Additive closed symmetric monoidal structures on
R-modules
Mark Hovey
Department of Mathematics
Wesleyan University
Middletown, CT 06459
____________________________________________________________________
Abstract
In this paper, we classify additive closed symmetric monoidal structures on
the category of left R-modules by using Watts' theorem. An additive closed
symmetric monoidal structure is equivalent to an R-module A,B equipped with
two commuting right R-module structures represented by the symbols A and
B, an R-module K to serve as the unit, and certain isomorphisms. We use this
result to look at simple cases. We find rings R for which there are no additive
closed symmetric monoidal structures on R-modules, for which there is exactly
one (up to isomorphism), for which there are exactly seven, and for which there
are a proper class of isomorphism classes of such structures. We also prove some
general structual results; for example, we prove that the unit K must always be
a finitely generated R-module.
Key words: symmetric monoidal, closed symmetric monoidal, module
2000 MSC: 18D10, 16D90
____________________________________________________________________
Introduction
It is well known that the category of left R-modules becomes closed symmet-
ric monoidal under the tensor product A R B if and only if R is commutative.
However, there are many other cases when the category of R-modules is closed
symmetric monoidal. For example, if k is a field and G is a group, the cate-
gory of k[G]-modules (that is, representations of G on k-vector spaces) is clos*
*ed
symmetric monoidal under A k B, even though k[G] is not commutative in
general. This is explained by the fact that k[G] is a Hopf algebra. But there a*
*re
other examples where R is not a Hopf algebra, such as the category of perverse
R-modules considered in [Hov08 ].
So a natural question to ask is just what one needs to know about R in
order to produce a closed symmetric monoidal structure on the category of
R-modules. Of course, we do not really want an arbitrary closed symmetric
____________________________
Email address: hovey@member.ams.org (Mark Hovey)
Preprint submitted to Elsevier June 5, 2009
monoidal structure; we require that the monoidal product be an additive functor
in both variables. We would like to be able to answer basic questions such as t*
*he
following. Are there rings R where the category of R-modules cannot be given
an additive closed symmetric monoidal structure? Are there rings R where the
category of R-modules possesses a unique additive closed symmetric monoidal
structure?
At first glance, such problems seem completely intractable because closed
symmetric monoidal structures are so complicated, involving the entire category
of R-modules. The key ingredient, though, is Watts' theorem [Wat60 ]. This
theorem says that any additive functor F from R-modules to abelian groups
that is right exact and commutes with direct sums is naturally isomorphic to
R (-) for some R-bimodule . After some work, we then see that a closed
symmetric monoidal structure A ^ B on left R-modules must be given by
A ^ B ~=((R ^ R) R A) R B,
so that the functor - ^ - is determined by R ^ R, as a 2-fold bimodule (one left
module structure, and two right module structures).
Then the natural thing to do is try to determine which 2-fold bimodules A,B
actually arise as R^R for some closed symmetric monoidal structure -^-. This
is more complicated than it seems because one must deal with the coherence
isomorphisms of a closed symmetric monoidal structure, but of course it can be
done. We also determine when two symmetric monoidal structures determined
by and , respectively, are equivalent as symmetric monoidal functors. This
involves an isomorphism
X X -!X
of 2-fold bimodules, where X is an element of the bimodule Picard group of R.
We establish some basic structural results, though we think there is much
more to say. For example, we show that the unit K of an additive closed
symmetric monoidal structure on left R-modules must be a finitely generated
R-module with a commutative endomorphism ring. To proceed further along
these lines, it might be worthwhile to develop a theory of flatness for an addi-
tive symmetric monoidal structure ^ on R-modules, and concentrate on those
additive symmetric monoidal structures for which projectives are flat.
We also consider examples. For example, if R is a field or a principal ideal
domain that does not contain a field, then there is exactly one additive closed
symmetric monoidal structure on R-modules (up to symmetric monoidal equiva-
lence). If R is a division ring that is not a field, there are no additive clos*
*ed sym-
metric monoidal structures on R-modules. If R is the group ring F2[Z=2], there
are precisely 7 different closed symmetric monoidal structures on R-modules,
though only three different underlying functors. If R is the group ring k[Z=2]
where the characteristic is not 2, however, there are a proper class of inequiv*
*alent
additive closed symmetric monoidal structures on the category of R-modules.
Most of these cannot come from Hopf algebra structures on R.
2
Throughout this paper, the symbol will denote the tensor product over
the ring R unless otherwise stated. Furthermore, all functors will be assumed
to be additive, even if not explicitly stated to be so.
