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

$R = k[x_{11}, x_{12}, x_{21}, x_{22}, y_{11}, y_{12}, y_{21}, y_{22}]/ \mathfrak a,$

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

$\left( \begin{matrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{matrix} \right) \left( \begin{matrix} y_{11} & y_{12} \\ y_{21} & y_{22} \end{matrix} \right).$

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

$\left\{ (X, Y) \in \text{Mat}(2, k)\times \text{Mat}(2, k) \mid X = \left( \begin{matrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{matrix} \right), Y= \left( \begin{matrix} y_{11} & y_{12} \\ y_{21} & y_{22} \end{matrix} \right) \right\}$

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

$(g_1, g_2, g_3) \times (X, Y) \mapsto ((g_1Xg_2^{-1}, g_2Yg_3^{-1})).$

Here, also observe that the algebraic set

$\text{GL}(2, k)\times \text{GL}(2, k)\times \text{GL}(2, k) \subset \text{Mat}(2, k)\times \text{Mat}(2, k) \times \text{Mat}(2, k)$

is irreducible since it is the max spectrum of the domain

$k[x_{11}, x_{12}, \ldots , z_{21}, z_{22}, (x_{11}x_{22}-x_{12}x_{21})^{-1} , (y_{11}y_{22}-y_{12}y_{21})^{-1}, (z_{11}z_{22}-z_{12}z_{21})^{-1}].$

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:

1. $\exists (g_1, g_2)$ 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.

2. $\exists (g_1, g_2)$ 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 “rank 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.

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

Comment #6 by Pieter Belmans on

Some minor issues with this tag:

• the label should be example-product-matrices-zero, fixing the spelling of matrixes
• the sentence "we shall also describe Spec(R)" is kind of lost in the flow
• at the end of the second paragraph I would write "point sets of the $y$- and $x$-axis."
• the instances of Mat should be encoded as \text{Mat}, or at least something that doesn't have the current kerning :)
• the same applies to GL

Comment #8 by Pieter Belmans on

There is no missing item, I just messed up.

Comment #18 by Johan on

Thanks. Fixed. This example can still be explained better however.

There are also:

• 4 comment(s) on Section 10.34: Jacobson rings

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