Timeline for Multiplicative identity of determinant of multiplicative action of a polynomial on a quotient ring (companion matrices)
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 21, 2021 at 12:41 | comment | added | Arrow | perhaps you'd like to edit your answer, so that people are not misled? | |
| Sep 21, 2021 at 0:02 | comment | added | Mohan | @Arrow If $F=M\oplus N $ and $\phi$ is an endomorphism of $F$ such that $\phi(M)\subset M$ then $\phi$ induces an endomorphism of $M$ and $F/M=N$. One can easily check that $\phi$ can be put in the desired form. | |
| Sep 20, 2021 at 23:40 | comment | added | Arrow | Dear @Mohan, is there a way to see the block triangular decomposition from the splitting of the exact sequence? I do not see a way to circumvent the use of the basis above. | |
| Sep 20, 2021 at 23:35 | comment | added | Mohan | @Arrow Yes, I was sloppy, it is block matrix with zeroes on one block. The basis you need is $1, x,x^2,\ldots, x^{n-1}$, where $\deg f_2=n$ and $f_2,f_2x,\ldots, f_2x^{m-1}$ where $\deg f_1=m$. | |
| Sep 20, 2021 at 22:59 | comment | added | Arrow | Also, how does the direct sum decomposition force the multiplication by $g$ in the middle to be blockwise diagonal multiplication by $g$ on the factors? I'd have thought the matrix would at least be block-triangular. | |
| Sep 20, 2021 at 21:30 | comment | added | Arrow | Dear Mohan, do you by any chance see an explicit basis of $A[x]/\langle f_1f_2\rangle$ for which the matrix representation of the multiplication map has the asserted block diagonal form? | |
| Sep 20, 2021 at 21:13 | vote | accept | Arrow | ||
| Sep 20, 2021 at 19:58 | vote | accept | Arrow | ||
| Sep 20, 2021 at 21:10 | |||||
| Sep 20, 2021 at 19:36 | vote | accept | Arrow | ||
| Sep 20, 2021 at 19:58 | |||||
| Sep 20, 2021 at 14:56 | history | answered | Mohan | CC BY-SA 4.0 |