Timeline for Multivariable multilinear homogeneous polynomials with co-prime coefficients representing 1
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| May 20, 2019 at 19:56 | comment | added | Lenny Fukshansky | @alpoge Thank you very much, very helpful remark! | |
| May 20, 2019 at 17:55 | comment | added | alpoge | [Oh! Sorry, 5x+2 isn’t a problem since you of course wrote that it is homogeneous.] | |
| May 20, 2019 at 17:46 | comment | added | alpoge | I’m too lazy to try to deal with the general case unless it’s strictly necessary (note that some condition is required because e.g. 5x+2 has coprime coefficients but will never represent 1) —- probably having a solution locally everywhere is sufficient, but I haven’t thought about it. | |
| May 20, 2019 at 17:45 | comment | added | alpoge | Evidently m\leq n. Wlog F depends on all x_i. Let me assume that for all p, F mod p also depends on all x_i (i.e. for all i, F is not of the form P(x_1, ..., x_{i-1}, k x_i, x_{i+1}, ..., x_n) for P integral). Then write F(x_1, ..., x_n) =: x_n G(x_1, ..., x_{n-1}) + H(x_1, ..., x_{n-1}). Note that G is in n-1 variables and of degree m-1 and satisfies the hypothesis of the question. It suffices to find a solution to G(x_1, ..., x_{n-1}) = 1, since then we may simply solve for x_n. Thus by induction we reduce to the case m = 1, i.e. F is affine linear, in which case it is evident. | |
| May 20, 2019 at 16:31 | comment | added | Lenny Fukshansky | Let us continue this discussion in chat. | |
| May 20, 2019 at 15:48 | comment | added | Lenny Fukshansky | All I mean is that degree of $F$ in each of the variables is $1$. Sorry for the confusion -- I made the edit to the question. | |
| May 20, 2019 at 15:47 | history | edited | Lenny Fukshansky | CC BY-SA 4.0 |
added 5 characters in body
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| May 20, 2019 at 7:08 | comment | added | pavl0 | I am most likely mistaken, but shouldn't we have that $F(x+1,y,z)=F(x,y,z)+F(1,y,z)$, where $F(x,y,z)=xy+xz+yz$? | |
| May 20, 2019 at 5:57 | comment | added | Lenny Fukshansky | No, for example the quadratic form $xy+xz+yz$ is linear in each variable, and still assumes the value $1$ at the point $(1,1,0)$. | |
| May 20, 2019 at 5:45 | comment | added | pavl0 | From the above it follows that $F(x_1,\ldots,x_n)=ax_1\dots x_n$, where $a \in \mathbb{Z}$. This is because $F(0,x_2,\ldots,x_n)$ is homogeneous of degree m with n-1 variables, and still linear in each variable, thus $F(0,x_2,\ldots,x_n)=0$. Therefore, $F(x_1,\ldots,x_n)=1$ iff $a=\pm 1$. | |
| May 20, 2019 at 5:23 | comment | added | pavl0 | By the way, would linearity imply that $m=n$, as $F(x,\ldots,x)=x^nF(1,\ldots,1)$? | |
| May 20, 2019 at 4:44 | comment | added | pavl0 | they are dealing homogeneous forms in $n-1$ variables of degree $n(n-1)/2$. | |
| May 20, 2019 at 3:48 | comment | added | Lenny Fukshansky | @pavl0 can you please clarify? The papers you are referring to all deal with one variable polynomials. | |
| May 19, 2019 at 8:10 | comment | added | pavl0 | A special case of this relates to deciding whether a number field is monogenic (in this case $m=n/(n-1)/2$). This was question 6 in Narkiewicz and answered by Uchita, plus given an effective method to check it by Györy. Denis Simons also has a paper for this special case. (This ignores linearity of variables). | |
| May 16, 2019 at 23:32 | history | asked | Lenny Fukshansky | CC BY-SA 4.0 |