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Ed. by Blomer, Valentin / Cohen, Frederick R. / Droste, Manfred / Duzaar, Frank / Echterhoff, Siegfried / Frahm, Jan / Gordina, Maria / Shahidi, Freydoon / Sogge, Christopher D. / Takayama, Shigeharu / Wienhard, Anna


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Volume 31, Issue 1

Issues

Infinite-dimensional triangularizable algebras

Zachary Mesyan
Published Online: 2018-08-11 | DOI: https://doi.org/10.1515/forum-2018-0109

Abstract

Let Endk(V) denote the ring of all linear transformations of an arbitrary k-vector space V over a field k. We define XEndk(V) to be triangularizable if V has a well-ordered basis such that X sends each vector in that basis to the subspace spanned by basis vectors no greater than it. We then show that an arbitrary subset of Endk(V) is strictly triangularizable (defined in the obvious way) if and only if it is topologically nilpotent. This generalizes the theorem of Levitzki that every nilpotent semigroup of matrices is triangularizable. We also give a description of the triangularizable subalgebras of Endk(V), which generalizes a theorem of McCoy classifying triangularizable algebras of matrices over algebraically closed fields.

Keywords: Triangular matrix; linear transformation; simultaneous triangularization; triangularizable algebra; nilpotent semigroup; endomorphism ring; function topology

MSC 2010: 15A04; 16S50; 16W80; 47D03

1 Introduction

Given a field k and a k-vector space V, we denote by Endk(V) the k-algebra of all linear transformations of V. We define a transformation TEndk(V) to be triangularizable if V has a well-ordered basis (,) such that T sends each vector v to the subspace spanned by {uuv}. If V is finite-dimensional, then this notion is clearly equivalent to T being representable as an upper-triangular matrix with respect to some basis for V, which in turn is equivalent to T being representable as a lower-triangular matrix with respect to some basis for V. The above definition of triangularizability was first introduced in [7], where various equivalent characterizations of transformations of this sort are given, along with other facts about them.

In this paper we focus on subsets of Endk(V) that are (simultaneously) triangularizable, that is, ones where all elements are triangularizable with respect to a common well-ordered basis for V. We give a general tool for showing sets of transformations to be triangularizable (Lemma 3.2), which is analogous to the so-called “Triangularization Lemma”, proved for bounded linear operators on Banach spaces by Radjavi and Rosenthal [8]. Using this tool, we show that a subset X of Endk(V) is triangularizable if and only if there exists a well-ordered (by inclusion) set of X-invariant subspaces of V, which is maximal as a well-ordered set of subspaces of V (Proposition 3.3).

Defining a transformation TEndk(V) to be strictly triangularizable if V has a well-ordered basis (,) such that T sends each vector v to the subspace spanned by {uu<v}, we then characterize the strictly (simultaneously) triangularizable subsets of Endk(V) in Theorem 4.1. Specifically, XEndk(V) is strictly triangularizable if and only if X is topologically nilpotent (i.e., given any infinite list T1,T2,T3,X and any finite-dimensional subspace W of V, one has TnT2T1(W)=0 for some positive integer n). As immediate consequences of this theorem, we obtain the well-known result of Levitzki [5] that every nilpotent multiplicative semigroup of matrices is triangularizable, and the statement that the nonunital k-algebra generated by a topologically nilpotent subset of Endk(V) is also topologically nilpotent (Corollary 4.4). Additionally, this theorem generalizes a recent result of Chebotar [1], who used our approach to triangularization to answer a question of Greenfeld, Smoktunowicz, and Ziembowski. Theorem 4.1 can also be viewed as an analogue of the result [10, Corollary 11] of Turovskii that a quasinilpotent semigroup (i.e., one where the spectrum of each of its operators is contained in {0}) of compact operators on a Banach space is chain-triangularizable (that is, possesses a maximal chain of invariant subspaces).

Most of the paper, starting with Section 5, is devoted to describing triangularizable subalgebras of Endk(V). In Theorem 6.4 we show that a k-subalgebra R of Endk(V) is triangularizable if and only if R is contained in a k-subalgebra A of Endk(V) such that A/rad(A)kΩ as topological k-algebras for some set Ω, and rad(A) is topologically nilpotent. We then give another equivalent condition to these in the case where the field k is algebraically closed (Corollary 7.3). A consequence of Theorem 6.4 is that a subalgebra R of a matrix ring 𝕄n(k) is triangularizable if and only if R/rad(R)km as k-algebras, for some positive integer m, and moreover, if k is algebraically closed, then R being triangularizable is also equivalent to R/rad(R) being commutative (Corollary 7.4). This generalizes a well-known result of McCoy [6]. Various examples illustrating our results are given along the way.

2 Preliminaries

All rings will be assumed to be unital, unless stated otherwise. We denote the set of the integers by and the set of the positive integers by +. Also, given a ring R and n+, we denote by 𝕄n(R) the ring of all n×n matrices over R.

In the following subsections, we review the standard topology on Endk(V), some basics of the theory of topological rings, and a few facts about triangularizable elements of Endk(V).

2.1 Topological rings

Let X and Y be sets, and let YX denote the set of all functions XY. The function (or finite, or pointwise) topology on YX has a base of open sets of the following form:

{fYXf(x1)=y1,,f(xn)=yn}(x1,,xnX,y1,,ynY).

It is straightforward to see that this coincides with the product topology on YX=XY, where each component set Y is given the discrete topology. As a product of discrete spaces, this space is Hausdorff.

Now let V be a vector space over a field k. Then Endk(V)VV inherits a topology from the function topology on VV, which we shall also call the function topology. Under this topology, Endk(V) is a topological ring (see, e.g., [11, Theorem 29.1]), i.e., a ring R equipped with a topology that makes +:R×RR, -:RR, and :R×RR continuous. Alternatively, we may describe the function topology on Endk(V) as the topology having a base of open sets of the following form:

{SEndk(V)S|W=T|W}(TEndk(V),WVa finite-dimensional subspace),

where S|W and T|W denote the restrictions of S and T, respectively, to W. Observe that when V is finite-dimensional, Endk(V) is discrete in this topology.

