Definition 10.35.1. Let $R$ be a ring. We say that $R$ is a *Jacobson ring* if every radical ideal $I$ is the intersection of the maximal ideals containing it.

## 10.35 Jacobson rings

Let $R$ be a ring. The closed points of $\mathop{\mathrm{Spec}}(R)$ are the maximal ideals of $R$. Often rings which occur naturally in algebraic geometry have lots of maximal ideals. For example finite type algebras over a field or over $\mathbf{Z}$. We will show that these are examples of Jacobson rings.

Lemma 10.35.2. Any algebra of finite type over a field is Jacobson.

**Proof.**
This follows from Theorem 10.34.1 and Definition 10.35.1.
$\square$

Lemma 10.35.3. Let $R$ be a ring. If every prime ideal of $R$ is the intersection of the maximal ideals containing it, then $R$ is Jacobson.

**Proof.**
This is immediately clear from the fact that every radical ideal $I \subset R$ is the intersection of the primes containing it. See Lemma 10.17.2.
$\square$

Lemma 10.35.4. A ring $R$ is Jacobson if and only if $\mathop{\mathrm{Spec}}(R)$ is Jacobson, see Topology, Definition 5.18.1.

**Proof.**
Suppose $R$ is Jacobson. Let $Z \subset \mathop{\mathrm{Spec}}(R)$ be a closed subset. We have to show that the set of closed points in $Z$ is dense in $Z$. Let $U \subset \mathop{\mathrm{Spec}}(R)$ be an open such that $U \cap Z$ is nonempty. We have to show $Z \cap U$ contains a closed point of $\mathop{\mathrm{Spec}}(R)$. We may assume $U = D(f)$ as standard opens form a basis for the topology on $\mathop{\mathrm{Spec}}(R)$. According to Lemma 10.17.2 we may assume that $Z = V(I)$, where $I$ is a radical ideal. We see also that $f \not\in I$. By assumption, there exists a maximal ideal $\mathfrak m \subset R$ such that $I \subset \mathfrak m$ but $f \not\in \mathfrak m$. Hence $\mathfrak m \in D(f) \cap V(I) = U \cap Z$ as desired.

Conversely, suppose that $\mathop{\mathrm{Spec}}(R)$ is Jacobson. Let $I \subset R$ be a radical ideal. Let $J = \cap _{I \subset \mathfrak m} \mathfrak m$ be the intersection of the maximal ideals containing $I$. Clearly $J$ is a radical ideal, $V(J) \subset V(I)$, and $V(J)$ is the smallest closed subset of $V(I)$ containing all the closed points of $V(I)$. By assumption we see that $V(J) = V(I)$. But Lemma 10.17.2 shows there is a bijection between Zariski closed sets and radical ideals, hence $I = J$ as desired. $\square$

Lemma 10.35.5. Let $R$ be a ring. If $R$ is not Jacobson there exist a prime $\mathfrak p \subset R$, an element $f \in R$ such that the following hold

$\mathfrak p$ is not a maximal ideal,

$f \not\in \mathfrak p$,

$V(\mathfrak p) \cap D(f) = \{ \mathfrak p\} $, and

$(R/\mathfrak p)_ f$ is a field.

On the other hand, if $R$ is Jacobson, then for any pair $(\mathfrak p, f)$ such that (1) and (2) hold the set $V(\mathfrak p) \cap D(f)$ is infinite.

**Proof.**
Assume $R$ is not Jacobson. By Lemma 10.35.4 this means there exists an closed subset $T \subset \mathop{\mathrm{Spec}}(R)$ whose set $T_0 \subset T$ of closed points is not dense in $T$. Choose an $f \in R$ such that $T_0 \subset V(f)$ but $T \not\subset V(f)$. Note that $T \cap D(f)$ is homeomorphic to $\mathop{\mathrm{Spec}}((R/I)_ f)$ if $T = V(I)$, see Lemmas 10.17.7 and 10.17.6. As any ring has a maximal ideal (Lemma 10.17.2) we can choose a closed point $t$ of space $T \cap D(f)$. Then $t$ corresponds to a prime ideal $\mathfrak p \subset R$ which is not maximal (as $t \not\in T_0$). Thus (1) holds. By construction $f \not\in \mathfrak p$, hence (2). As $t$ is a closed point of $T \cap D(f)$ we see that $V(\mathfrak p) \cap D(f) = \{ \mathfrak p\} $, i.e., (3) holds. Hence we conclude that $(R/\mathfrak p)_ f$ is a domain whose spectrum has one point, hence (4) holds (for example combine Lemmas 10.18.2 and 10.25.1).

