| Dec 1, 2014 at 14:22 | comment | added | David Lampert | @DavidSpeyer: Neat answer! Are there "Castelonuovo theorem" (unirational => rational) results on surfaces with rational points over finite fields that would work for n=3? Your answer seems to suggest that you don't think so... | |
| Dec 1, 2014 at 12:44 | history | edited | David E Speyer | CC BY-SA 3.0 |
added 287 characters in body
|
| Dec 1, 2014 at 12:42 | comment | added | David E Speyer | @PeterMueller I think the $h_i$ should be rational functions, not polynomials. So $(f_1(a), f_2(a))$ can assume three values, not two. But that is good enough, and otherwise I agree. Thanks! | |
| Dec 1, 2014 at 11:51 | comment | added | Peter Mueller | ... continued: The proof of the background result in Schinzel is quite elementary, like the direct field theoretic proofs of Lüroth's Theorem. | |
| Dec 1, 2014 at 11:49 | comment | added | Peter Mueller | This proof can be elementarized quite a bit: If $f_1$ and $f_2$ are algebraically dependent, then $\mathbb F_2(f_1,f_2)\subseteq\mathbb F_2(x_1,x_2)$ has transcendence degree $1$ over $\mathbb F_2$. By Theorem 4, Chapter 1 in Schinzel's book `Polynomials with special regard to reducibility', there is a polynomial $g\in\mathbb F_2[x_1,x_2]$ with $\mathbb F_2(f_1,f_2)=\mathbb F_2(g)$. So $f_i=h_i(g(x_1,x_2))$ for univariate polynomials $h_i$. In particular, the pairs $(f_1(a),f_2(a))$ assume at most $2$ different values, while the assumption requires $4$ values. | |
| Dec 1, 2014 at 2:48 | history | answered | David E Speyer | CC BY-SA 3.0 |