Let us also review some standard facts about topological rings that will be needed later on. Given any topological ring R and an ideal I of R, the quotient R/I may be viewed as a topological ring in the quotient topology, where a subset of R/I is open if and only if its preimage under the canonical projection RR/I is open in R. (See, e.g., [11, Theorem 5.4] for details.) Also, two topological rings R1 and R2 are isomorphic as topological rings if there is a topological isomorphism R1R2, i.e., a function that is both a ring isomorphism and a homeomorphism. The concept isomorphic as topological k-algebras is defined analogously.

Finally, recall that a map ϕ:XY of topological spaces is open if ϕ(U) is open whenever U is an open subset of X, and it is closed if ϕ(U) is closed whenever U is a closed subset of X.

2.2 Triangular transformations

Recall that a binary relation on a set X is a partial order if it is reflexive, antisymmetric, and transitive. If, in addition, xy or yx for all x,yX, then is a total order. If is a total order and, moreover, every nonempty subset of X has a least element with respect to , then is a well order.

Given a subset X of a vector space, we denote by X the subspace spanned by X.

Definition 2.1.

Let k be a field, V a k-vector space, a basis for V, and a partial ordering on .

We say that TEndk(V) is triangular with respect to (,) if T(v){uuv} for all v, and that T is strictly triangular with respect to (,) if T(v){uu<v} for all v.

We say that XEndk(V) is triangular with respect to (,), respectively strictly triangular with respect to (,), if each TX is triangular with respect to (,), respectively strictly triangular with respect to (,).

If TEndk(V) (or XEndk(V)) is triangular, respectively strictly triangular, with respect to some well-ordered basis for V, then we say that T (or X) is triangularizable, respectively strictly triangularizable.

Sometimes we shall find it more convenient to index bases with ordered sets rather than ordering the bases themselves, when dealing with triangularization.

It is easy to see that in the case where V is a finite-dimensional k-vector space, TEndk(V) is triangularizable in the above sense if and only if there is a basis for V with respect to which T is upper-triangular as a matrix, if and only if there is a basis for V with respect to which T is lower-triangular as a matrix. A more detailed discussion of the choices made in the above definition can be found in [7, Section 3], and various characterizations of triangularizable transformations are given in [7, Theorem 8].

Next, we recall a couple of results from [7] that will be used frequently. Given k-vector spaces WV and a transformation TEndk(V), we say that W is T-invariant if T(W)W.

Lemma 2.2 ([7, part of Proposition 16]).

Let k be a field, V a k-vector space and TEndk(V) triangular with respect to some well-ordered basis (B,) for V. Also, for each vB, let πvEndk(V) be the projection onto v with kernel B{v}. Then T is invertible if and only if πvTπv0 for all vB.

Definition 2.3.

Let k be a field, V a k-vector space, and XEndk(V). We say that X is topologically nilpotent if the sequence (TiT2T1)i=1 converges to 0 in the function topology on Endk(V) for every infinite list T1,T2,T3,X of not necessarily distinct transformations. Moreover, if {T} is topologically nilpotent for some TEndk(V), then we shall refer to T as topologically nilpotent.

Proposition 2.4 ([7, Proposition 20]).

Let k be a field and V a nonzero k-vector space. The following are equivalent for any TEndk(V).

  • (1)

    T is topologically nilpotent.

  • (2)

    V=i=1ker(Ti).

  • (3)

    T is strictly triangularizable.

  • (4)

    T is triangularizable, and if (,) is a well-ordered basis for V with respect to which T is triangular, then T is strictly triangular with respect to (,).

  • (5)

    T is triangularizable, and for all ak , we have ker(T-a1){0} if and only if a=0.

We conclude this section with a basic fact about triangularizable sets.

Lemma 2.5.

Let k be a field, V a nonzero k-vector space, (B,) a partially ordered basis for V, XEndk(V), and cl(X) the closure of X in the function topology. Then X is triangular with respect to B if and only if cl(X) is.

Proof.

Suppose that X is triangular with respect to . If X=, then cl(X)=. So we may assume that X. Letting Tcl(X) and v be arbitrary, there is some SX such that T|v=S|v, by properties of the function topology. Hence T(v)=S(v){uuv}. Since v was arbitrary, it follows that T is triangular with respect to . Since Tcl(X) was arbitrary, we conclude that cl(X) is triangular with respect to .

Conversely, if cl(X) is triangular with respect to , then the same holds for any subset of cl(X), and for X in particular. ∎

3 Triangularization lemma

Let k be a field, V a k-vector space, XEndk(V), and WV an X-invariant subspace (i.e., X(W)W). We shall denote by X¯ the subset of Endk(V/W) consisting of the transformations T¯ defined by T¯(v+W)=T(v)+W, for TX. (It is routine to check that any such T¯ is well-defined and k-linear.) We make the next definition following Radjavi and Rosenthal [8].

Definition 3.1.

A property P on sets of k-vector space transformations is inherited by quotients if given any k-vector spaces WV, and any XEndk(V) such that X(W)W and X has property P, then X¯Endk(V/W) also has property P.

The following is an analogue of the Triangularization Lemma [8, Lemma 1] (alternatively [9, Lemma 7.1.11]), proved for bounded linear operators on Banach spaces by Radjavi and Rosenthal. It will be our main tool for showing sets of transformations to be triangularizable. The (short) proof is essentially the same as that of [7, Theorem 15] (which says that any finite commutative subset of Endk(V) consisting of triangularizable transformations is triangular with respect to a common well-ordered basis), but we give the details here for completeness.

Lemma 3.2.

Let k be a field, and let P be a property on sets of k-vector space transformations that is inherited by quotients. Suppose that for every nonzero k-vector space V and every subset X of Endk(V) that satisfies property P there exists a 1-dimensional X-invariant subspace of V. Then given any k-vector space V, every subset of Endk(V) that satisfies property P is triangularizable.

Proof.

Let V be a k-vector space and XEndk(V) be such that X satisfies property P. We shall show that X is triangularizable. We may assume that V{0}, since otherwise every subset of Endk(V) is triangularizable.