1. n-fold bimodules
Throughout this paper, we will be working with the category of left R-
modules, but we will frequently have to work with left R-modules that have
multiple different commuting right R-module structures. This necessitates some
complicated notation. We will denote an R Z (Rop) Zn-module by 1,2,...,n,
where the subscripts denote the commuting right R-actions. If we need elements,
we will write x ir for the ith multiplication. We will call such an object an
n-fold bimodule, and denote the category of such things as Bimod n(R). Now
the symmetric group n acts on Bimod n(R) by permuting the right module
structures. Indeed, if oe 2 n is a permutation, then
(oe )1,2,...,n= oe(1),...,oe(n).
We note that in practice, as we will see below, it is usually easier to denote *
*the
different right module structures by letters, such as A,B,C. If oe denotes the
permutation (132), for example, then we would write oe as C,A,B. A map
f : A,B,C-! C,A,B
would then be a map of left R-modules such that f(x 1r) = f(x) 2r (matching
up the position of the A's), f(x 2 r) = f(x) 3 r, and f(x 3 r) = f(x) 1 r.
Now, note that the tensor product over R defines a bifunctor
Bimod n(R) x Bimodm (R) -!Bimod n+m-1 (R),
obtained by tensoring the nth right module structure on the first factor with
the left module structure on the second factor. For example, we will have
expressions like
f : C, B,A -! ,A C,B.
This denotes a map of 3-fold bimodules. In the domain, we tensor the second
right R-module structure on with the left module structure on , whereas
in the range we tensor the first bimodule structure on with the left module
structure on . Furthermore, we have f((x 1r) y) = f(x y) 2r, where the
second right module structure on the range comes from the first right module
structure on . Similarly, we have f(x (y 1 r)) = f(x y) 3 r. We also
have f(x (y 2 r)) = f(x y) 1 r, where the first right module structure
on the range comes from the second right module structure on , because we
have used the first one to form the tensor product in the range.
3
2. Closed symmetric monoidal structures
In this section, we prove our main classification result. We remind the read*
*er
that all tensor products are over R unless otherwise stated.
Theorem 2.1. Suppose the category of R-modules admits an additive closed
symmetric monoidal structure - ^ -. Then R ^ R is a 2-fold bimodule, and
there is a natural isomorphism of bifunctors
(R ^ R)B,A A B ~=A ^ B.
This theorem is written using the notation of the previous section, so that,
in the domain of this isomorphism
(x 1 r) y z = x y (rz) and (x 2 r) y z = x (ry) z.
Also note that if R is a k-algebra, for a commutative ring k, we can look at
k-linear closed symmetric monoidal structures. This would mean that multipli-
cation by x 2 k on A would induce multiplication by x on A ^ B and on B ^ A
for all B. In this case, the three potentially different actions of k on R^R wo*
*uld
in fact all be the same.
Proof. Fix an R-module A, and consider the functor B 7! A ^ B from left R-
modules to left R-modules. This functor preserves direct sums and is right exact
(it is a left adjoint, because of the closed structure). Watt's theorem [Wat60 ]
then implies that A ^ R is an R-bimodule, and
(A ^ R) B ~=A ^ B
naturally in B. To see that this is also natural in A, recall that the map
ffA,B: (A ^ R) B -!A ^ B
is defined as follows, in the proof of Watt's theorem. Given b 2 B, let OEb: R *
*-!B
denote the map of R-modules that takes 1 to b. There is then an induced map
A ^ OEb, and ffA,B(x b) is defined to be (A ^ OEb)(x). From this, it is easy *
*to
check that ffA,B is natural in A as well.
Now let G(A) = A ^ R. Then G(A) is a right exact functor from R-modules
to R-bimodules (or R Z Rop-modules) that preserves direct sums. We can *
*__
therefore apply Watt's theorem [Wat60 ] again to give us the desired result. |*
*__|
The natural question then arises as to which 2-fold bimodules B,A define a
closed symmetric monoidal structure on the category of R-modules.
We first point out that the closed structure always exists.
Lemma 2.2. Suppose B,A is a 2-fold R-bimodule. Then we have a natural
isomorphism
R-mod( B,A A B, P ) ~=R-mod(A, Bimod( B,A, Z-mod(B, P ))),
where we use the right R-module structure on denoted by the subscript B to
form Bimod ( B,A, Z-mod(B, P )), and we use the one denoted by the subscript
A to make this abelian group into a left R-module.
4
Here Hom Z(B, P ) is an R-module via the R-action on P , and a right R-
module via the R-action on B.
Proof. This is really just an exercise in adjointness of tensor and Hom, though
one has to be careful to keep track of all the actions. It is easiest to work m*
*ore
generally. Suppose M is a bimodule. Then we have a natural isomorphism
OE: R-mod (M B, P ) ~=Bimod (M, Z-mod (B, P )).