Conversely, suppose that $R$ is Jacobson and $(\mathfrak p, f)$ satisfy (1) and (2). If $V(\mathfrak p) \cap D(f) = \{ \mathfrak p, \mathfrak q_1, \ldots , \mathfrak q_ t\} $ then $\mathfrak p \not= \mathfrak q_ i$ implies there exists an element $g \in R$ such that $g \not\in \mathfrak p$ but $g \in \mathfrak q_ i$ for all $i$. Hence $V(\mathfrak p) \cap D(fg) = \{ \mathfrak p\} $ which is impossible since each locally closed subset of $\mathop{\mathrm{Spec}}(R)$ contains at least one closed point as $\mathop{\mathrm{Spec}}(R)$ is a Jacobson topological space. $\square$

Lemma 10.35.6. The ring $\mathbf{Z}$ is a Jacobson ring. More generally, let $R$ be a ring such that

$R$ is a domain,

$R$ is Noetherian,

any nonzero prime ideal is a maximal ideal, and

$R$ has infinitely many maximal ideals.

Then $R$ is a Jacobson ring.

**Proof.**
Let $R$ satisfy (1), (2), (3) and (4). The statement means that $(0) = \bigcap _{\mathfrak m \subset R} \mathfrak m$. Since $R$ has infinitely many maximal ideals it suffices to show that any nonzero $x \in R$ is contained in at most finitely many maximal ideals, in other words that $V(x)$ is finite. By Lemma 10.17.7 we see that $V(x)$ is homeomorphic to $\mathop{\mathrm{Spec}}(R/xR)$. By assumption (3) every prime of $R/xR$ is minimal and hence corresponds to an irreducible component of $\mathop{\mathrm{Spec}}(R/xR)$ (Lemma 10.26.1). As $R/xR$ is Noetherian, the topological space $\mathop{\mathrm{Spec}}(R/xR)$ is Noetherian (Lemma 10.31.5) and has finitely many irreducible components (Topology, Lemma 5.9.2). Thus $V(x)$ is finite as desired.
$\square$

Example 10.35.7. Let $A$ be an infinite set. For each $\alpha \in A$, let $k_\alpha $ be a field. We claim that $R = \prod _{\alpha \in A} k_\alpha $ is Jacobson. First, note that any element $f \in R$ has the form $f = ue$, with $u \in R$ a unit and $e\in R$ an idempotent (left to the reader). Hence $D(f) = D(e)$, and $R_ f = R_ e = R/(1-e)$ is a quotient of $R$. Actually, any ring with this property is Jacobson. Namely, say $\mathfrak p \subset R$ is a prime ideal and $f \in R$, $f \not\in \mathfrak p$. We have to find a maximal ideal $\mathfrak m$ of $R$ such that $\mathfrak p \subset \mathfrak m$ and $f \not\in \mathfrak m$. Because $R_ f$ is a quotient of $R$ we see that any maximal ideal of $R_ f$ corresponds to a maximal ideal of $R$ not containing $f$. Hence the result follows by choosing a maximal ideal of $R_ f$ containing $\mathfrak p R_ f$.

Example 10.35.8. A domain $R$ with finitely many maximal ideals $\mathfrak m_ i$, $i = 1, \ldots , n$ is not a Jacobson ring, except when it is a field. Namely, in this case $(0)$ is not the intersection of the maximal ideals $(0) \not= \mathfrak m_1 \cap \mathfrak m_2 \cap \ldots \cap \mathfrak m_ n \supset \mathfrak m_1 \cdot \mathfrak m_2 \cdot \ldots \cdot \mathfrak m_ n \not= 0$. In particular a discrete valuation ring, or any local ring with at least two prime ideals is not a Jacobson ring.

