On the radical of an ideal in the polynomial ring in 4 variables over complex field

Question

Consider the following ideal in $\mathbb C[X_1,X_2,X_3,X_4]$.

$I=\langle X_1^2+X_2^2-1, X_3^2+X_4^2-1 , X_1X_3+X_2X_4, X_1X_4-X_2X_3-1\rangle$.

Then, is $\sqrt I$ a prime ideal (i.e. due to Hilbert Nullstellensatz, is $V(I)$ an irreducible affine algebraic variety ) ?

[NOTE : For an ideal $I$ in a commutative ring with unity, $\sqrt I$ is the set of all those eleemnts $a\in R$ such that $a+I$ is Nilpotent in $R/I$. It is easy to check that $\sqrt I$ is alwyas an ideal. ]

Answer 1

Computation

The following singular computation:

> LIB "primdec.lib";
...
> ring R=0,(X1,X2,X3,X4),dp;
> ideal I=X1^2+X2^2-1,X3^2+X4^2-1,X1*X3+X2*X4,X1*X4-X2*X3-1;
> radical(I);
_[1]=X2+X3
_[2]=X1-X4
_[3]=X3^2+X4^2-1
> list pr=primdecGTZ(I);
> testPrimary(pr,I);
1

shows that the radical of this ideal is \begin{align*} J &= \sqrt I = ( X_2+X_3, X_1-X_4, X_2^2+X_4^2 - 1) \end{align*} which defines the variety \begin{align*} S_1 &= \left\{ \begin{pmatrix} s & t \\ -t & s \end{pmatrix} ~\middle\vert~ s^2+t^2=1 \right \}. \end{align*}

Math

Consider the variety $C:=Z(x^2+y^2-1)\subseteq\Bbb C^2$. It is defined by the single polynomial $f:=x^2+y^2-1\in\Bbb C[x,y]$. Since $f$ is irreducible, the variety $C$ is also irreducible. Furthermore, the map \begin{align*} \phi:C&\longrightarrow S_1\\ (x,y) &\longmapsto \begin{pmatrix} x&y\\-y&x\end{pmatrix} \end{align*} is an isomorphism of algebraic varieties, therefore $S_1$ is irreducible.

In your question, you gave the ideal $I$ which defines the variety $$S_2:=Z(I)=\{ A\in\Bbb C^ {2\times2} \mid A^TA=I\land\det(A)=1 \},$$ and you asked whether it is irreducible. We will show that $S_1=S_2$ and therefore conclude that it is. The inclusion $S_1\subseteq S_2$ is trivial. Let us assume $$A=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)\in S_2.$$ Then, the points $(a,b),(c,d)\in\Bbb C^2$ both lie on the circle and are orthogonal to each other. Since the orthogonal complement of the line spanned by $(a,b)$ is one-dimensional, there are only two choices for $(c,d)$, differing in sign, because $(c,d)$ is on the intersection of said orthogonal complement with the circle. Hence, $(c,d)=\pm(-b,a)$. The determinant condition yields the desired conclusion.

Remark

The variety $S_2$ can also be written as $S_1=\operatorname{SL}_2(\Bbb C)\cap\operatorname O_2(\Bbb C)$. I understand there is some controversy about calling this group $\operatorname O_2$, however I would use the notation $\operatorname U_2$ when referring to the unitary group and always mean $\operatorname O_2$ to be the group of matrces $A$ with $A^TA=I$. This definition makes sense over any field. Consequently, I would refer to $S_1$ as $\mathrm{SO}_2(\Bbb C)$. Regardless of what you call it however, it is a connected, smooth, abelian curve.