Example 10.34.23. Let $k$ be a field. The space $\mathop{\mathrm{Spec}}(k[x, y]/(xy))$ has two irreducible components: namely the $x$-axis and the $y$-axis. As a generalization, let

where $\mathfrak a$ is the ideal in $k[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12}, y_{21}, y_{22}]$ generated by the entries of the $2 \times 2$ product matrix

In this example we will describe $\mathop{\mathrm{Spec}}(R)$.

To prove the statement about $\mathop{\mathrm{Spec}}(k[x, y]/(xy))$ we argue as follows. If $\mathfrak p \subset k[x, y]$ is any ideal containing $xy$, then either $x$ or $y$ would be contained in $\mathfrak p$. Hence the minimal such prime ideals are just $(x)$ and $(y)$. In case $k$ is algebraically closed, the $\text{max-Spec}$ of these components can then be visualized as the point sets of $y$- and $x$-axis.

For the generalization, note that we may identify the closed points of the spectrum of $k[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12}, y_{21}, y_{22}])$ with the space of matrices

at least if $k$ is algebraically closed. Now define a group action of $\text{GL}(2, k)\times \text{GL}(2, k)\times \text{GL}(2, k)$ on the space of matrices $\{ (X, Y)\} $ by

Here, also observe that the algebraic set

is irreducible since it is the max spectrum of the domain

Since the image of irreducible an algebraic set is still irreducible, it suffices to classify the orbits of the set $\{ (X, Y)\in \text{Mat}(2, k)\times \text{Mat}(2, k)|XY = 0\} $ and take their closures. From standard linear algebra, we are reduced to the following three cases:

$\exists (g_1, g_2)$ such that $g_1Xg_2^{-1} = I_{2\times 2}$. Then $Y$ is necessarily $0$, which as an algebraic set is invariant under the group action. It follows that this orbit is contained in the irreducible algebraic set defined by the prime ideal $(y_{11}, y_{12}, y_{21}, y_{22})$. Taking the closure, we see that $(y_{11}, y_{12}, y_{21}, y_{22})$ is actually a component.

$\exists (g_1, g_2)$ such that

\[ g_1Xg_2^{-1} = \left( \begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix} \right). \]This case occurs if and only if $X$ is a rank 1 matrix, and furthermore, $Y$ is killed by such an $X$ if and only if

\[ x_{11}y_{11}+x_{12}y_{21} = 0; \quad x_{11}y_{12}+x_{12}y_{22} = 0; \]\[ x_{21}y_{11}+x_{22}y_{21} = 0; \quad x_{21}y_{12}+x_{22}y_{22} = 0. \]Fix a rank 1 $X$, such non zero $Y$'s satisfying the above equations form an irreducible algebraic set for the following reason($Y = 0$ is contained the previous case): $0 = g_1Xg_2^{-1}g_2Y$ implies that

\[ g_2Y = \left( \begin{matrix} 0 & 0 \\ y_{21}' & y_{22}' \end{matrix} \right). \]With a further $\text{GL}(2, k)$-action on the right by $g_3$, $g_2Y$ can be brought into

\[ g_2Yg_3^{-1} = \left( \begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix} \right), \]and thus such $Y$'s form an irreducible algebraic set isomorphic to the image of $\text{GL}(2, k)$ under this action. Finally, notice that the â€śrank 1" condition for $X$'s forms an open dense subset of the irreducible algebraic set $\det X = x_{11}x_{22} - x_{12}x_{21} = 0$. It now follows that all the five equations define an irreducible component $(x_{11}y_{11}+x_{12}y_{21}, x_{11}y_{12}+x_{12}y_{22}, x_{21}y_{11} +x_{22}y_{21}, x_{21}y_{12}+x_{22}y_{22}, x_{11}x_{22}-x_{12}x_{21})$ in the open subset of the space of pairs of nonzero matrices. It can be shown that the pair of equations $\det X = 0$, $\det Y = 0$ cuts $\mathop{\mathrm{Spec}}(R)$ in an irreducible component with the above locus an open dense subset.

$\exists (g_1, g_2)$ such that $g_1Xg_2^{-1} = 0$, or equivalently, $X = 0$. Then $Y$ can be arbitrary and this component is thus defined by $(x_{11}, x_{12}, x_{21}, x_{22})$.

## Comments (3)

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