This isomorphism is defined as usual by OE(f)(m)(b) = f(m b). The reader
must check that OE(f) is a map of bimodules. To see that OE is an isomorphism,
one constructs its inverse _, where _(g)(m b) = g(m)(b). Again, there are
many details to check, which we leave to the reader. Applying this isomorphism
to A = B,A A, we get
R-mod ( B,A A B, P ) ~=Bimod ( B,A A, Z-mod (B, P )).
Now suppose N is a general bimodule. Then there is a natural isomorphism
oe :Bimod ( B,A A, N) -!R-mod (A, Bimod( B,A, N)).
Once again, we have oe(f)(a)(~) = f(~ a), and, for the inverse o of oe, we ha*
*ve
o(g)(~ a) = g(a)(~). We leave it to the reader to check the details. Taking_
N = Hom (B, P ) completes the proof. |__|
Naturally, the other conditions necessary for a symmetric monoidal structure
are considerably more complicated. The basic idea, however, is simple. In order
to get symmetry of the product A ^ B, we need the two right module structures
on to be isomorphic. In order to get associativity of A ^ B, we need the
three different right module structures on to be isomorphic. These two
together, of course, will imply that all of the different permutations of the n*
* + 1
different right module structures on n will be isomorphic. Then we also need
a unit.
Theorem 2.3. Let B,A be a 2-fold bimodule used to define - ^ - on R-
modules. There is a one-to-one correspondence between additive closed symmet-
ric monoidal structures on R-modules with - ^ - as the monoidal product and
the following data:
(a)An associativity isomorphism
a: C, B,A -! ,A C,B.
This can be remembered by noting that the subscripts on the first in the
target are the second subscripts on the two 's in the domain, and the
subscripts on the second in the target are the first subscripts on the t*
*wo
's in the domain.
(b)A left R-module K and a unit isomorphism `: B,K K ~=RB of bimod-
ules.
5
(c)A commutativity isomorphism c: B,A -! A,B.
This data must satisfy the following coherence conditions.
1.(Associativity pentagon) Let = = for notational clarity. Then the
following diagram commutes.
D, C, B,A -a-1-! , D,C B,A-1-T-! , B,A D,C
? ?
1 a?y ?ya 1
D, ,A C,B ---!a 1 ,A D, C,B---!1 a ,A ,B D,C
Here 1 T switches the last two factors using the commutativity isomorphism*
* of
R, but also reverses the symbols and , which after all both mean . Th*
*is
necessitates changing the subscripts on as well.
2.(Compatibility of left and right unit) The following diagram commutes:
B, K,A K -a-1---! ,A B,K K-1-`---! B,A
flfl fl
fl flfl
B, K,A K -----!1 cB1, A,K K-----!1 `B,A.
3.(Commutativity-associativity hexagon) The following diagram commutes.
C, B,A -c--! C, A,B -a--! ,B C,A--c-! ,B A,C
flfl fl
fl flfl
C, B,A ---!a ,A C,B ---!c A, C,B---!a ,B A,C.
4.The composite
B,A c-! A,B c-! B,A
is the identity.
If R is a k-algebra for a commutative ring k, and we are looking at k-linear
closed symmetric monoidal structures, then this theorem remains true as long
as the three different k-module structures on are the same.
Proof. Just using the usual associativity and commutativity isomorphisms for
R , we find natural isomorphisms
(A ^ B) ^ C ~= C. B,A A B C
and
A ^ (B ^ C) ~= ,A C,B A B C.
Given a, it is now clear how to define a natural associativity isomorphism aA,B*
*,C
for - ^ -, simply as a 1 1 1. On the other hand, given the natural
associativity isomorphism aA,B,C, we let A = B = C = R to get a. One can see
that a then respects the given right module structures by using naturality with
6
respect to the right multiplication by x maps rx: R -!R, in the A, B, and C
slots.
There is a similar equivalence between the isomorphism `: B,K K -!RB
and a natural left unit isomorphism `B :K ^ B -! B. There is also a similar
equivalence between c and a natural commutativity isomorphism cA,B.
An excellent reference for the coherence diagrams needed to make aA,B,C,
`B , and cA,B part of a symmetric monoidal structure is [JS93], particularly
Propositions 1.1 and 2.1. They show that the only coherence diagrams needed
are the associativity pentagon, the compatibility between the right and left
unit, the commutativity-associativity hexagon, the fact that the right unit is
rB = `B cB,K, and the fact that c2 is the identity. Given cA,B, this means we
do not need rB , so we have omitted it. One must now merely translate these
coherence diagrams into analogous facts about a, `, and c to complete the proof.
The associativity pentagon is perhaps the most confusing, so we will discuss
that one in some detail, and leave the others to the reader. Here is the standa*
*rd
associativity pentagon.