We begin by constructing recursively for each ordinal α an X-invariant subspace VαV, and for each successor ordinal α a vector vαV. Set V0={0}. Now let α>0 be an ordinal, and assume that Vγ has been defined for every γ<α. If α is a limit ordinal, then let Vα=γ<αVγ. Since each Vγ is assumed to be X-invariant, their union Vα will also be X-invariant. Next, if α is a successor ordinal, then let β be its predecessor. By hypothesis, the set X¯ of transformations on V/Vβ induced by X satisfies property P. Thus there is a 1-dimensional X¯-invariant subspace W/Vβ of V/Vβ (assuming that VVβ). Let vαV be such that {vα+Vβ} is a basis for W/Vβ, and define Vα=Vβ{vα}. Then Vα must be X-invariant, because of the invariance of Vβ and W/Vβ. We proceed in this fashion until V=αΛVα for some ordinal Λ.

Now let

Γ={αΛαis a successor ordinal},

and let ={vααΓ}. As a subset of a well-ordered set, Γ is itself well-ordered. Since we introduced new vectors only at successor steps in our construction,

V=αΓVα=αΓ{vγγα,γΓ},

and hence V=. Since Vα/Vβ=vα+Vβ is 1-dimensional for all αΓ with predecessor β, we conclude that is a basis for V. Also, since Vα={vγγα,γΓ} is X-invariant for all αΓ, it follows that X(vα){vγγα,γΓ} for all αΓ. Thus X is triangular with respect to , a basis for V indexed by the well-ordered set Γ. ∎

Lemma 3.2 gives the following alternative description of triangularizable sets, which additionally generalizes [7, Lemma 5].

Proposition 3.3.

Let k be a field, V a nonzero k-vector space and XEndk(V). Then X is triangularizable if and only if there exists a well-ordered (by inclusion) set of X-invariant subspaces of V, which is maximal as a well-ordered set of subspaces of V.

Proof.

Suppose that X is triangularizable. Then there is a well-ordered set (Ω,) and a basis ={vααΩ} for V such that T(vα){vββα} for all αΩ and all TX. Since every well-ordered set is order-isomorphic to an ordinal, we may assume that Ω is an ordinal. For each αΩ set Vα={vββ<α}, where V0 is understood to be the zero space (0 being the least element of Ω). Then, for all α1,α2Ω, we have Vα1Vα2 if and only if α1α2. Since (Ω,) is well-ordered, it follows that U={VααΩ+} is well-ordered by set inclusion, where Ω+=Ω{Ω} is the successor of Ω and V=VΩ. Moreover, X(vβ)Vα for all α,βΩ satisfying β<α, from which it follows that each element of U is X-invariant. It remains to show that U is maximal. First, note that for each αΩ+ we have β<αVβVα, with equality if α is a limit ordinal, and Vα/(β<αVβ) 1-dimensional otherwise.

Now, let WV be a subspace, which we may assume to be nonzero, that is comparable under set inclusion to Vα for each αΩ+. Since Ω+ is well-ordered, there is a least αΩ+ such that WVα. Since W is comparable to each element of U, from the choice of Vα it follows that VβW for all β<α. If α is a limit ordinal, then Vα=β<αVβW, and hence W=VαU. Otherwise, there is a βΩ+ such that α is the successor of β, and VβWVα. But in this case, Vα/Vβ is 1-dimensional, and therefore W=VαU once again. Thus U is a maximal well-ordered set of subspaces of V.

Conversely, suppose that there exists an ordinal Ω and a set U={VααΩ} of X-invariant subspaces of V, such that Vα1Vα2 if and only if α1α2 (for all α1,α2Ω), and U is maximal as a well-ordered set of subspaces of V. Since V{0}, the set U contains at least one nonzero subspace of V. Then letting αΩ be the least element such that Vα{0}, the maximality of U implies that Vα is an X-invariant subspace that is necessarily 1-dimensional.

By Lemma 3.2, to conclude that X is triangularizable, it suffices to show that the property of having a maximal well-ordered set of invariant subspaces is inherited by quotients. To that end, keeping U as before, and letting WV be an X-invariant subspace, we shall show that U¯={(Vα+W)/WαΩ} is a maximal well-ordered set of X¯-invariant subspaces of V/W.

For each αΩ, it follows immediately from Vα and W being X-invariant that (Vα+W)/W is X¯-invariant. Next, let α,βΩ. If VαVβ, then (Vα+W)/W(Vβ+W)/W. Conversely, if (Vα+W)/W(Vβ+W)/W, then Vα+WVβ+W, and hence VαVβ (since U being well-ordered implies that either VαVβ or VβVα). Thus, upon removing any repeated terms from U¯, sending (Vα+W)/W to Vα gives an order-embedding of U¯ into U, from which it follows that U¯ is well-ordered.

Finally, to show that U¯ is maximal, let Y be a subspace of V containing W, such that Y/W is comparable to each element of U¯. Then we can find a least αΩ such that Y/W(Vα+W)/W. If α is a limit ordinal, then Vα=β<αVβ, by the maximality of U, and hence

(Vα+W)/W=β<α(Vβ+W)/WY/W

implies that Y/W=(Vα+W)/W. Thus let us assume that α is a successor ordinal, with predecessor β. Then Vβ+WYVα+W. Writing Y=YVα, we have VβYVα, since VβVα. Hence Y=Vα, by the maximality of U, and therefore Y/W=(Vα+W)/W once again. Thus U¯ is maximal as a well-ordered set of subspaces of V/W. ∎

We conclude this section with a list of properties that are inherited by quotients, which will be useful later. These are all standard or easy to prove.

Lemma 3.4.

Let k be a field, V a k-vector space, XEndk(V), and W an X-invariant subspace of V. For each of the following properties, if X satisfies it, then so does X¯Endk(V/W).

  • (1)

    The set is closed under addition.

  • (2)

    The set is closed under multiplication (i.e., composition of transformations).

  • (3)

    The set is closed under scalar multiplication.

  • (4)

    The set is topologically nilpotent.

Proof.

We may assume that X, since otherwise the conditions above are vacuously true for both X and X¯. Let T,SX. Then for all vV, we have

(T¯+S¯)(v+W)=T¯(v+W)+S¯(v+W)=(T(v)+W)+(S(v)+W)=(T(v)+S(v))+W=(T+S)(v)+W=(T+S¯)(v+W),

showing that T¯+S¯=T+S¯. Analogous computations also show that T¯S¯=TS¯ and aT¯=aT¯ for all ak. From this it follows immediately that if X is closed under addition, multiplication, or scalar multiplication, then so is X¯.

