Timeline for Recommendations for binomial system solver
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 21, 2013 at 3:46 | vote | accept | ssquidd | ||
| Feb 20, 2013 at 8:29 | comment | added | Thomas Kahle | Yes it is this product. The reason is also in the Eisenbud Sturmfels paper. In their terminology you are asking: How many saturations does the character defined by your Laurent lattice ideal have. The answer is: The order of the finite group $\text{sat}(L)/L$ and the Smith normal form computes a canonical embedding of L into $ZZ^d$ from which you extract it. | |
| Feb 19, 2013 at 17:00 | comment | added | ssquidd | Thank you for your answer! I'm reading your paper right now. About the last paragraph, I know I can find the parameterization (and hence dimension) from Smith normal form easily. But how can I find the number of component? First since I only want nonzero solutions, I can turn each binomial equation into the form "Laurent monomial = constant". With this, let $A$ be the integer matrix made from the exponents. After turning it into Smith Normal form i.e. $PAQ$ is diagonal, is the absolute value of the product of the nonzero entries on the diagonal the number of components? | |
| Feb 19, 2013 at 7:37 | history | answered | Thomas Kahle | CC BY-SA 3.0 |