App. 1. Algebras without Unit Element

Appendix 1 algebras without unit element

In this appendix, $k$ is a commutative ring; the $k$ -algebras are assumed associative but not necessarily unital.

1. Regular Ideals

Let A be a $k$ -algebra. Recall (III, §1, No. 2, p. 430) that a left ideal of A is a $k$ -submodule $\mathfrak{a}$ of A such that the relations $a\in A, x\in \mathfrak{a}$ imply $ax\in \mathfrak{a}$ . We define the notions of right ideal and two-sided ideal likewise. If A is a $k$ -algebra and $\mathfrak{a}$ a two-sided ideal of A, then when passing to the quotients, the multiplication in A defines a $k$ -algebra structure on the $k$ -module $A/\mathfrak{a}$ (loc. cit.).

A left ideal of A is called maximal if it is a maximal element of the set of proper left ideals of A for the inclusion.

Definition 1 00P4

Let A be a $k$ -algebra and $\mathfrak{a}$ a left ideal of A. An element $u$ of A such that $au-a$ belongs to $\mathfrak{a}$ for every $a\in A$ is called a right unit modulo $\mathfrak{a}$ . We say that the ideal $\mathfrak{a}$ is regular if there exists a right unit modulo $\mathfrak{a}$ .

Cites 1

Cites Theorem 2 ("Zorn's lemma"), E III § 2

Likewise, we say that a right ideal $\mathfrak{b}$ of A is regular if A has a left unit modulo $\mathfrak{b}$ , that is, an element $v$ such that $va-a\in \mathfrak{b}$ for every $a\in A$ . When an ideal is two-sided, one needs to specify whether it is regular as a left ideal or as a right ideal.

When the algebra A is unital, the unit element of A is a right unit modulo $\mathfrak{a}$ for every left ideal $\mathfrak{a}$ of A; in this case, every left (or right) ideal of A is regular. On the other hand, if A is a $k$ -algebra whose elements are all nilpotent, then A is the only (left or right) ideal of A that is regular.

Let $\mathfrak{a}$ be a regular left ideal of A and $u$ a right unit modulo $\mathfrak{a}$ . If $\mathfrak{b}$ is a left ideal of A containing $\mathfrak{a}$ , then $u$ is a right unit modulo $\mathfrak{a}$ , so $\mathfrak{b}$ is regular. So the left ideals of A that are maximal and regular are the maximal elements of the set of regular proper left ideals of A. Moreover, a left ideal $\mathfrak{b}$ of A containing $\mathfrak{a}$ is proper if and only if $u$ does not belong to $\mathfrak{b}$ . The set of proper left ideals $\mathfrak{b}$ of A containing $\mathfrak{a}$ , ordered by inclusion, is therefore an inductive set. Theorem 2 of Set Theory, III, §2, No. 4, p. 154 applied to this set gives the following result.

Proposition 1 00P5

Every regular proper left (resp. right) ideal of A is contained in a regular maximal left (resp. right) ideal.

Proposition 2 00P6

Let A be a commutative $k$ -algebra. An ideal $\mathfrak{a}$ of A is a regular maximal ideal if and only if the pseudoring $A/\mathfrak{a}$ (I, §8, No. 1, p. 98) is a field.

Cites 2

Cites I § 8. Rings, Theorem 1, I § 9

An element $u$ of A is a (right or left) unit modulo $\mathfrak{a}$ if and only if the canonical image of $u$ in $A/\mathfrak{a}$ is a unit element of the algebra $A/\mathfrak{a}$ . Therefore, the ideal $\mathfrak{a}$ is regular if and only if the algebra $A/\mathfrak{a}$ is unital. Suppose that these conditions are satisfied.

The mapping $\mathfrak{b}\mapsto \mathfrak{b}/\mathfrak{a}$ is a bijection from the set of ideals of the algebra A containing $\mathfrak{a}$ to the set of ideals of the algebra $A/\mathfrak{a}$ . These are the ideals of the ring $A/\mathfrak{a}$ because this algebra is unital. Finally, by Theorem 1 of I, §9, No. 1, p. 115, the ring $A/\mathfrak{a}$ is a field if and only if it is nonzero and its only ideals are 0 and itself. The proposition follows.

