Consider the polynomial ring $R=\mathbb C[x_1,x_2,...,x_{16}]$, and set

$$X=\begin{pmatrix} x_1 &x_2&x_3 &x_4\\ x_5&x_6& x_7&x_8\\x_9&x_{10}&x_{11}&x_{12}\\x_{13}&x_{14}&x_{15}&x_{16}\end{pmatrix}.$$

Now, using these three matrices

$$L=\begin{pmatrix}0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0 \end{pmatrix}$$ $$M=\begin{pmatrix}0&0&0&-1\\0&0&-1&0\\0&1&0&0\\1&0&0&0\end{pmatrix}$$ $$N=\begin{pmatrix}0&0&-1&0\\0&0&0&1\\1&0&0&0\\0&-1&0&0\end{pmatrix}$$

we create polynomials $f_i, g_i,$ and $h_i$ in the following way:

$$XLX^t-L=\begin{pmatrix} f_1 &f_2&f_3 &f_4\\ f_5&f_6& f_7&f_8\\f_9&f_{10}&f_{11}&f_{12}\\f_{13}&f_{14}&f_{15}&f_{16}\end{pmatrix}$$

$$XMX^t-M=\begin{pmatrix} g_1 &g_2&g_3 &g_4\\ g_5&g_6& g_7&g_8\\g_9&g_{10}&g_{11}&g_{12}\\g_{13}&g_{14}&g_{15}&g_{16}\end{pmatrix}$$

$$XNX^t-N=\begin{pmatrix} h_1 &h_2&h_3 &h_4\\ h_5&h_6& h_7&h_8\\h_9&h_{10}&h_{11}&h_{12}\\h_{13}&h_{14}&h_{15}&h_{16}\end{pmatrix}$$

Finally, let $I = (f_i, g_i, h_i)$ be the ideal generated by these $48$ polynomials. Then how to show that the radical of $I$, i.e. $\sqrt I$, is generated by twelve linear polynomials and one quadratic polynomial ?

Nullstellensatz may be of help ... but I can't quite see it ...

NOTE : All the matrices $L,M,N$ are orthogonal , so the three defining equations can be written as $(XL)(LX)^t=(XM)(MX)^t=(XN)(NX)^t=Id$. Now if we can find some pattern in $XL,LX,MX,XM,NX,XN$ then it could be helpful to find the zero set of the ideal $I$ ... Also $L,M,N$ are skew symmetric matrices and , $LM=-N$ ... this means $L,M,N$ works as the $i,j,k$ in the Quaternion ring ...

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    $\begingroup$ Could you give some motivation for why you believe that the radical should be generated in the degrees you list? $\endgroup$ – dhy Sep 5 '18 at 3:45
  • $\begingroup$ @dhy: well ... no motivation ...I was just told by my professor that it is true ... he asked me to try it out saying it would be interesting ... and he also asked me to see if I can say anything about $I$ itself ... unfortunately I haven't even been able to do even the radical ... $\endgroup$ – user521337 Sep 5 '18 at 4:07
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    $\begingroup$ In M2 associatedPrimes yields $\langle {x}_{12}+{x}_{15},{x}_{11}-{x}_{16},{x}_{10}+{x}_{13},{x}_{9}-{x}_{14},{x}_{8}+{x}_{14},{x}_{7}-{x}_{13},{x}_{6}-{x}_{16},{x}_{5}+{x}_{15},{x}_{4}+{x}_{13},{x}_{3}+{x}_{14},{x}_{2}-{x}_{15},{x}_{1}-{x}_{16},{x}_{13}^{2}+{x}_{14}^{2}+{x}_{15}^{2}+{x}_{16}^{2}-1\rangle$. $\endgroup$ – Jan-Magnus Økland Sep 6 '18 at 5:06

First, let us do all the calculations over $\mathbb{R}$ onstead of over $\mathbb{C}$. It will facilitate things, and will not change the result.

Let us rewrite the equations a bit: You have $$X\cdot (LX^tL^t)=Id$$ and similarly for $M$ and $N$. In particular, your system is now equivalent to the sent of equations; $$LX^tL^t = MX^tM^t = NX^tN^t= X^{-1}.$$ The first two equalities mean that $X^t$ commutes with the matrices $M^{-1}L$ and $N^{-1}L$ (since for these matrices the transpose is the inverse). This is a set of linear equations on $X^t$ (and hence on $X$). Since $LM=-N$ and so on, this is equivalent to the fact that $X^t$ commutes with $L$, $M$, and $N$. Let now $H$ be the four dimensional quaternion algebra with basis $\{1,i,j,k\}$. Then $L$, $M$, and $N$ represent the multiplication from the left by $i$, $j$, and $k$ respectively. 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. Let us write such an element as $a+bi+cj+dk$. Then you get that your matrix $X$ has the form $$\begin{pmatrix} a & -b & -c & -d \\ b & a & d& -c \\ c & -d& a& b \\ d & c& -b& a \\ \end{pmatrix} $$ Writing the original entries of $X$ in terms of these numbers gives you the desired 12 equations. We are now left with the equation $$XLX^tL^t=Id$$ which is quadratic. Notice that since $X$ commutes with $L$, and since $L$ is orthogonal, this is equivalent to $$XX^t=Id.$$ But this can easily be seen, by a direct calculation, to be equivalent to $a^2+b^2+c^2+d^2 = 1$. Since the polynomial $a^2+b^2+c^2+d^2-1$ is irreducible, you get that the ideal $I$ is radical, and tha it is generated by 12 linear polynomials and one quadratic polynomial.

  • $\begingroup$ 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 ? $\endgroup$ – user521337 Sep 6 '18 at 23:01
  • $\begingroup$ 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 ? $\endgroup$ – user521337 Sep 6 '18 at 23:01
  • $\begingroup$ 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. $\endgroup$ – Ehud Meir Sep 7 '18 at 0:49
  • $\begingroup$ 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. $\endgroup$ – Ehud Meir Sep 7 '18 at 0:51
  • $\begingroup$ 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 ? $\endgroup$ – user521337 Sep 7 '18 at 14:07

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