Timeline for Is Multilinear Hilbert's tenth problem version undecidable?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2019 at 22:46 | comment | added | Will Sawin | @MattF. The coefficients of $x,y,z$ in the linear form you get by fixing the other variables are all zero mod $2$ in that case. My method only finds solutions where at least one of those coefficients is nonzero. | |
| Sep 7, 2019 at 21:49 | comment | added | user44143 | I don't understand how that works, e.g. for $f(x,y,z)=xy+yz+xz+2x+1$. Then $(1,1,1)$ is a root mod $2$ which does not lift to any roots mod $4$. (That is not a counterexample to the claim, since $(0,1,3)$ is a root mod $4$, but it makes me realize that I don't understand how the proof works.) | |
| Sep 6, 2019 at 12:30 | comment | added | Will Sawin | @MattF. Yes if the coefficients do not all together have a common factor. | |
| Sep 6, 2019 at 12:29 | comment | added | user44143 | Does this give an argument that if there is a solution mod $p$, then there is also a solution mod $p^k$? | |
| Aug 23, 2019 at 16:11 | comment | added | user44143 | Thanks! I am glad to see the lemma and the impression of plausibility. | |
| Aug 23, 2019 at 15:16 | history | answered | Will Sawin | CC BY-SA 4.0 |