Timeline for Does linear independence imply algebraic independence for partitioned homogeneous polynomials?
Current License: CC BY-SA 4.0
12 events
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| Jan 28, 2019 at 7:06 | history | edited | Turbo | CC BY-SA 4.0 |
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| Jan 28, 2019 at 7:01 | history | edited | Turbo | CC BY-SA 4.0 |
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| Jan 20, 2019 at 10:27 | history | edited | Turbo | CC BY-SA 4.0 |
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| Jan 19, 2019 at 20:02 | comment | added | Laurent Moret-Bailly | Oh, this I don't know. | |
| Jan 19, 2019 at 19:25 | comment | added | Turbo | @LaurentMoret-Bailly I am asking if linear implies algebraic independence is easily seen? | |
| Jan 19, 2019 at 19:03 | comment | added | Laurent Moret-Bailly | Well, any $d$-linear form is a linear combination of forms of the shape given in Wojowu's comment. | |
| Jan 18, 2019 at 9:02 | comment | added | Turbo | @LaurentMoret-Bailly Yes that is correct. | |
| Jan 18, 2019 at 8:23 | comment | added | Laurent Moret-Bailly | In other words, it is a $d$-linear form on $\mathbb{Z}^n$, right? | |
| Jan 17, 2019 at 21:03 | history | edited | Turbo | CC BY-SA 4.0 |
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| Jan 17, 2019 at 20:51 | comment | added | Turbo | @Wojowu Yes that is correct. | |
| Jan 17, 2019 at 20:37 | comment | added | Wojowu | I'm not sure I understand your notation - do you mean that the polynomial is a linear combination of monomials of the form $x_{1a_1}x_{2a_2}\dots x_{da_d}$ with $1\leq a_k\leq n$? | |
| Jan 17, 2019 at 20:20 | history | asked | Turbo | CC BY-SA 4.0 |