Example 1 00P7

Let V be an infinite-dimensional vector space over a commutative field K. Let End $^f_K(V)$ be the K-subalgebra of End $_K(V)$ consisting of the endomorphisms of finite rank. Let W be a linear subspace of V, and let $\mathfrak{a}_W$ be the set of elements of End $^f_K(V)$ whose kernel contains W; it is a left ideal of End $^f_K(V)$ . An element $u$ of End $^f_K(V)$ is a right unit modulo $\mathfrak{a}_W$ if and only if we have $u(x) =x$ for every $x\in W$ . Such an element $u$ exists, that is, $\mathfrak{a}_W$ is regular, if and only if W is finite-dimensional. The ideal $\mathfrak{a}_W$ is maximal and regular if and only if W has dimension 1.

Cited by 1

Cited by Example 1

Example 2 00P8

Let T be a locally compact space, and let $\mathscr{C}_0(T)$ be the commutative $\mathbf{C}$ -algebra of continuous mappings from T to $\mathbf{C}$ that tend to 0 at infinity (Gen. Top., X, §4, No. 4, p. 316); it is unital if and only if T is compact. Let F be a closed subset of T, and let $\mathfrak{a}_F$ be the set of elements of $\mathscr{C}_0(T)$ whose restriction to F is the zero function; it is an ideal of $\mathscr{C}_0(T)$ . An element $u$ of $\mathscr{C}_0(T)$ is a unit modulo $\mathfrak{a}_F$ if and only if we have $u(t) = 1$ for every $t\in F$ . Such an element $u$ exists, that is, $\mathfrak{a}_F$ is regular, if and only if F is compact. The mapping $t\mapsto \mathfrak{a}_{\{t\}}$ is a bijection from T to the set of regular maximal ideals of $\mathscr{C}_0(T)$ (TS, I, §3, n $^o2$ , p. 32, corollaire 1). Suppose that T is not compact, and denote by $\mathfrak{a}$ the subset of $\mathscr{C}_0(T)$ consisting of the functions with compact support; then $\mathfrak{a}$ is an ideal of $\mathscr{C}_0(T)$ that is not contained in any regular ideal of $\mathscr{C}_0(T)$ .

Example 3 00P9

Let $L^1(\mathbf{R})$ be the convolution algebra of the locally compact group $\mathbf{R}$ . Recall (cf. Int., VIII, §4, No. 5, p. 38) that $L^1(\mathbf{R})$ is the space of classes of functions on $\mathbf{R}$ that are integrable for the Lebesgue measure. The product of the classes of two functions $f$ and $g$ is the class of the function $f*g$ defined by the formula

(f*g)(s) =\int_{-\infty}^{+\infty}f(t)g(s-t)dt

for almost every $s\in \mathbf{R}$ . The algebra $L^1(\mathbf{R})$ is not unital. For any $a$ in $\mathbf{R}$ , denote by $\mathfrak{m}_a$ the set of elements $f$ of $L^1(\mathbf{R})$ satisfying

\int_{-\infty}^{+\infty}f(t)e^{-iat}dt= 0

By TS, II, §3, n $^o2$ , p. 252, théorème 1, the mapping $a\mapsto \mathfrak{m}_a$ is a bijection from $\mathbf{R}$ to the set of regular maximal ideals of the algebra $L^1(\mathbf{R}).*$

2. Adjunction of a Unit Element

Let A be a $k$ -algebra. In III, §1, No. 2, p. 431, we defined the unital algebra $\widetilde{A}$ deduced from A by the adjunction of a unit element $e$ . We identify A with a two-sided ideal of $\widetilde{A}$ . The $k$ -module $\widetilde{A}$ is the direct sum of the submodules $ke$ and A.

Proposition 3 00PA

a) Let $\widetilde{\mathfrak{a}}$ be a left ideal of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ . We set $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$ . There exists an element $u$ of A such that $u-e$ belongs to $\widetilde{\mathfrak{a}}$ . If $u$ is such an element, then it is a right unit of A modulo $\mathfrak{a}$ , and we have $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$ ; in particular, the ideal $\mathfrak{a}$ is regular.

Cited by 2

Cited by Exercise 3, Appendix 1, Proposition 4

b) Conversely, let $\mathfrak{a}$ be a regular left ideal of A and $u$ a right unit of A modulo $\mathfrak{a}$ . Set $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$ . Then $\widetilde{\mathfrak{a}}$ is a left ideal of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ , and we have $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$ .

c) If $k$ is a field, then the condition $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ is equivalent to saying that $\widetilde{\mathfrak{a}}$ is not contained in A.

