radical of a certain ideal of sixteen variable polynomial ring, generated by the entries of certain matrices

Question

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 ...

Answer 1

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.