Finally, suppose that X is topologically nilpotent, let T1,T2,T3,X, and let U be a subspace of V containing W, such that U/W is finite-dimensional. Also, let Y be a finite-dimensional subspace of V such that U+W=Y+W. Then TnT2T1(Y)=0 for some n+. Hence

Tn¯T2¯T1¯(U/W)=TnT2T1¯(U/W)=TnT2T1(Y)+W=W,

from which it follows that X¯ is topologically nilpotent. ∎

4 Topologically nilpotent sets

Using the triangularization lemma of the previous section, we can give an alternative characterization of an arbitrary strictly triangularizable subset of Endk(V), which generalizes the equivalence of (1) and (3) in Proposition 2.4.

Theorem 4.1.

Let k be a field, V a nonzero k-vector space, and XEndk(V). Then X is strictly triangularizable if and only if X is topologically nilpotent.

Proof.

We may assume that X, since otherwise X is both strictly triangularizable and topologically nilpotent, by the definitions of those terms.

Suppose that X is strictly triangular with respect to a well-ordered basis (,) for V, let T1,T2,T3,X, and let v. Seeking a contradiction, suppose also that TnT2T1(v)0 for all n+. Since T1 is strictly triangular with respect to , we have T1(v)=wu1aww for some awk and u1,w, where u1<v and au10. Similarly,

T2T1(v)=wu1awT2(w)=wu2bww

for some bwk and u2,w, where u2<u1 and bu20. Continuing in this fashion, for each n+, we can write TnT2T1(v)=wuncww for some cwk and un,w, where un<<u2<u1, and cun0. Thus we obtain an infinite descending chain <u3<u2<u1 of elements of , which contradicts the hypothesis that is well-ordered. Therefore, it must be the case that TnT2T1(v)=0 for some n+. Since v was arbitrary and the Ti are linear, it follows that, for every finite-dimensional subspace W of V, there exists n+ such that TnT2T1(W)=0. Since the transformations TiX were arbitrary, this shows that X is topologically nilpotent.

Conversely, suppose that X is topologically nilpotent. It suffices to show that X is triangularizable, since by hypothesis, every element of X is topologically nilpotent, and hence strictly triangular with respect to any well-ordered basis for V with respect to which it is triangular, by Proposition 2.4. Since, according to Lemma 3.4 (4), the property of being topologically nilpotent is inherited by quotients, by Lemma 3.2, in turn, it suffices to show that there is a 1-dimensional X-invariant subspace of V. To find such a 1-dimensional subspace, we shall show that TXker(T){0}. For, every (1-dimensional) subspace of TXker(T) is necessarily X-invariant.

Seeking a contradiction, suppose that

TXker(T)={0}.

Then necessarily X{0}, since we have assumed that V is nonzero. Hence we can find T1X and vV such that T1(v)0. Again, since TXker(T)={0}, we can find T2X such that T2T1(v)0. Continuing in this fashion, we obtain T1,T2,T3,X such that TnT2T1(v)0 for all n+, which contradicts X being topologically nilpotent. Hence TXker(T){0}, as desired. ∎

Next, we note that Theorem 4.1 generalizes [1, Theorem 2]. The algebra R in this statement is necessarily nonunital, unless it is zero.

Corollary 4.2 (Chebotar).

Let k be a field, V a nonzero k-vector space, and R a finite-dimensional nilpotent k-subalgebra of Endk(V). Then R is strictly triangularizable.

Proof.

Since R is nilpotent, it is trivially topologically nilpotent, and hence strictly triangularizable, by Theorem 4.1. ∎

Another immediate consequence of Theorem 4.1 is a well-known result from [5].

Corollary 4.3 (Levitzki).

Let k be a field, nZ+, and X a nilpotent multiplicative subsemigroup of Mn(k). Then X is triangularizable.

We conclude this section with one more application of the theorem.

Corollary 4.4.

Let k be a field, V a nonzero k-vector space, XEndk(V), and R the nonunital k-subalgebra of Endk(V) generated by X. If X is topologically nilpotent, then so is R.

Proof.

We may assume that X, since otherwise R={0}. Supposing that X is topologically nilpotent, by Theorem 4.1, there is a well-ordered basis (,) for V with respect to which X is strictly triangular. Now let TR be any element. Then T can be expressed as T=i=1naiSi,1Si,mi for some aik, Si,jX, and n,mi+. Since each Si,j is strictly triangular with respect to , the same is true of the products Si,1Si,mi, and hence also of T. Thus R is strictly triangular with respect to , and therefore topologically nilpotent, by Theorem 4.1. ∎

5 Radicals

The remainder of the paper is concerned with triangularizable subrings of Endk(V). The main goal of this section is to show that the Jacobson radical rad(R) of such a subring R of Endk(V) is topologically nilpotent. We make the following definition to facilitate the discussion.

Definition 5.1.

Let k be a field, V a nonzero k-vector space, and R a subring of Endk(V). Set

TNil(R)={TRTis topologically nilpotent}.

The next lemma is an analogue of the standard fact that every nil left (or right) ideal of a ring is contained in its Jacobson radical (see, e.g., [4, Lemma 4.11]).

Lemma 5.2.

Let k be a field, V a nonzero k-vector space, R a subring of Endk(V) that is closed in the function topology, and J a left (or right) ideal of R. If JTNil(R), then Jrad(R).

Proof.

We shall only treat the case where J is a left ideal, since the right ideal version is entirely analogous.

Let TJ, and let SR. We shall show that 1-ST is invertible in R. Since SR is arbitrary, this implies that Trad(R) (see, e.g., [4, Lemma 4.1]), and hence Jrad(R).

To show that 1-ST is invertible, we first note that STJ, and hence ST is topologically nilpotent, assuming that JTNil(R). Thus for any vV, there exists n+ such that (ST)i(v)=0 for all in. It follows that i=0(ST)i converges to transformation in the function topology on Endk(V). Since i=0m(ST)iR for all m+, since every open neighborhood of i=0(ST)i contains such a finite sum, and since R is assumed to be closed, we have i=0(ST)iR.

