Skip to main content
7 events
when toggle format what by license comment
Dec 3, 2023 at 16:12 vote accept Oleksandr Kulkov
Dec 3, 2023 at 0:29 answer added Oleksandr Kulkov timeline score: 2
Dec 1, 2023 at 14:36 comment added Aleksei Kulikov While above comments are of course correct, life is much more prosaic: $A_d$ is simply the characteristic polynomial of $A^2$. So, it has integer coefficients. Then, $A_s^2 = (A*A)/A_d$ also has integer coefficients, thus $A_s$ has integer coefficients (by Gauss's lemma, say).
Dec 1, 2023 at 13:36 comment added Dave Benson $A_s$ is the characteristic polynomial of the exterior square, and $A_dA_s$ is the characteristic polynomial of the symmetric square. They multiply up to the tensor square, and their quotient is $A_d$.
Dec 1, 2023 at 13:33 comment added Martin Rubey If the degree of $A$ is $n$, then $A_d(x) = \sum_{k=0}^n m_{2^k}(\lambda_1,\dots,\lambda_n) x^{n-k}$, where $m_{2^k}$ is the monomial symmetric function for the partition $(2,\dots,2)$. The monomial symmetric functions are integer linear combinations of the elementary symmetric functions. Finally, the elementary symmetric functions are the coefficients of $A$.
Dec 1, 2023 at 13:22 history edited YCor
edited tags
Dec 1, 2023 at 12:27 history asked Oleksandr Kulkov CC BY-SA 4.0