Under the assumption of a), the element $e$ of $\widetilde{A}$ can be written as $(e-u)+u$ with $u\in A$ and $u-e\in \widetilde{\mathfrak{a}}$ . Let $x=\lambda e+a$ be an element of $\widetilde{A}$ , where $\lambda \in k$ and $a\in A$ ; if $x$ belongs to $\widetilde{\mathfrak{a}}$ , then the element $y=a+\lambda u=x+\lambda (u-e)$ of A belongs to $\mathfrak{a}$ , and we have $x=y-\lambda (u-e)$ ; this proves the equality $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$ . Moreover, for every $a$ in A, the element $au-a=a(u-e)$ of A belongs to $\mathfrak{a}$ , so $u$ is a right unit modulo $\mathfrak{a}$ . This proves assertion a).

Let $\mathfrak{a}$ and $u$ be as in b). Since $\widetilde{A}$ is the direct sum of A and $k(u-e)$ , we have $\widetilde{\mathfrak{a}}+ A =\widetilde{A}$ and $\widetilde{\mathfrak{a}}\cap A =\mathfrak{a}$ . Every element of $\widetilde{\mathfrak{a}}$ is of the form $x+\lambda (u-e)$ with $x\in \mathfrak{a}$ and $\lambda \in k$ . For every $a\in A$ , we have

a(x+\lambda (u-e)) =ax+\lambda (au-a)

Since $u$ is a right unit of A modulo $\mathfrak{a}$ , the sum $ax+\lambda (au-a)$ belongs to $\mathfrak{a}$ , so that $\widetilde{\mathfrak{a}}$ is a left ideal of A. This proves b).

Finally, if $k$ is a field, then A is a hyperplane in $\widetilde{A}$ , and $\widetilde{A}$ is the sum of $\widetilde{\mathfrak{a}}$ and A if and only if $\widetilde{\mathfrak{a}}$ is not contained in A.

Corollary 00PB

The regular left ideals of A are the ideals of the form $A\cap \widetilde{\mathfrak{a}}$ , where $\widetilde{\mathfrak{a}}$ is a left ideal of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ .

Proposition 4 00PC

a) Let $\mathfrak{a}$ be a regular maximal left ideal of A. There exists a unique left ideal $\widetilde{\mathfrak{a}}$ of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ and $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$ . This ideal is maximal and does not contain A.

Cites 2 · Cited by 1

Cites Proposition 3, Lemma 2, VIII § 1

Cited by Proposition 7

b) The mapping $\widetilde{\mathfrak{a}}\mapsto \widetilde{\mathfrak{a}}\cap A$ is a bijection from the set of maximal left ideals of $\widetilde{A}$ not containing A to the set of regular maximal left ideals of A.

Let $\mathfrak{a}$ be a regular maximal left ideal of A. By Proposition 3, a), the left ideals $\mathfrak{b}$ of $\widetilde{A}$ such that $\mathfrak{b}+ A =\widetilde{A}$ and $\mathfrak{b}\cap A =\mathfrak{a}$ are the ideals $\mathfrak{a}+k(u-e)$ , where $u$ is a right unit modulo $\mathfrak{a}$ . To prove the uniqueness of $\widetilde{\mathfrak{a}}$ , it therefore suffices to prove that two right units $u$ and $u'$ of A modulo $\mathfrak{a}$ are congruent modulo $\mathfrak{a}$ . Let us reason by contradiction and suppose that $u-u'$ does not belong to $\mathfrak{a}$ . The formula $x(u-u') = (xu-x)-(xu'-x)$ shows that we have $A(u-u')\subset \mathfrak{a}$ ; it follows that $\mathfrak{a}+k(u-u')$ is a left ideal of A containing $\mathfrak{a}$ and distinct from $\mathfrak{a}$ . Since $\mathfrak{a}$ is maximal, we therefore have $\mathfrak{a}+k(u-u') = A$ , and so AA $\subset \mathfrak{a}$ . For every $x\in A$ , we have $x\equiv xu$ (mod $\mathfrak{a}$ ), and therefore $x\in \mathfrak{a}$ by the above, which contradicts the assumption $\mathfrak{a}\not= A$ .