((A ^ B) ^ C) ^ DaA^B,C,D-------!(A ^ B) ^a(CA^,D)B,C^D-------!A ^ (B ^ (C ^ *
*D))
? fl
aA,B,C^1?y flfl
(A ^ (B ^ C)) ^ D-------!aA,B^C,DA ^ ((B ^-C)-^-D)----!1^aB,C,DA ^ (B ^ (C ^ *
*D))
Using the standard commutativity and associativity isomorphisms of , the first
term
((A ^ B) ^ C) ^ D
is represented by D, C, B,A (tensored with A B C D). Here B,A
tensors A and B, C, tensors (A ^ B) and C, and D, tensors ((A ^ B) ^ C)
and D. The map aA^B,C,D treats A ^ B as a single object, and will therefore
leave the factor B,A unchanged, so is represented by
a 1: D, C, B,A -! , D,C B,A.
The map aA,B,C^D treats C ^D as a single object, and will therefore leave D,C
unchanged and apply a to the other two. In order to do this, we would like to
switch the order of D,C and B,A, and then apply a 1. It turns out to be
notationally much easier later if we also reverse the names of and , so that
the map aA,B,C^D is represented by the composite
, D,C B,A 1-T--! , B,A D,C a-1-! ,A ,B D,C.
This completes the clockwise half of the associativity pentagon. The counter-
clockwise part is simpler. The first map aA,B,C^ 1 leaves D alone, so will leave
D, alone. It is therefore represented by
1 a: D, C, B,A -! D, ,A C,B.
7
The next map aA,B^C,D treats B ^C as a single entity, so will leave C,B alone.
It is then represented by
a 1: D, ,A C,B -! ,A D, C,B.
Finally, the last map 1 aB,C,D leaves A alone, so will leave ,A alone. It is
reprensented by
a 1: ,A D, C,B -! ,A ,B D,C.
This completes the construction of the associativity pentagon. |___|
We now point out that Watt's theorem can also be used to classify additive
symmetric monoidal equivalences between additive symmetric monoidal struc-
tures on R-modules. In an attempt to make the various bimodule structures
clear, we have used Y and Z as alternative names for X in the theorem below.
We have also used T for the usual commutativity isomorphism of the tensor
product and for a general permutation of tensor factors.
Theorem 2.4. Suppose ^ and are additive symmetric monoidal structures
on the category of R-modules with units K and K0, respectively, and repre-
sented by the 2-fold bimodules and , respectively. Then an additive symmet-
ric monoidal functor from ^ to that has a right adjoint is equivalent to an
bimodule X, an isomorphism j :K0-! X K, and an isomorphism
m: Y,X X Y -! X
of 2-fold bimodules, such that the following diagrams commute:
1.(Unit)
Y,K0 K0 Y 1-j-1---! Y,X XK K Y1-1-T---! Y,X XK Y K
? ?
` 1?y ?ym
X _______ X -----1 ` X Y,K K
2.(Commutativity)
Y,X X Y --m---!X
? ?
c T?y ?y1 c
X,Y Y X-----!mX
3.(Associativity)
Z, Y,X X Y Z -m-! Z,X (X ) Z-T-! Z,X X Z
? ?
a T?y ?ym 1
,X Z,Y Y Z X X
? ?
1 m 1?y ?y1 a
X,X (X ) X --!T X,X X X --!m X
8
Composition of additive symmetric monoidal functors corresponds to the
tensor product of bimodules, and the identity functor corresponds to the bi-
module RR . Thus, additive symmetric monoidal equivalences of additive sym-
metric monoidal structures are given by tensoring with a bimodule that lies in
the bimodule Picard group (see [Yek99]). In fact, if X lies in the bimodule Pi-
card group, then tensoring with X loses no information. In this case, then, the
compatibility diagrams above show that the isomorphisms `, c, and a for are
determined by the corresponding isomorphisms for ^, m, and j. Thus we can
think of the bimodule Picard group as acting on symmetric monoidal structures
with fixed unit K, though there is also an action by the automorphisms of K,
and, if is fixed, the 2-fold bimodule automorphisms of .
It is important to realize that the tensor product Y,X X Y does not use
the right module structures on X and Y , only the left module structures. Thus
these right module structures are still available to make Y,X X Y into a
2-fold bimodule.
One could similarly prove that natural transformations between additive
symmetric monoidal functors represented by X1 and X2 are induced by maps
of bimodules X1 -!X2.
Proof. Suppose F is a symmetric monoidal functor from ^ to with a right
adjoint. Then Watts' theorem implies that there is a bimodule X and a natural
isomorphism X M -! F M. Because F is symmetric monoidal, we have a
natural isomorphism
mM,N -! F M F N -! F (M ^ N).
