Timeline for radical of a certain ideal of sixteen variable polynomial ring, generated by the entries of certain matrices
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 14, 2018 at 0:24 | vote | accept | user521337 | ||
| Sep 10, 2018 at 6:14 | comment | added | user521337 | thanks ... though I still don't understand why $I$ is radical ... by determining $X$ etc. you are actually finding the "Zero Set" of $I$ hence those 12 restrictions actually give the generating polynomials of the idealizer of the zero set of $I$ which is $rad(I)$ ... I don't see why those polynomials should generate $I$ ... | |
| Sep 9, 2018 at 15:22 | comment | added | Ehud Meir | that's more or less the definition of being a module map. X is a module map if $X(Lv) = L(Xv)$, and same for $M$ and $N$. Then, since $1,i,j,k$ span the quaternion algebra, it is a module map. | |
| Sep 8, 2018 at 18:13 | comment | added | user521337 | ah that makes sense ... thanks ... then why $X$ commuting with $L,M,N$ implies that the map on $H$ is a $H$ module homomorphism ? | |
| Sep 8, 2018 at 12:42 | comment | added | Ehud Meir | Write $H = \mathbb{R}^4$. Call the standard basis elements $\{1,i,j,k\}$. Then $X,L,M$ and $N$ are $4\times 4$ matrices, and therefore can be considered as linear maps $H\to H$. | |
| Sep 7, 2018 at 22:31 | comment | added | user521337 | because if you already know $X$ is a linear map $H\to H$, then apriori $X=X.1\in H$ ... so how do you know that $X$ maps every element of $H$ into $H$ itself ? | |
| Sep 7, 2018 at 21:56 | comment | added | user521337 | Yes but if say $LX$ or $XL$ is not in $H$, then how can you think of $X$ as a linear map $H \to H$ ? If you are considering the map $h\to Xh, \forall h\in H$, then in that case it should only be a linear map from $H $ to $M_4(K)$ ... am I missing something obvious here ... ? | |
| Sep 7, 2018 at 15:13 | comment | added | Ehud Meir | You do not need that $LX$ is in $H$. We think of $X$ as a linear transformation $H\to H$. The fact that this linear transformation commutes with $L$ $M$ and $N$ is equivalent to this being a homomorphism of $H$-modules. About $I$ being a radical ideal: what we actually do is the following. Write $J$ for the subideal of $I$ generated by the linear polynomials in $I$. Then you can consider the image of $I$ in the quotient ring $R/J$. It is enough to prove that $I/J$ is a radical ideal there, and this follows from the explicit description of the equations of $XX^t=Id$ | |
| Sep 7, 2018 at 14:07 | comment | added | user521337 | but $LX$ is not in $H$ ... so how do you get a homomorphism $H \to H$ ? Please could you elaborate ... I can't understand this point ... and how did you conclude at the last step that $I$ is a radical ideal ? | |
| Sep 7, 2018 at 0:51 | comment | added | Ehud Meir | About the second part: Let us write $\mathbb{R}^4 = H$ for the Quaternions. Then we can think of $L$, $M$ and $N$ in the way described. A matrix $X$ which commutes with $L$, $M$ and $N$ is the same as a homomorphism $H\to H$ of $H$ modules. But such a maps always has the form $x\mapsto xy$ for some fixed $y$. Then a direct calculation shows that for $y=a+bi+cj+dk$ the map we get is of the above form. Again, this can also be seen by a direct calculation, using $X=LXL^{-1}$ et cetera. | |
| Sep 7, 2018 at 0:49 | comment | added | Ehud Meir | Since the entries of the matrices are all real, you can show easily that the generating set you give for I is a collection of polynomials with real coefficients. so $\mathbb{R}[x_1,\ldots, x_{16}]/\tilde{I}\otimes_{\mathbb{R}}\mathbb{C}$ is the same as the ring you are interested in. In any case, viewing this over $\mathbb{R}$ is just good for a point of view. The fact that if $X$ commutes with $L$, $M$ and $N$ it must be of the specific form follows also from a direct calculation. Notice that when I say that $X$ must be of this form, I just say that $X$ satisfies some linear equations. | |
| Sep 6, 2018 at 23:01 | comment | added | user521337 | I don't under stand where you say "Since the quaternions form a division algebra, the only thing which commute with multiplication from the left with L, M and N is multiplication from the right with someone from the quaternion algebra" .. what do you mean by this ? Could you please elaborate ? | |
| Sep 6, 2018 at 23:01 | comment | added | user521337 | So if I understand correctly ... you are first trying to find the zero set of $I$ right ? Um why doesn't working with $\mathbb R$ change the result ? Also , I could follow your argument upto the point that $X$ commutes with $L,M,N$ which can be considered as the $\{i,j,k\}$ basis elements of the Quaternion algebra ... but then I don't get what do you say about why $X$ has to be of that special form ? | |
| Sep 6, 2018 at 11:09 | history | answered | Ehud Meir | CC BY-SA 4.0 |