Hence there exists a unique left ideal $\widetilde{\mathfrak{a}}$ of $\widetilde{A}$ such that $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ and $\mathfrak{a}=\widetilde{\mathfrak{a}}\cap A$ . Let $\mathfrak{b}$ be a proper left ideal of $\widetilde{A}$ containing $\widetilde{\mathfrak{a}}$ . Then $\mathfrak{b}\cap A$ is a proper left ideal of A containing $\mathfrak{a}$ . It is therefore equal to $\mathfrak{a}$ because $\mathfrak{a}$ is maximal, which implies $\mathfrak{b}=\widetilde{\mathfrak{a}}$ by Lemma 2 of VIII, p. 4. This proves that $\widetilde{\mathfrak{a}}$ is a maximal ideal of $\widetilde{A}$ ; for such an ideal, the condition $\widetilde{A} =\widetilde{\mathfrak{a}}+ A$ means that $\widetilde{\mathfrak{a}}$ does not contain A.

It remains to prove that if $\widetilde{\mathfrak{a}}$ is a maximal left ideal of $\widetilde{A}$ that does not contain A, then the left ideal $\mathfrak{a}= A\cap \widetilde{\mathfrak{a}}$ of A is maximal. Let $\mathfrak{b}$ be a proper left ideal of A containing $\mathfrak{a}$ . Let $u$ be a right unit modulo $\mathfrak{a}$ such that $\widetilde{\mathfrak{a}}=\mathfrak{a}+k(u-e)$ (Proposition 3). It is also a right unit modulo $\mathfrak{b}$ ; denote by $\widetilde{\mathfrak{b}}$ the ideal $\mathfrak{b}+k(u-e)$ of $\widetilde{A}$ . By Proposition 3, b), we have $\mathfrak{b}= A\cap \widetilde{\mathfrak{b}}$ . The ideal $\widetilde{\mathfrak{a}}$ , equal to $\mathfrak{a}+k(u-e)$ , is contained in $\widetilde{\mathfrak{b}}$ and is therefore equal to $\widetilde{\mathfrak{b}}$ because $\widetilde{\mathfrak{b}}$ is distinct from $\widetilde{A}$ and $\widetilde{\mathfrak{a}}$ is maximal. Consequently, we have $\mathfrak{a}=\mathfrak{b}$ , and $\mathfrak{a}$ is maximal.

We leave it to the reader to translate Propositions 3 and 4 to right ideals.

3. The Radical of an Algebra

Let A be a $k$ -algebra. A left pseudomodule over A is a $k$ -module M endowed with the structure of a left pseudomodule over the pseudoring A (II, Appendix, No. 2, p. 378) such that we have $a(\lambda x) =\lambda (ax) = (\lambda a)x$ for $\lambda \in k$ , $a\in A$ , and $x\in M$ . We define right pseudomodules over A likewise.

Let M be a left pseudomodule over A; we define a left $\widetilde{A}$ -module structure on M called canonical by setting

$(\lambda e+a)x=\lambda x+ax$ for $\lambda \in k, a\in A, x\in M$ .

Conversely, every left $\widetilde{A}$ -module is canonically endowed, by restricting the ring of operators to $k$ and A, with the structure of a left pseudomodule over A. Thus, the species of structures of left pseudomodules over A and of modules over $\widetilde{A}$ are equivalent (Set Theory, IV, §1, No. 7, p. 262).

Let M be a left pseudomodule over A, and let N be a $k$ -submodule of M. Then N is an $\widetilde{A}$ -submodule of M if and only if it is stable under the action of A; we then say that N is a sub-pseudomodule of M.

As in the case of rings, we define the left pseudomodule $A_s$ over A and the right pseudomodule $A_d$ . The left (resp. right) ideals of A are the sub-pseudomodules of $A_s$ (resp. $A_d$ ).

Let $\mathfrak{a}$ be a regular left ideal of A and $u$ a right unit modulo $\mathfrak{a}$ . Set $M = A_s/\mathfrak{a}$ , and denote the image of $u$ in M by $z$ . We have $M = Az$ , and $\mathfrak{a}$ is the annihilator of $z$ .

Conversely, let M be a left pseudomodule over A, and let $z$ be an element of M such that $M = Az$ . Then there exists an element $u$ of A such that $z=uz$ . For every $a\in A$ , we have $(au-a)z= 0$ , so $au-a$ belongs to the annihilator $\mathfrak{a}$ of $z$ . Consequently, $\mathfrak{a}$ is a regular left ideal of A, the element $u$ is a right unit modulo $\mathfrak{a}$ , and when passing to the quotient, the mapping $a\mapsto az$ defines an isomorphism from $A_s/\mathfrak{a}$ to M.