This translates to a natural isomorphism
mM,N : Y,X (X M) (Y N) -!X ( N,M M N),
where Y is just another name for X. Taking M = N = R gives us the desired
isomorphism m. The unit isomorphism j :K0-! F K is just the map j :K0-!
X K. On the other hand, given X, m, and j, we define F M = X M, j in
the obvious way, and mM,N by naturality from m. We leave to the reader the
translation between the compatibility diagrams of F and the diagrams_in_the
theorem. |__|
3. Examples
In this section, we consider some examples of additive closed symmetric
monoidal categories on R-modules. In particular, we find rings R where there
are no such structures, where there is eactly one (up to additive symmetric
monoidal equivalence), where there are exactly seven, and where there are a
proper class.
The most obvious case is when R is a commutative ring, where - ^ - is the
usual tensor product. This corresponds to B,A = R with a 1 r = a 2 r =
9
ar = ra. The maps a and c are both identity maps, the unit K is R itself, and
the map ` is multiplication.
Now suppose R is a cocommutative Hopf algebra over a field K, with diago-
nal and counit ffl. As is well-known, the category of R-modules then becomes
a closed symmetric monoidal category under the functor - K -, where R acts
by taking the diagonal and having it then act on each factor. This corresponds
B,A = RB K RA . The right R-module structures are just right multiplication
on the two factors, and the left R-module structure is, as we mentioned above,
the composite
R K (R K R) --1-!R K R K R K R 1-T-1---!R K R K R K R ~-~--!R K R,
where denotes the diagonal and ~ denotes the multiplication. The commu-
tativity isomorphism is just the twist map c: RB K RA -! RA K RB . For
this to be a map of left R-modules, we need R to be cocommutative. The unit
isomorphism is the obvious isomorphism
`: (RB K RK ) R K ~=RB K K ~=RB .
The associativity isomorphism
a: (RC K RR ) R (RB K RA ) -!(RR K RA ) R (RC K RB )
is more confusing. It is pretty clear that we should define
a(x 1 z w) = 1 w x z.
But then this forces us to define
X X
a(x y z w) = a(x 1 y0z y00w) = 1 y00w x y0z.
Coassociativity then implies of the diagonal on R then implies, after some
painful checking, that this is a map of left R-modules. We leave to the ex-
cessively diligent reader the check that all the required coherence diagrams
commute.
We would now like to classify all the additive closed symmetric monoidal
structures on R-modules, up to additive symmetric monoidal equivalence, for
various R. The easiest case is when the unit of the symmetric monoidal structure
is R itself. This forces R to be commutative, and in this case there is only one
such closed symmetric monoidal structure.
Proposition 3.1. Suppose R is a ring equipped with an additive closed sym-
metric monoidal structure on the category of R-modules with unit isomorphic
to R. Then R is commutative and this closed symmetric monoidal structure is
additively symmetric monoidal equivalent to the usual one.
We point out as a general rule that if the unit K0 is isomorphic to some R-
module K, we can always assume that the unit is K, up to symmetric monoidal
10
equivalence. Indeed, we construct a new additive closed symmetric monoidal
structure by leaving everything the same except the unit isomorphism `, which
we modify by the isomorphism so that the unit is K. The coherence diagrams
still commute, so this is a closed symmetric monoidal structure. A symmetric
monoidal equivalence between this new structure and the old one is given by
the identity functor, with the unit map K0-! K given by the isomorphism.
This proposition is saying that the coherence isomorphisms a, `, and c of
the standard symmetric monoidal structure are determined up to symmetric
monoidal equivalence. In fact, the proof shows that a and c are exactly deter-
mined, though there is some room for flexibility in `.
There is a quick proof that R must be commutative, since the endomorphism
ring of the unit in a symmetric monoidal category must always be commutative,
and if the unit is isomorphic to R that endomorphism ring is R as well. However,
this also falls out of the coherence isomorphisms, so we re-prove this fact in *
*the
proof below.
Proof. The unit isomorphism shows that 1,2~=R as a bimodule, using the first
right module structure on . We can then assume it is R using a symmetric
monoidal equivalence. Define oe :R -!R by oe(x) = 1 2 x. Note that
z 2 x = (z . 1) 2 x = z(1 2 x)zoe(x),
so that oe gives us complete information on the bimodule 1,2. A similar com-
putation shows that oe is a ring homomorphism.
To see that oe is in fact an isomorphism, consider the commutativity isomor-
phism c: A,B -! B,A. This has the property that
c(y) = c(1 1 y) = c(1) 2 y = c(1)oe(y).
Since c is an isomorphism, we conclude that oe is an isomorphism.
We claim that associativity forces R to be commutative and oe to be the
identity. Indeed, we have
a: C, B,A -! ,A C,B.