Now, for any vV, we can find n+ such that (ST)n(v)=0=(ST)n(1-ST)(v). Then

(1-ST)i=0(ST)i(v)=(1-ST)i=0n-1(ST)i(v)=(1-(ST)n)(v)=v,

and

i=0(ST)i(1-ST)(v)=i=0n-1(ST)i(1-ST)(v)=(1-(ST)n)(v)=v.

Hence (1-ST)-1=i=0(ST)iR. ∎

The next example shows the necessity of assuming that R is closed in this lemma.

Example 5.3.

Let k be a field, and let V be a k-vector space with basis ={vii+}. Define TEndk(V) by

T(vi)={vi-1ifi>1,0ifi=1,

and extend linearly to all of V. Also, let R be the k-subalgebra of Endk(V) generated by T. It is easy to see that p(T)0 for any nonzero polynomial p(x)k[x]. (If p(x)=i=0naixi for some n+ and aik, then 0=p(T)(vn+1)=i=0naivn+1-i would imply that a0==an=0, since {v1,v2,,vn+1} is linearly independent.) Hence Rk[x] as k-algebras, via a map that sends T to x, and in particular rad(R)=0. Now let J be the ideal of R generated by T. Then, for all i=1naiTiJ and for all m+, we have (i=1naiTi)m(vm)=0. Hence JTNil(R), but Jrad(R). (In fact, J=TNil(R).)

We conclude by noting that R is not closed in the function topology. This follows from the fact that i=0Ti converges to a transformation in Endk(V) (by the same reasoning as in the proof of Lemma 5.2), and every open neighborhood of i=0Ti contains an element of R (of the form i=0nTi), but i=0TiR.

If a subring R of Endk(V) is triangularizable, then we can say much more about TNil(R) and its relationship with rad(R) than we did in Lemma 5.2.

Proposition 5.4.

Let k be a field, V a nonzero k-vector space, and R a subring of Endk(V) triangular with respect to a well-ordered basis (B,) for V. Then the following hold.

  • (1)

    TNil(R) is the ideal consisting precisely of the transformations in R that are strictly triangular with respect to (,).

  • (2)

    TNil(R) is topologically nilpotent.

  • (3)

    rad(R)TNil(R).

Moreover, if R is closed in the function topology, then TNil(R)=rad(R).

Proof.

(1) It follows from Proposition 2.4 that TNil(R) consists precisely of the transformations in R that are strictly triangular with respect to (,), and in particular, TNil(R). To show that this set is an ideal of R, let TTNil(R) and S1,S2R. Since T is strictly triangular with respect to , for any v, we have

S1TS2(v)S1T({uuv})S1({uu<v}){uu<v},

and therefore S1TS2 is also strictly triangular with respect to . Since the sum of any two transformations in R that are strictly triangular with respect to is also strictly triangular with respect to , we see that TNil(R) is an ideal of R.

(2) Since TNil(R) is strictly triangularizable, by (1), Theorem 4.1 implies that it is topologically nilpotent.

(3) By (1), it suffices to show that every element of rad(R) is strictly triangular with respect to . Seeking a contradiction, suppose that Trad(R) is not strictly triangular with respect to . Thus πvT(v)=av for some v and ak{0}, where πvEndk(V) is the projection onto v with kernel {v}. Hence

πv(1-a-1T)(v)=πv(v)-a-1πvT(v)=v-v=0,

and so πv(1-a-1T)πv=0. By Lemma 2.2, this shows that 1-a-1T is not invertible (in Endk(V), and hence also in R), contradicting Trad(R). Thus every element of rad(R) is strictly triangular with respect to , and therefore rad(R)TNil(R).

Finally, if R is closed, then TNil(R)rad(R), by Lemma 5.2, since TNil(R) is an ideal of R, by (1). Hence in this case TNil(R)=rad(R), by (3). ∎

The next example shows that TNil(R) need not be topologically nilpotent in general.

Example 5.5.

Let k be a field, let V be a countably infinite-dimensional k-vector space, and identify Endk(V) with the set of column-finite infinite matrices (indexed by the positive integers). Also, let XEndk(V) be the set of all strictly lower-triangular matrices with only finitely many nonzero entries. Thus an arbitrary element of X has the following form:

(00000*0000**000***00000).

It is easy to see that X is a nonunital k-subalgebra of Endk(V), where every element is nilpotent. Thus, in particular, each element of X is strictly triangularizable, by Proposition 2.4. Letting R be the unital k-subalgebra of Endk(V) generated by X, we have X=TNil(R), and this set is clearly an ideal of R.

Next, we wish to show that X=TNil(R) is not topologically nilpotent. For all i,j+, let Ei,j be the matrix unit with a 1 in the i-th row and j-th column, and zeros elsewhere. Then Ei,i-1TNil(R) for all i2, and it is easy to see that En,n-1E3,2E2,1=En,1 for all n2. Hence letting

v=(100),

we have En,n-1E3,2E2,1(v)=En,1(v)0 for all n2, and therefore TNil(R) is not topologically nilpotent.

It follows from Proposition 5.4 that R is not triangularizable (as we have defined the term). This can also be shown directly, by noting that there is no 1-dimensional R-invariant subspace of V. (If R were triangular with respect to a well-ordered basis (,) for V, then the subspace spanned by the least element of would necessarily be R-invariant.) For suppose, seeking a contradiction, that there exists vV{0} such that R(v)v. We can write

v=(a1a2an0),

where a1,,ank and an0. Then En+1,n(v) has an in the (n+1)-th coordinate and zeros elsewhere. Thus En+1,nR, but En+1,n(v)v, giving the desired contradiction.

We conclude by noting that X=rad(R). Since R/Xk as k-algebras, X is a maximal left ideal, and hence rad(R)X. Also, for any TX and any SR,

1-ST=(M00𝟏),

where M is a lower-triangular (finite) matrix with ones on the main diagonal, and 𝟏 is an infinite matrix having ones on the main diagonal and zeros elsewhere. Since M is invertible, by finite-dimensional linear algebra, so is 1-ST, and hence Trad(R). It follows that rad(R)=X=TNil(R).

6 Triangularizable algebras

This section is devoted to describing the triangularizable subalgebras of Endk(V). We begin with a technical lemma.

Lemma 6.1.