Lemma 10.35.9. Let $R \to S$ be a ring map. Let $\mathfrak m \subset R$ be a maximal ideal. Let $\mathfrak q \subset S$ be a prime ideal lying over $\mathfrak m$ such that $\kappa (\mathfrak q)/\kappa (\mathfrak m)$ is an algebraic field extension. Then $\mathfrak q$ is a maximal ideal of $S$.

**Proof.**
Consider the diagram

We see that $\kappa (\mathfrak m) \subset S/\mathfrak q \subset \kappa (\mathfrak q)$. Because the field extension $\kappa (\mathfrak m) \subset \kappa (\mathfrak q)$ is algebraic, any ring between $\kappa (\mathfrak m)$ and $\kappa (\mathfrak q)$ is a field (Fields, Lemma 9.8.10). Thus $S/\mathfrak q$ is a field, and a posteriori equal to $\kappa (\mathfrak q)$. $\square$

Lemma 10.35.10. Suppose that $k$ is a field and suppose that $V$ is a nonzero vector space over $k$. Assume the dimension of $V$ (which is a cardinal number) is smaller than the cardinality of $k$. Then for any linear operator $T : V \to V$ there exists some monic polynomial $P(t) \in k[t]$ such that $P(T)$ is not invertible.

**Proof.**
If not then $V$ inherits the structure of a vector space over the field $k(t)$. But the dimension of $k(t)$ over $k$ is at least the cardinality of $k$ for example due to the fact that the elements $\frac{1}{t - \lambda }$ are $k$-linearly independent.
$\square$

Here is another version of Hilbert's Nullstellensatz.

Theorem 10.35.11. Let $k$ be a field. Let $S$ be a $k$-algebra generated over $k$ by the elements $\{ x_ i\} _{i \in I}$. Assume the cardinality of $I$ is smaller than the cardinality of $k$. Then

for all maximal ideals $\mathfrak m \subset S$ the field extension $\kappa (\mathfrak m)/k$ is algebraic, and

$S$ is a Jacobson ring.

**Proof.**
If $I$ is finite then the result follows from the Hilbert Nullstellensatz, Theorem 10.34.1. In the rest of the proof we assume $I$ is infinite. It suffices to prove the result for $\mathfrak m \subset k[\{ x_ i\} _{i \in I}]$ maximal in the polynomial ring on variables $x_ i$, since $S$ is a quotient of this. As $I$ is infinite the set of monomials $x_{i_1}^{e_1} \ldots x_{i_ r}^{e_ r}$, $i_1, \ldots , i_ r \in I$ and $e_1, \ldots , e_ r \geq 0$ has cardinality at most equal to the cardinality of $I$. Because the cardinality of $I \times \ldots \times I$ is the cardinality of $I$, and also the cardinality of $\bigcup _{n \geq 0} I^ n$ has the same cardinality. (If $I$ is finite, then this is not true and in that case this proof only works if $k$ is uncountable.)

To arrive at a contradiction pick $T \in \kappa (\mathfrak m)$ transcendental over $k$. Note that the $k$-linear map $T : \kappa (\mathfrak m) \to \kappa (\mathfrak m)$ given by multiplication by $T$ has the property that $P(T)$ is invertible for all monic polynomials $P(t) \in k[t]$. Also, $\kappa (\mathfrak m)$ has dimension at most the cardinality of $I$ over $k$ since it is a quotient of the vector space $k[\{ x_ i\} _{i \in I}]$ over $k$ (whose dimension is $\# I$ as we saw above). This is impossible by Lemma 10.35.10.