Both of these are isomorphic to R as left modules. In the domain, we have
x y = x(1 y) = x(oe(y) 1) = xoe(y)(1 1),
and in the target we have
z w = z(1 w) = z(w 1) = zw(1 1).
Thus a is determined by a(1 1), which must be fl 1 for some unit fl 2 R.
We will then have
a(x y) = xoe(y)(fl 1).
We first show that R is commutative. Choose an arbitrary r, s 2 R. Find y
such that oe(y) = sfl-1, using the fact that oe is an isomorphism. Then we have
a((1 C r) y) = a(1 y) C r.
11
But we have
a((1 C r) y) = a(r y) = roe(y)fl(1 1) = rs(1 1),
and
a(1 y) C r = oe(y)fl(1 r) = s(r 1) = sr(1 1).
We conclude that rs = sr, so R is commutative.
We must also have
a(1 (1 A z)) = a(1 1) A z.
This means that
a(1 oe(z)) = (fl A z) 1 sooe2(z)fl(1 1) = floe(z)(1 1).
Thus oe2(z) = oe(z) for all z. Since oe is necessarily one-to-one, we conclude *
*that
oe(z) = z.
We now know that 1,2is isomorphic to R with both right module struc-
tures, and the left module structure, equal to the canonical one. Then the
associativity isomorphism a of Theorem 2.3 is just an isomorphism of left R-
modules from R to itself, so must be right multiplication by some unit r. But
then the associativity pentagon shows that r2 = r3, so r = 1. Similarly, the
commutativity isomorphism is right multiplication by a unit s, and, since we
now know a is the identity, the commutativity-associativity hexagon says that
s2 = s, so s = 1. Finally, ` must also be multiplication by some unit t, but the
coherence diagrams will commute no matter what t is. However, we can define a
symmetric monoidal equivalence from the usual symmetric monoidal structure
to the one with ` = t by letting F be the identity functor, letting the natural
isomorphism m be the usual one, and letting j :R -!R be right multiplication_
by t. |__|
There are some simple cases where R is the only possible unit of a closed
symmetric monoidal structure on the category of R-modules.
Theorem 3.2. Let n be an integer. There is a unique additive closed symmetric
monoidal category structure on the category of Z=nZ-modules, up to symmetric
monoidal equivalence.
This theorem was proved in case n = 0 by Foltz, Lair, and Kelly [FLK80 ].
Proof. A right or left Z=nZ-module structure on an abelian group is unique;
we must have nx = xn = x + x + . .+.x for n 0 and the negative of this
for n < 0. Thus the 2-fold bimodule needed to define a closed symmetric
monoidal structure on Z=nZ-modules is simply a Z=nZ-module, with all of the
module structures being the same. The unit isomorphism guarantees that is
in the Picard group of Z=nZ, which is trivial (see [Lam99 , Example 2.22D]).
Hence there is an isomorphism f : -! Z=nZ. Proposition 3.1 completes_the_
proof. |__|
12
The other simple case is when R is a division ring.
Theorem 3.3. Suppose k is a division ring. If k is not a field, then there is
no additive closed symmetric monoidal structure on the category of k-modules.
If k is a field, there is a unique additive closed symmetric monoidal category
structure on the category of k-modules, up to symmetric monoidal equivalence.
Proof. Suppose we have a closed symmetric monoidal structure induced by
B,A. The unit isomorphism
B,K k K ~=k
shows that K has to be a one-dimensional vector space, so is isomorphic to_k.
Proposition 3.1 completes the proof. |__|
Since the axioms for an additive closed symmetric monoidal structure on
the category of R-modules do not actually mention R itself, the existence and
number of such structures are both Morita invariant. Hence we get the following
corollary.
Corollary 3.4. Suppose R is a simple artinian ring, so that R ~=Mn(D) for
some division ring D and some integer n. If D is commutative, there is a unique
additive closed symmetric monoidal structure on the category of R-modules, up
to symmetric monoidal equivalence. If D is not commutative, then there is no
additive closed symmetric monoidal structure on the category of R-modules.
The unit of the closed symmetric monoidal structure on Mn(k)-modules, for
k a field, is the unique simple left Mn(k)-module kn.
To find a case where the additive closed symmetric monoidal structure is
not unique, we consider the group ring k[Z=2]. Even in this simple case, the
classification of additive closed symmetric monoidal structures is quite involv*
*ed,
and will take the rest of this section and many lemmas. The ring k[Z=2] is both*
* a
commutative ring and a Hopf algebra, so we know there are at least two closed
symmetric monoidal structures. The behavior of this group ring depends on
whether the characteristic of k is 2, so we begin with this case.
We start by identifying the Hopf algebra structures on k[Z=2].