Let k be a field, V a nonzero k-vector space, R a k-subalgebra of Endk(V), and I a topologically nilpotent ideal of R. Suppose that there exists {EααΩ}R such that the following conditions are satisfied.

  • (a)

    EαEα-EαI for all αΩ , and EαEβI whenever αβ.

  • (b)

    For every TR and vV , there exist α1,,αnΩ, aα1,,aαnk , and SI such that

    T(v)=i=1naαiEαi(v)+S(v).

Then R is triangularizable.

Proof.

Since I is topologically nilpotent, it is (strictly) triangularizable, by Theorem 4.1. Hence there exists a 1-dimensional I-invariant subspace WV (e.g., by Proposition 3.3), where necessarily I(W)={0}. It follows that U=TIker(T){0}. Moreover, for all TR we have T(U)U, since STI for all SI, and hence ST(U)={0}.

We also note that Eα(U){0} for some αΩ. For if it were the case that Eα(U)={0} for all αΩ, then the fact that I(U)={0} and condition (b) would imply that R(U)={0}. Since R is assumed to be unital, this would mean that U={0}, contrary to construction. Thus we may choose uU{0} and βΩ such that v:=Eβ(u)0. We claim that v is R-invariant.

Since I(U)={0}, the first relation in condition (a) gives 0=EβEβ(u)-Eβ(u), and hence Eβ(v)=v. Likewise, the second relation in condition (a) gives 0=EαEβ(u)=Eα(v) for all αΩ{β}. Moreover, I(v)=0, since v=Eβ(u)U, as noted in the first paragraph. Now let TR, and let α1,,αnΩ, aα1,,aαnk, and SI be such that T(v)=i=1naαiEαi(v)+S(v), using condition (b). Then T(v)=aβEβ(v)=aβv for some aβk (where aβ=0 if β{α1,,αn}), showing that the 1-dimensional space v is R-invariant.

Finally, it follows from Lemma 3.4 that the property of being a k-algebra with a topologically nilpotent ideal is inherited by quotients (see Definition 3.1). Moreover, it is easy to see that the property of having a subset satisfying conditions (a) and (b), with respect to the relevant topologically nilpotent ideal, is also inherited by quotients. Hence R is triangularizable, by Lemma 3.2. ∎

In the results that follow, we view subrings of Endk(V) as topological rings via the subspace topology induced by the function topology on Endk(V), and we view the product kΩ as a topological ring via the product topology, where each component copy of k is taken to be discrete. Also, see Section 2.1 for a review of the quotient topology and isomorphisms of topological algebras.

Proposition 6.2.

Let k be a field, V a nonzero k-vector space, and R a k-subalgebra of Endk(V). Suppose there exist a topologically nilpotent ideal I of R and a set Ω such that R/IkΩ as topological k-algebras. Then R is triangularizable, and I=TNil(R).

Proof.

Let π:RR/I be the canonical projection. Then, by the definition of the quotient topology, π is continuous. Moreover, it is a standard fact that π is necessarily an open map. (For if 𝒰R is an open set, then 𝒰+I=iI(𝒰+i), and each 𝒰+i is open in R, by the continuity of addition. Hence π-1(π(𝒰))=𝒰+I is open, and therefore so is π(𝒰), again by the definition of the quotient topology.)

Since R/IkΩ as topological k-algebras, there must exist {EααΩ}R, such that {Eα+IαΩ}R/I is a set of orthogonal idempotents, and R/I is the closure of the k-vector space spanned by this set. Now let TR, and let 𝒰R be an open neighborhood of T. Then, by the previous paragraph, π(𝒰) is an open neighborhood of π(T), and hence there exist α1,,αnΩ and aα1,,aαnk such that i=1naαiEαi+Iπ(𝒰). It follows that i=1naαiEαi+S𝒰 for some SI. Hence every open neighborhood of TR contains an element of the form i=1naαiEαi+S. In particular, for all vV, there exist α1,,αnΩ, aα1,,aαnk, and SI such that T(v)=i=1naαiEαi(v)+S(v). Thus {EααΩ} satisfies the conditions (a) and (b) of Lemma 6.1, and therefore R is triangularizable.

Next, we note that ITNil(R), since every element of I is topologically nilpotent. To prove the opposite inclusion, let TTNil(R). Then for every open neighborhood 𝒰R of 0, there is some n+ such that Tn𝒰. We claim that T+IR/I enjoys the same property. Letting 𝒰+IR/I be an open neighborhood of I, where 𝒰R, the continuity of the canonical projection π:RR/I implies that 𝒰+I is an open neighborhood of 0 in R. Hence there is some n+ such that Tn𝒰+I, and therefore

(T+I)n=Tn+I𝒰+IR/I.

But in R/IkΩ, the only element r having this property (that for every open neighborhood 𝒰 of 0, there is some n+ such that rn𝒰) is 0. Thus T+I=I, and hence TI. It follows that I=TNil(R). ∎

We observe that Example 5.5 shows that the conclusion of the previous proposition would cease to hold if we were to remove the hypothesis that I is topologically nilpotent. In that example, we constructed a subalgebra R of Endk(V) such that R/rad(R)k as k-algebras, but R is not triangularizable and rad(R) is not topologically nilpotent. (Note that R/rad(R) and k in Example 5.5 are both discrete, and so the isomorphism between them is necessarily topological.)

We now turn to giving necessary conditions for subalgebras of Endk(V) to be triangularizable.

Proposition 6.3.

Let k be a field, V a nonzero k-vector space, and R a k-subalgebra of Endk(V) triangular with respect to a well-ordered basis (B,). Define ϕ:RkB by ϕ(T)=(av)vB, where avk is such that πvT(v)=avv, and πvEndk(V) denotes the projection onto v with kernel B{v}. Then the following hold.

  • (1)

    The map ϕ:Rk is a continuous k -algebra homomorphism.

  • (2)

    TNil(R)=ker(ϕ) , and this ideal is closed in the induced topology on R.

  • (3)

    If R is the k -subalgebra of Endk(V) consisting of all the transformations triangular with respect to (,) , then ϕ is open and surjective.

In particular, R/TNil(R) is commutative.

Proof.