Definition 2 00PD

We say that a pseudomodule M over A is simple if we have AM $\not= 0$ and if 0 and M are the only sub-pseudomodules of M.

Cites 1 · Cited by 1

Cites Definition 1, VIII § 3

Cited by Proposition 5

This corresponds to saying that the $\widetilde{A}$ -module M is simple and that its annihilator does not contain A. When A is a ring and M is an A-module, we have AM = M, so Definition 2 coincides with Definition 1 of VIII, p. 45.

Proposition 5 00PE

a) Let $\mathfrak{m}$ be a regular maximal left ideal of A. Then the pseudomodule $A_s/\mathfrak{m}$ is simple, and there exists a nonzero element of $A_s/\mathfrak{m}$ with annihilator $\mathfrak{m}$ .

Cites 1 · Cited by 1

Cites Definition 2

Cited by Example 2

b) Let M be a simple pseudomodule, and let $x$ be a nonzero element of M. We have $M = Ax$ , the annihilator $\mathfrak{m}$ of $x$ is a regular maximal left ideal of A, and when passing to the quotient, the mapping $a\mapsto ax$ defines an isomorphism from $A_s/\mathfrak{m}$ to M.

c) Let M be a pseudomodule not reduced to 0. Then M is simple if and only if we have $M = Ax$ for every nonzero element $x$ of M.

Let $\mathfrak{m}$ be a regular maximal left ideal of A. We have seen that there exists a nonzero element $z$ of $A_s/\mathfrak{m}$ with annihilator equal to $\mathfrak{m}$ and such that $Az= A_s/\mathfrak{m}$ ; in particular, we have $A(A_s/\mathfrak{m})\not= 0$ . Moreover, every sub-pseudomodule of $A_s/\mathfrak{m}$ is of the form $\mathfrak{n}/\mathfrak{m}$ , where $\mathfrak{n}$ is a left ideal of A containing $\mathfrak{m}$ . Since $\mathfrak{m}$ is maximal, the only possibilities are $\mathfrak{n}=\mathfrak{m}$ and $\mathfrak{n}= A$ , so that the pseudomodule $A_s/\mathfrak{a}$ is simple. This proves a).

Under the assumptions of b), the set of elements $y$ of M such that $Ay= 0$ is a sub-pseudomodule of M different from M and therefore reduced to 0. Consequently, $Ax$ is a nonzero sub-pseudomodule of M, which implies $M = Ax$ . Assertion b) then follows from the remarks before Definition 2.

Let M be a nonzero pseudomodule. Suppose that we have $Ax= M$ for every nonzero element $x$ of M. In particular, we have AM $\not= 0$ . Let N be a nonzero sub-pseudomodule of M, and let $x$ be a nonzero element of N. We have $Ax= M$ , and therefore N = M. Hence M is simple. Conversely, if M is simple, then we have $M = Ax$ for every $x\not= 0$ by b).

Definition 3 00RT

The radical of the $k$ -algebra A, denoted by $\Re (A)$ , is the intersection of the regular maximal left ideals of A.

Cites 1

Cites Definition 2, VIII § 9

When A is a ring, every left ideal of A is regular, so the definition of the radical coincides with Definition 2 of VIII, p. 154.

Example 1 00PF

The radical of the $k$ -algebra End $^f_k(V)$ is reduced to 0 (VIII, p. 436, Example 1) $.*$ The same holds for the algebras $\mathscr{C}_0(T)$ and $L^1(\mathbf{R}).*$

Cites 1

Cites Example 1

Example 2 00PG

Let A be a pseudoring in which all elements are nilpotent. The radical of A is equal to A because A is the only regular left ideal.

Proposition 5 immediately implies the following result.

Cites 1

Cites Proposition 5

Proposition 6 00PH

The radical of the algebra A is the intersection of the annihilators of the simple pseudomodules. It is, in particular, a two-sided ideal of A.

Proposition 7 00PI

The radical of A is the trace on A of the radical of $\widetilde{A}$ ; it is also equal to the radical of the opposite algebra $A^o($ that is, to the intersection of the regular maximal right ideals of A). If the ring $k$ is without radical, then the radical of A is equal to that of $\widetilde{A}$ .

