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 ...

`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