(1) Let S,TR and v. Since S is triangular with respect to , we can write S(v)=bvv+u<vbuu for some bv,buk and u. Since T is triangular with respect to , we have

πvTS(v)=πvT(bvv)+πvT(u<vbuu)=bvπvT(v)=avbvv,

where avk is defined by πvT(v)=avv. It follows that ϕ(TS)=ϕ(T)ϕ(S). It is also straightforward to show that ϕ is k-linear, and therefore ϕ is a k-algebra homomorphism.

Now let 𝒰k be a basic open set. Thus there exist v1,,vn and bv1,,bvnk such that

𝒰={(cv)vcvi=bvifor 1in}.

Then

ϕ-1(𝒰)={SRπviS(vi)=bvivifor 1in}=i=1nw{uu<vi}{SRS(vi)=bvivi+w}

is open in the induced topology on R, from which it follows that ϕ is continuous.

(2) For any TR, we have

ϕ(T)=0πvT(v)=0for allvT(v){uu<v}for allvTTNil(R)(by Proposition 2.4),

and hence TNil(R)=ker(ϕ). Since {0} is closed in k (which, as a product of discrete spaces, is Hausdorff), ker(ϕ)=ϕ-1({0}), and ϕ is continuous, we conclude that ker(ϕ) is closed in the induced topology on R.

(3) Suppose that R consists of all the transformations in Endk(V) triangular with respect to , and let 𝒰R be an open subset. We shall show that ϕ(𝒰) is open.

We may assume that

𝒰={SRS(vi)=wifor 1in}

for some v1,,vn and w1,,wnV, since it is easy to see that every open set in R is a union of sets of this form. Write πvi(wi)=bvivi (1in) for some bvik. Given a cvk for each v, such that cvi=bvi for 1in, define TEndk(V) by T(vi)=wi for 1in, and T(v)=cvv for all v{v1,,vn}. Since T agrees with certain elements of R on {v1,,vn}, and each element of {v1,,vn} is an eigenvector of T, it is triangular with respect to , and hence TR. Thus T𝒰 and ϕ(T)=(cv)v. It follows that

ϕ(𝒰)={(cv)vkcvi=bvifor 1in},

which is open in k. Hence ϕ is open, and it is surjective, by a simpler version of the same argument.

Finally, R/TNil(R) is isomorphic to a subalgebra of k, by (1) and (2), and is hence commutative. ∎

Combining the previous two propositions gives our main result.

Theorem 6.4.

Let k be a field, V a nonzero k-vector space, and R a k-subalgebra of Endk(V). Then the following are equivalent.

  • (1)

    R is triangularizable.

  • (2)

    R is contained in a k -subalgebra A of Endk(V) such that A/rad(A)kΩ as topological k -algebras for some set Ω , and rad(A) is topologically nilpotent.

Moreover, if (2) holds and R is closed in the function topology, then rad(R)=Rrad(A).

Proof.

Suppose that R is triangular with respect to some well-ordered basis (,) for V. Let A be the k-subalgebra of Endk(V) consisting of all the transformations triangular with respect to (,), and let ϕ:Ak be as in Proposition 6.3. Then, by that proposition, ϕ is a surjective open continuous k-algebra homomorphism, such that ker(ϕ)=TNil(A). Hence A/TNil(A)k as topological k-algebras (see, e.g., [11, Theorem 5.11]). Now, by Lemma 2.5, the closure cl(A) of A in the function topology on Endk(V) is triangular with respect to . Hence cl(A)A, and so A is closed in Endk(V). Therefore, by Proposition 5.4, TNil(A)=rad(A), and this ideal is topologically nilpotent, showing (2).

Now suppose that (2) holds. Then, by Proposition 6.2, A is triangularizable, and hence so is R, as a subset of A, proving (1). Moreover, by the same proposition, rad(A)=TNil(A), and if R is closed, then also rad(R)=TNil(R), by Proposition 5.4. Hence in this situation,

rad(R)=TNil(R)=RTNil(A)=Rrad(A),

proving the final claim. ∎

The next example shows that for a closed triangularizable subalgebra R of Endk(V) it is not necessarily the case that R/rad(R)kΩ (as topological k-algebras) for some set Ω.

Example 6.5.

Let k be a countable field, Ω=+{}, and V a k-vector space with basis ={viiΩ}. Then is well-ordered by the relation , where vivj if and only if either i,j+ and ij, or j=. For each i+, define EiEndk(V) by

Ei(vj)={viifj{i,},0otherwise,

and extend linearly to all of V. Then {Eii+} is a set of orthogonal idempotents, which is clearly triangular with respect to (,). It follows that the k-subalgebra R of Endk(V) generated by this set is also triangular with respect to (,).

Every element of R can be expressed in the form a1+i=1naiEi for some n+ and a,aik. Notice that

(a1+i=1naiEi)(v)=av+i=1naivi,

and so each element of R is completely determined by its action on v. This implies that R is discrete in the topology induced on it by the function topology on Endk(V). The above computation also shows that if a0, then (a1+i=1naiEi)m(v)0 for all m+. Moreover, if a=0 but an0, then

(a1+i=1naiEi)m(vn)=anmvn0

for all m+. Thus R has no nonzero topologically nilpotent elements, and hence rad(R)=0, by Proposition 5.4.

Define ϕ:RkΩ as in Proposition 6.3. Specifically, ϕ(a1+i=1naiEi)=(bi)iΩ, where bi=ai+a for 1in and bi=a for all j>n. Thus ϕ is an injective k-algebra homomorphism (which is also continuous, by Proposition 6.3), with ϕ(R)=k(+){1}, where k(+) is the direct sum of copies of k indexed by the elements of + (with each element of this subring of kΩ understood to have 0 in the coordinate). Therefore

R/rad(R)Rk(+){1}

as k-algebras. In particular, R/rad(R)kΔ for any set Δ, since such a Δ would need to be infinite, making kΔ uncountable, in contrast to R (given that k was assumed to be countable).

Next, let us show that R is closed. Let cl(R) denote the closure of R in the function topology on Endk(V), and let Tcl(R). Then there exist n+ and a,a1,,ank such that a1+i=1naiEiR agrees with T on v. Since each element of R is completely determined by its action on v, and since every neighborhood of T must contain an element of R, it follows that T=a1+i=1naiEi, which belongs to R, and hence R=cl(R).