Cites 2 · Cited by 2

Cites Proposition 4, Corollary 1, VIII § 9

Cited by Exercise 4, Appendix 1, Remark

The equality $\Re (\widetilde{A})\cap A =\Re (A)$ follows from Proposition 4 of VIII, p. 438, b). Since $\widetilde{A}$ and $\widetilde{A}^o$ have the same radical (VIII, p. 156, Corollary 1), the equality $\Re (A) =\Re (A^o)$ follows. If $k$ is without radical, then the intersection of the maximal left ideals of $\widetilde{A}$ containing A is equal to A. Consequently, $\Re (\widetilde{A})$ is contained in A and therefore equal to $\Re (A)$ .

Remark 00PJ

Let $x$ and $y$ be elements of A; we say that $x$ is a left adverse of $y$ , or that $y$ is a right adverse of $x$ , if, in $\widetilde{A},x-e$ is a left inverse of $y-e$ , that is, if we have $x+y=xy$ . By Proposition 7 and Jacobson’s theorem (VIII, p. 156, Theorem 1), the radical of A consists of the elements $x$ of A such that $ux-e$ is left invertible in $\widetilde{A}$ for every $u$ in $\widetilde{A}$ . Since we have $(a+\lambda e)x-e=ax+\lambda x-e$ (for $a\in A,\lambda \in k$ ), the radical of A is the set of elements $x$ of A such that $ax+\lambda x$ has a left adverse in A for every $a\in A$ and $\lambda \in k$ .

Cites 2

Cites Proposition 7, Theorem 1 (Jacobson), VIII § 9

4. Density Theorem

Let A be a $k$ -algebra. A left pseudomodule M is called semisimple if it is the direct sum of a family of simple left pseudomodules.

Lemma 1 00PK

Let M be a semisimple left A-pseudomodule. Let B be the bicommutant of the $\widetilde{A}$ -module M. Then every A-sub-pseudomodule of M is a B-submodule of M.

Cites 1 · Cited by 1

Cites Corollary 2, VIII § 4

Cited by Theorem 1 (Jacobson’s density theorem)

The $\widetilde{A}$ -module M is semisimple. Let N be an A-sub-pseudomodule of M. Then N is an $\widetilde{A}$ -submodule of M. By Corollary 2 of VIII, p. 56, there exists a projector $p$ of the A-module M with image N. Since we have the relation $pb=bp$ for every $b\in B$ , we obtain that N is a B-submodule of M.

Lemma 2 00PL

Let M be a semisimple left A-pseudomodule, and let $x$ be an element of M. There exists an element $a\in A$ such that $ax=x$ .

Cites 1 · Cited by 1

Cites Corollary 3, VIII § 4

Cited by Theorem 1 (Jacobson’s density theorem)

Let N be the left A-pseudomodule $\widetilde{A}x/Ax$ . It satisfies AN $=\{0\}$ . By Corollary 3 of VIII, p. 56 applied to the $\widetilde{A}$ -modules M and $\widetilde{A}x$ , the A-pseudomodule N is semisimple. By definition, every simple sub-pseudomodule S of N satisfies AS $\not=\{0\}$ . Consequently, the pseudomodule N is zero, and we obtain that $x\in Ax$ .

Theorem 1 (Jacobson’s density theorem) 00SD

Let M be a semisimple left A-pseudomodule. Let $b$ be an element of the bicommutant of the $\widetilde{A}$ -module M. Let $F =\{x_1, . . . , x_n\}$ be a finite subset of M. Then there exists an element $a\in A$ such that $bx_i=ax_i$ for every $i\in [1, n]$ .

Cites 3

Cites Lemma 1, Lemma 2, Proposition 2, VIII § 5

Let B be the bicommutant of the $\widetilde{A}$ -module M. The A-pseudomodule $M^n$ is semisimple. Let $\boldsymbol{x}= (x_1, . . . , x_n)\in M^n$ . It follows from Lemma 2 that $\boldsymbol{x}\in$ $A\boldsymbol{x}$ . The bicommutant of the $\widetilde{A}$ -module $M^n$ coincides with the homotheties of the B-module $M^n$ (VIII, p. 79, Proposition 2). By Lemma 1, the A-sub-pseudomodule $A\boldsymbol{x}$ of $M^n$ is therefore a B-submodule of $M^n$ . We therefore have the inclusion $B\boldsymbol{x}\subset A\boldsymbol{x}$ . So there exists an $a\in A$ such that $b\boldsymbol{x}=a\boldsymbol{x}$ . The result follows.

Exercises

See the exercises for Appendix 1.