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 ... 



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