To show that $S$ is Jacobson we argue as follows. If not then there exists a prime $\mathfrak q \subset S$ and an element $f \in S$, $f \not\in \mathfrak q$ such that $\mathfrak q$ is not maximal and $(S/\mathfrak q)_ f$ is a field, see Lemma 10.35.5. But note that $(S/\mathfrak q)_ f$ is generated by at most $\# I + 1$ elements. Hence the field extension $(S/\mathfrak q)_ f/k$ is algebraic (by the first part of the proof). This implies that $\kappa (\mathfrak q)$ is an algebraic extension of $k$ hence $\mathfrak q$ is maximal by Lemma 10.35.9. This contradiction finishes the proof. $\square$

Lemma 10.35.12. Let $k$ be a field. Let $S$ be a $k$-algebra. For any field extension $K/k$ whose cardinality is larger than the cardinality of $S$ we have

for every maximal ideal $\mathfrak m$ of $S_ K$ the field $\kappa (\mathfrak m)$ is algebraic over $K$, and

$S_ K$ is a Jacobson ring.

**Proof.**
Choose $k \subset K$ such that the cardinality of $K$ is greater than the cardinality of $S$. Since the elements of $S$ generate the $K$-algebra $S_ K$ we see that Theorem 10.35.11 applies.
$\square$

Example 10.35.13. The trick in the proof of Theorem 10.35.11 really does not work if $k$ is a countable field and $I$ is countable too. Let $k$ be a countable field. Let $x$ be a variable, and let $k(x)$ be the field of rational functions in $x$. Consider the polynomial algebra $R = k[x, \{ x_ f\} _{f \in k[x]-\{ 0\} }]$. Let $I = (\{ fx_ f - 1\} _{f\in k[x] - \{ 0\} })$. Note that $I$ is a proper ideal in $R$. Choose a maximal ideal $I \subset \mathfrak m$. Then $k \subset R/\mathfrak m$ is isomorphic to $k(x)$, and is not algebraic over $k$.

Lemma 10.35.14. Let $R$ be a Jacobson ring. Let $f \in R$. The ring $R_ f$ is Jacobson and maximal ideals of $R_ f$ correspond to maximal ideals of $R$ not containing $f$.

**Proof.**
By Topology, Lemma 5.18.5 we see that $D(f) = \mathop{\mathrm{Spec}}(R_ f)$ is Jacobson and that closed points of $D(f)$ correspond to closed points in $\mathop{\mathrm{Spec}}(R)$ which happen to lie in $D(f)$. Thus $R_ f$ is Jacobson by Lemma 10.35.4.
$\square$

Example 10.35.15. Here is a simple example that shows Lemma 10.35.14 to be false if $R$ is not Jacobson. Consider the ring $R = \mathbf{Z}_{(2)}$, i.e., the localization of $\mathbf{Z}$ at the prime $(2)$. The localization of $R$ at the element $2$ is isomorphic to $\mathbf{Q}$, in a formula: $R_2 \cong \mathbf{Q}$. Clearly the map $R \to R_2$ maps the closed point of $\mathop{\mathrm{Spec}}(\mathbf{Q})$ to the generic point of $\mathop{\mathrm{Spec}}(R)$.

Example 10.35.16. Here is a simple example that shows Lemma 10.35.14 is false if $R$ is Jacobson but we localize at infinitely many elements. Namely, let $R = \mathbf{Z}$ and consider the localization $(R \setminus \{ 0\} )^{-1}R \cong \mathbf{Q}$ of $R$ at the set of all nonzero elements. Clearly the map $\mathbf{Z} \to \mathbf{Q}$ maps the closed point of $\mathop{\mathrm{Spec}}(\mathbf{Q})$ to the generic point of $\mathop{\mathrm{Spec}}(\mathbf{Z})$.

Lemma 10.35.17. Let $R$ be a Jacobson ring. Let $I \subset R$ be an ideal. The ring $R/I$ is Jacobson and maximal ideals of $R/I$ correspond to maximal ideals of $R$ containing $I$.

**Proof.**
The proof is the same as the proof of Lemma 10.35.14.
$\square$

Lemma 10.35.18. Let $R$ be a Jacobson ring. Let $K$ be a field. Let $R \subset K$ and $K$ is of finite type over $R$. Then $R$ is a field and $K/R$ is a finite field extension.