We also observe that ϕ is not open. Since R is discrete, {E1}R is open. On the other hand, the intersection of any open subset of kΩ with k(+){1} is infinite, and hence ϕ({E1}) cannot be open in the induced topology on k(+){1}.

Finally, let A be the subalgebra of Endk(V) consisting of all the transformations triangular with respect to (,), and extend ϕ to a map ϕ:AkΩ (again, defined as in Proposition 6.3). Then ϕ is open and surjective, by Proposition 6.3(3), but ϕ is not closed, since it sends the closed set R to k(+){1}, whose closure in kΩ is kΩ.

7 Algebraically closed fields

Next, we wish to add another condition to Theorem 6.4, in the case where the field is algebraically closed. This will require defining more topological terms and recalling an earlier result.

Definition 7.1.

Let k be a field and R a topological k-algebra. Then the ring R is called pseudocompact if the topology on R is complete, is Hausdorff, and has a basis of neighborhoods of 0 consisting of ideals I such that R/I has finite length both on the left and on the right (i.e., R/I is a two-sided artinian ring). We also say that R is k-pseudocompact if it is pseudocompact and every open ideal of R has finite k-codimension.

Proposition 7.2 ([3, part of Proposition 3.15]).

Let k be a field and R a topological k-algebra. Then the following are equivalent.

  • (1)

    R is k -pseudocompact and rad(R)={0}.

  • (2)

    RαΩ𝕄nα(Dα) as topological k -algebras for some set Ω, nα+ , and Dα finite-dimensional division k -algebras, where αΩ𝕄nα(Dα) is given the product topology with each 𝕄nα(Dα) discrete.

We are now ready for the extension of Theorem 6.4 in the case where the field is algebraically closed.

Corollary 7.3.

Let k be an algebraically closed field, V a nonzero k-vector space, and R a k-subalgebra of Endk(V). Then the following are equivalent.

  • (1)

    R is triangularizable.

  • (2)

    R is contained in a k -subalgebra A of Endk(V) such that A/rad(A)kΩ as topological k -algebras for some set Ω , and rad(A) is topologically nilpotent.

  • (3)

    R is contained in a k -subalgebra A of Endk(V) such that A/rad(A) is k -pseudocompact and commutative, and rad(A) is topologically nilpotent.

Proof.

The equivalence of (1) and (2) follows from Theorem 6.4. Also, by Proposition 7.2, if A/rad(A)kΩ as topological k-algebras, then A/rad(A) is k-pseudocompact, and hence (2) implies (3).

Finally, let us suppose that (3) holds and prove (2). By Proposition 7.2, A/rad(A) being k-pseudocompact implies that A/rad(A)αΩ𝕄nα(Dα) as topological k-algebras, where each Dα is a finite-dimensional division algebra over k. Since we are assuming that A/rad(A) is commutative, each nα=1, and each Dα must be a finite-dimensional field extension of k. Since k is algebraically closed, each Dα must be isomorphic to k, and hence A/rad(A)kΩ as topological k-algebras, from which (2) follows. ∎

We can now give a generalization of the well-known result, proved by McCoy [6], that a subalgebra of a matrix ring over an algebraically closed field is triangularizable if and only if the subalgebra is commutative modulo its Jacobson radical.

Corollary 7.4.

Let k be a field, nZ+, and R a k-subalgebra of Mn(k). Then the following are equivalent.

  • (1)

    R is triangularizable.

  • (2)

    R/rad(R)km as k -algebras, for some m+.

If k is algebraically closed, then these are also equivalent to the following.

  • (3)

    R/rad(R) is commutative.

Moreover, if (2) holds, then mn.

Proof.

Suppose that (1) holds. Since 𝕄n(k) is finite-dimensional as a k-algebra, it and R are discrete in the function topology, and in particular, R is closed. Hence, by Theorem 6.4, R/rad(R) is isomorphic to a subalgebra of kr for some r+. It is well known that every subalgebra of kr is of the form km for some m+. (See [3, Corollary 4.8] for a proof of a more general version of this statement.) Therefore R/rad(R)km as k-algebras, proving (2).

Next suppose that (2) holds. Since R is finite-dimensional, and hence left artinian, rad(R) is (topologically) nilpotent (see, e.g., [2, Proposition IX.2.13] or [4, Theorem 4.12]). Moreover, since R is discrete in the function topology and km is discrete in the product topology, R/rad(R)km is a topological isomorphism. Therefore R is triangularizable, by Theorem 6.4, showing (1).

Clearly, (2) implies (3). Let us next assume that k is algebraically closed and that (3) holds, and show that (2) must also hold. Since R/rad(R) is left artinian, and since it has zero Jacobson radical, R/rad(R) is semisimple (see, e.g., [4, Theorem 4.14]). Thus R/rad(R) is isomorphic as a k-algebra to a finite direct product of matrix rings over division k-algebras (see, e.g., [2, Theorem IX.5.4]). Since R/rad(R) is commutative, each of these matrix rings must be 1×1, and each of the division k-algebras must be a field extension of k. Since R/rad(R) is a finite-dimensional k-algebra, and k is assumed to be algebraically closed, each of these fields must be isomorphic to k. Hence R/rad(R)km as k-algebras, for some m+.

Finally, if (2) holds, then, as mentioned above, rad(R)TNil(R). Since km has no nonzero nilpotent elements, it follows that rad(R)=TNil(R). Since R is triangularizable, by the equivalence of (1) and (2), Proposition 6.3 implies that R/rad(R) is isomorphic to a subalgebra of kn. Thus

m=dimk(km)=dimk(R/rad(R))dimk(kn)=n,

which shows that mn. ∎

Acknowledgements

I am grateful to George Bergman for a very enlightening conversation about this material, and for numerous comments on an earlier draft of this paper, which have led to significant improvements. I also would like to thank Greg Oman for a pointer to the literature, Manuel Reyes for a helpful example, and the referee for a careful reading of the manuscript.

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About the article


Received: 2018-05-07

Revised: 2018-06-27

Published Online: 2018-08-11

Published in Print: 2019-01-01


Citation Information: Forum Mathematicum, Volume 31, Issue 1, Pages 19–33, ISSN (Online) 1435-5337, ISSN (Print) 0933-7741, DOI: https://doi.org/10.1515/forum-2018-0109.

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