Lemma 3.5. Suppose k is a field of characteristic 2, and R = k[Z=2] ~=
k[x]=(x2). There are two different isomorphism classes of Hopf algebra struc-
tures on k, one represented by H0, in which (x) = 1 x + x 1, and one
represented by H1, in which (x) = 1 x + x 1 + x x.
We only use the associativity and unit axioms to prove this lemma, so it
follows that every bialgebra structure on k[Z=2] is a cocommutative Hopf algebra
structure.
Proof. The counit ffl of a Hopf algebra structure must have ffl(1) = 1 since it*
* is a
k-algebra map, and ffl(x) = 0 since x is nilpotent. We must have
(x) = a1(1 1) + a2(1 x) + a3(x 1) + a4(x x)
13
for some a1, a2, a3, a4 2 k. The fact that
0 = (x2) = (x)2
implies that a1 = 0. The fact that is counital implies that a2 = a3 = 1. One
then checks that is coassociative no matter what a4 is. It is of course also
cocommutative, and c(x) = x defines the only possible conjugation on k[Z=2].
Ler Ra denote the Hopf algebra where the coefficient of x x in (x) is a.
Any isomorphism f :Ra -! Rb of Hopf algebras must be compatible with the
counit, from which we conclude that f(x) = rx for some nonzero r 2 k. But
then compatibility with will hold if and only if ra = b. So if b and a are bo*
*th
nonzero, r = a=b will yield the desired isomorphism, but R0 is not isomorphic_
to any other Rb. |__|
In both of these two Hopf algebra structures on R = k[Z=2] (where k has
characteristic 2), the corresponding symmetric monoidal structure has = R k
R, freely generated as a left R-module by m = 1 1 and m 2 x = 1 x. We
also have m 1 x = x 1. However, in H0 we have
xm = 1 x + x 1 som 1 x = xm + m 2 x.
In H1, though, we have
xm = 1 x + x 1 + x x som 1 x = xm + (1 + x)m 2 x.
In both cases, the unit isomorphism `: k -!R has `(m) = 1 and `(m 2x) =
0. Also the commutativity isomorphism is defined by c(m) = m (and thus
c(m 2x) = m 1x). The associativity isomorphism has a(m m) = m m in
both cases, but in H0 we have
a(m 2 x) m) = 1 x 1 1 + 1 1 1 x = m 2 x m + m 2 x m,
whereas in H1 we have
a(m 2 m) = 1 x 1 1 + 1 1 1 x + 1 x 1 x
= m 2 m + m m 2 x + m 2 x m 2 x.
Let us refer to these k-linear closed symmetric monoidal structures as ^H0
and ^H1. Note that X ^ Y = X k Y as k-modules in either case, it is just the
action of Z=2 differs.
Theorem 3.6. Suppose k is a field of characteristic 2, and let R = k[Z=2] ~=
k[x]=(x2). Suppose - ^ - is a k-linear closed symmetric monoidal structure on
the category of R-modules with unit K. Then one of the following must hold.
1.K ~=R and - ^ - is k-linearly equivalent to - -.
2.K ~=k and -^- is k-linearly equivalent to -^H1- as a monoidal functor,
but not necessarily as a symmetric monoidal functor.
14
3.K k and - ^ - is k-linearly equivalent to - ^H0 - as a unital functor,
but not necessarily as a monoidal functor.
In addition, we have
1.The isomorphism classes (-^H1-, fi) of closed symmetric monoidal struc-
tures with underlying monoidal functor - ^H1 - are parametrized by ele-
ments fi 2 k, where c(m) = m + fix(m 2 x) in the symmetric monoidal
structure corresponding to fi.
2.The isomorphism classes (- ^H0 -, fl) of closed monoidal structures with
underlying unital functor - ^H0 - are parametrized by elements
fl 2 {0} [ kx =(kx )3,
where
a(m m) = m m + flx(m 2 x m 2 x)
in the monoidal structure corresponding to fl.
3.The isomorphism classes (- ^H0 -, fl, fi) of closed symmetric monoidal
structures with underlying monoidal functor (-^H0 -, fl) are parametrized
by elements fi, where
c(m) = m + fix(m 2 x)
in the symmetric monoidal structure corresponding to fi, as follows.
(a) If fl = 0, then fi 2 {0} \ kx =(kx )2.
(b) If fl 6= 0 and k does not have a primitive cube root of 1, then fi 2 *
*k.
(c) If fl 6= 0 and k does have a primitive cube root ! of 1, then fi = {0}
or a coset of the action of Z=3 on kx given by the action of !.
Just so we have a specific concrete example, this theorem says that when
k = Z=2, there are seven k-linear isomorphism classes of k-linear closed sym-
metric monoidal structures on k[Z=2]-modules, one corresponding to the usual
tensor product, two corresponding to different symmetric monoidal structures
on - ^H1 -, and four corresponding to different structures on the underlying
unital functor - ^H0 -.