**Proof.**
First note that $R$ is a domain. By Lemma 10.34.2 we see that $R_ f$ is a field and $K/R_ f$ is a finite field extension for some nonzero $f \in R$. Hence $(0)$ is a maximal ideal of $R_ f$ and by Lemma 10.35.14 we conclude $(0)$ is a maximal ideal of $R$.
$\square$

Proposition 10.35.19. Let $R$ be a Jacobson ring. Let $R \to S$ be a ring map of finite type. Then

The ring $S$ is Jacobson.

The map $\mathop{\mathrm{Spec}}(S) \to \mathop{\mathrm{Spec}}(R)$ transforms closed points to closed points.

For $\mathfrak m' \subset S$ maximal lying over $\mathfrak m \subset R$ the field extension $\kappa (\mathfrak m')/\kappa (\mathfrak m)$ is finite.

**Proof.**
Let $\mathfrak m' \subset S$ be a maximal ideal and $R \cap \mathfrak m' = \mathfrak m$. Then $R/\mathfrak m \to S/\mathfrak m'$ satisfies the conditions of Lemma 10.35.18 by Lemma 10.35.17. Hence $R/\mathfrak m$ is a field and $\mathfrak m$ a maximal ideal and the induced residue field extension is finite. This proves (2) and (3).

If $S$ is not Jacobson, then by Lemma 10.35.5 there exists a non-maximal prime ideal $\mathfrak q$ of $S$ and an $g \in S$, $g \not\in \mathfrak q$ such that $(S/\mathfrak q)_ g$ is a field. To arrive at a contradiction we show that $\mathfrak q$ is a maximal ideal. Let $\mathfrak p = \mathfrak q \cap R$. Then $R/\mathfrak p \to (S/\mathfrak q)_ g$ satisfies the conditions of Lemma 10.35.18 by Lemma 10.35.17. Hence $R/\mathfrak p$ is a field and the field extension $\kappa (\mathfrak p) \to (S/\mathfrak q)_ g = \kappa (\mathfrak q)$ is finite, thus algebraic. Then $\mathfrak q$ is a maximal ideal of $S$ by Lemma 10.35.9. Contradiction. $\square$

Lemma 10.35.20. Any finite type algebra over $\mathbf{Z}$ is Jacobson.

**Proof.**
Combine Lemma 10.35.6 and Proposition 10.35.19.
$\square$

Lemma 10.35.21. Let $R \to S$ be a finite type ring map of Jacobson rings. Denote $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(S)$. Write $f : Y \to X$ the induced map of spectra. Let $E \subset Y = \mathop{\mathrm{Spec}}(S)$ be a constructible set. Denote with a subscript ${}_0$ the set of closed points of a topological space.

We have $f(E)_0 = f(E_0) = X_0 \cap f(E)$.

A point $\xi \in X$ is in $f(E)$ if and only if $\overline{\{ \xi \} } \cap f(E_0)$ is dense in $\overline{\{ \xi \} }$.

**Proof.**
We have a commutative diagram of continuous maps

Suppose $x \in f(E)$ is closed in $f(E)$. Then $f^{-1}(\{ x\} )\cap E$ is nonempty and closed in $E$. Applying Topology, Lemma 5.18.5 to both inclusions

we find there exists a point $y \in f^{-1}(\{ x\} ) \cap E$ which is closed in $Y$. In other words, there exists $y \in Y_0$ and $y \in E_0$ mapping to $x$. Hence $x \in f(E_0)$. This proves that $f(E)_0 \subset f(E_0)$. Proposition 10.35.19 implies that $f(E_0) \subset X_0 \cap f(E)$. The inclusion $X_0 \cap f(E) \subset f(E)_0$ is trivial. This proves the first assertion.

