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Let there be ring $A = \mathbb{C}[x_{ij}, y_{ij}| i,j \in {1,2}] $ It's coordinate ring of 8 variables.

Consider $I$ generated by $4$ elements $x_{11}y_{11}+x_{12}y_{21} , x_{11}y_{12} + x_{12}y_{22}, x_{21}y_{11}+x_{22}y_{12}, x_{21}y_{12}+x_{22}y_{22}$

In other words, $I$ is generated by entries of the product $\begin{pmatrix}x_{11} & x_{12} \\ x_{21} & x_{22} \end{pmatrix} \cdot \begin{pmatrix}y_{11} & y_{12} \\ y_{21} & y_{22} \end{pmatrix}$

I want to find the equations that describe the irreducible components of vanishing set $V(I)$ where $V(I)$ is the set of $\mathbb{A_\mathbb{C}^8}$ where all the polynomials in $I$ vanish. Irreducible components of $V(I)$ are the components coinciding to minimal prime ideals.

Observation: after substitution of $x,y$ by an array from vanishing set, we obtain a zero product matrix, that means, that either all $x$s or all $y$s = 0, or both matrices have rank $1$. So, first matrix leaves a non-zero eigenvector that is going to zero after second matrix multiplication.

I get $ay_{11}+by_{12}=ay_{21}+by_{22}$ and $ax_{11}+bx_{12}=ax_{12}+bx_{22}=0$ for some $(a,b)$

However, I don't know how to go about it.

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  • $\begingroup$ Observation: If either of the two matrices is invertible, then the other must be $0$. If either matrix is $0$, the other matrix can be anything. That leaves the case when both matrices have rank $1$. In that case, two of the equations are redundant, as the rows of $X$ are proportional and the columns of $Y$ are proportional (although one can't say definitely which two equations are redundant, as for instance any of the rows of $X$ or any of the columns of $Y$ could be $0$). $\endgroup$ – Arthur Apr 25 at 16:23
  • $\begingroup$ Are you looking for a parametrization of $V(I)$? $\endgroup$ – Servaes Apr 25 at 17:05
  • $\begingroup$ @Servaes Edited, I'm looking for polynomials common zeroes of which are Irreducible components of $A(I)$ in Zariski topology. $\endgroup$ – Lada Dudnikova Apr 25 at 19:51
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    $\begingroup$ The quotient ring is an integral domain and in the fraction field you get $y_{2i} = \frac{x_{11}}{x_{12}}-\frac{x_{21}}{x_{22}}, y_{1i} = \frac{-x_{12}y_{21}}{x_{11}}$, you can assume the denominators are non-zero then add the few remaining cases $\endgroup$ – reuns Apr 25 at 20:00
  • $\begingroup$ @reuns, could you elaborate your statement? How would equations in field of fractions help me? $\endgroup$ – Lada Dudnikova Apr 27 at 20:32

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