Suppose that $\xi \in f(E)$. According to Lemma 10.30.2 the set $f(E) \cap \overline{\{ \xi \} }$ contains a dense open subset of $\overline{\{ \xi \} }$. Since $X$ is Jacobson we conclude that $f(E) \cap \overline{\{ \xi \} }$ contains a dense set of closed points, see Topology, Lemma 5.18.5. We conclude by part (1) of the lemma.

On the other hand, suppose that $\overline{\{ \xi \} } \cap f(E_0)$ is dense in $\overline{\{ \xi \} }$. By Lemma 10.29.4 there exists a ring map $S \to S'$ of finite presentation such that $E$ is the image of $Y' := \mathop{\mathrm{Spec}}(S') \to Y$. Then $E_0$ is the image of $Y'_0$ by the first part of the lemma applied to the ring map $S \to S'$. Thus we may assume that $E = Y$ by replacing $S$ by $S'$. Suppose $\xi $ corresponds to $\mathfrak p \subset R$. Consider the diagram

This diagram and the density of $f(Y_0) \cap V(\mathfrak p)$ in $V(\mathfrak p)$ shows that the morphism $R/\mathfrak p \to S/\mathfrak p S$ satisfies condition (2) of Lemma 10.30.4. Hence we conclude there exists a prime $\overline{\mathfrak q} \subset S/\mathfrak pS$ mapping to $(0)$. In other words the inverse image $\mathfrak q$ of $\overline{\mathfrak q}$ in $S$ maps to $\mathfrak p$ as desired. $\square$

The conclusion of the lemma above is that we can read off the image of $f$ from the set of closed points of the image. This is a little nicer in case the map is of finite presentation because then we know that images of a constructible is constructible. Before we state it we introduce some notation. Denote $\text{Constr}(X)$ the set of constructible sets. Let $R \to S$ be a ring map. Denote $X = \mathop{\mathrm{Spec}}(R)$ and $Y = \mathop{\mathrm{Spec}}(S)$. Write $f : Y \to X$ the induced map of spectra. Denote with a subscript ${}_0$ the set of closed points of a topological space.

Lemma 10.35.22. With notation as above. Assume that $R$ is a Noetherian Jacobson ring. Further assume $R \to S$ is of finite type. There is a commutative diagram

where the horizontal arrows are the bijections from Topology, Lemma 5.18.8.

**Proof.**
Since $R \to S$ is of finite type, it is of finite presentation, see Lemma 10.31.4. Thus the image of a constructible set in $X$ is constructible in $Y$ by Chevalley's theorem (Theorem 10.29.10). Combined with Lemma 10.35.21 the lemma follows.
$\square$

To illustrate the use of Jacobson rings, we give the following two examples.

Example 10.35.23. 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 $(x_{11}, x_{12}, x_{21}, x_{22})$.

Example 10.35.24. For another example, consider $R = k[\{ t_{ij}\} _{i, j = 1}^{n}]/\mathfrak a$, where $\mathfrak a$ is the ideal generated by the entries of the product matrix $T^2-T$, $T = (t_{ij})$. From linear algebra, we know that under the $GL(n, k)$-action defined by $g, T \mapsto gTg^{-1}$, $T$ is classified by the its rank and each $T$ is conjugate to some $\text{diag}(1, \ldots , 1, 0, \ldots , 0)$, which has $r$ 1's and $n-r$ 0's. Thus each orbit of such a $\text{diag}(1, \ldots , 1, 0, \ldots , 0)$ under the group action forms an irreducible component and every idempotent matrix is contained in one such orbit. Next we will show that any two different orbits are necessarily disjoint. For this purpose we only need to cook up polynomial functions that take different values on different orbits. In characteristic 0 cases, such a function can be taken to be $f(t_{ij}) = trace(T) = \sum _{i = 1}^ nt_{ii}$. In positive characteristic cases, things are slightly more tricky since we might have $trace(T) = 0$ even if $T \neq 0$. For instance, $char = 3$

Anyway, these components can be separated using other functions. For instance, in the characteristic 3 case, $tr(\wedge ^3T)$ takes value 1 on the components corresponding to $diag(1, 1, 1)$ and 0 on other components.

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