We will prove this theorem through a series of lemmas.
Lemma 3.7. Let R = k[x]=(x2) where k is a field, and suppose - ^ - is a k-
linear closed symmetric monoidal structure on the category of R-modules, with
unit K. Then either K ~=R or K ~=k. If K ~=R, then - ^ - is equivalent to
the usual tensor product.
Proof. Every R-module is equivalent to a direct sum of copies of k and R. Any
decomposition of K as a direct sum of R-modules induces a decomposition of
R as a direct sum of R-bimodules, via the unit isomorphism R,K K ~=RR .
Since R is indecomposable, K must also be indecomposable, so either K ~=k or_
K ~=R. The last statement follows from Proposition 3.1. |__|
15
Lemma 3.8. Let R = k[x]=(x2) where k is a field, and suppose - ^ - is a k-
linear closed symmetric monoidal structure on the category of R-modules, with
unit k. Let be the 2-fold bimodule inducing - ^ -. Then there is an element
m 2 such that
~=Rm R(m 2 x) ~=R R(m 2 x)
as left R-modules, where `(m 1) = 1 and `(m 2 1) = 0. Furthermore,
m 2 x 6= 0.
This lemma also says that is a principal bimodule under the right action
2, generated by m. We do not know yet whether is a free bimodule on m,
though we will prove this later.
Proof. Since B,k k ~=R, we have
=( 2 x) ~=R.
If we choose an m 2 with `(m 1) = 1, then for any ~ 2 , we have
~ = `(~ 1)m + n 2 x = `(~ 1)m + (`(n 1)m + d 2 x) 2 x
= `(~ 1)m + `(n 1)(m 2 x),
where n and d denote unknown elements of . Thus is generated as a left
R-module by m and m 2 x. Note also that
~ 2 x = `(~ 1)(m 2 x).
In particular, m 2 x 6= 0, since if it were then 2 x = 0. The commutativity
isomorphism then implies 1 x = 0, so
Rx = ( 1 x) k = 0,
which is a contradiction. |___|
At this point, we have not determined whether m 2x generates a copy of k
or a copy of R. This depends on whether x(m 2x) = 0 or not. Note, however,
that `(m 1 x 1) = x, so we must have
m 1 x = xm + b(m 2 x).
for some b 2 R. We will then have
(m 2 x) 1 x = (xm + b(m 2 x)) 2 x = xm 2 x.
It will be helpful in what follows if we write a = a0 + a1x for elements a 2 R,
with a0, a1 2 k.
16
Lemma 3.9. Suppose the characteristic of k is 2. In the situation of Lemma 3.8
and with the notation above, there is a commutativity involution c: A,B -!
B,A if and only if b0 = 1, and in that case we have
c(m) = (1 + ffx)m + (fix)(m 2 x)
for some ff, fi 2 k, which can be anything. This implies
c(m 2 x) = xm + (b + ffx)(m 2 x).
Proof. Write
c(m) = rm + s(m 2 x),
for some r, s 2 R. Then
c(m 2 x) = c(m) 1 x = rxm + rb(m 2 x) + sx(m 2 x).
In order for c to have the desired properties, we need
c(m 1 x) = c(m) 2 x and c2(m) = m.
In order for c2(m) = m, computation shows that we need
r2 + rsx = 1 and (rs + rsb + s2x)(m 2 x) = 0.
The first equation is equivalent to (r0)2 = 1 and 2r0r1 + r0s0 = 0, so r0 = 1
and s0 = -2r1. Since we are in characteristic 2, this means r0 = 1 and s0 = 0.
We leave the second equation aside for the moment. In order for c(m 1 x) =
c(m) 2 x, computation shows that we need
r(1 + b)x = 0 and (sx + rb2 + sbx)(m 2 x) = r(m 2 x).
Since we know that r0 = 1, this first equation implies that b0 = -1; since we
are in characterstic 2, b0 = 1. Computation then shows that, in characteristic *
*2,
these conditions guarantee that both equations involving m 2 x hold, whether_
x(m 2 x) = 0 or not. |__|
In fact, we can make ff = 0 by modifying our choice of m.
Lemma 3.10. In the situation of Lemma 3.9, so in particular when the char-
acteristic of k is 2, we can modify our choice of the bimodule generator m of
so that
c(m) = m + fix(m 2 x)
for some fi 2 k. In this case, c(m 2 x) = m 1 x.
From now on, we assume m is chosen as in Lemma 3.10.
17
Proof. Let n = m + ff(m 2 x). Then `(n 1) = 1, so n is a perfectly good
bimodule generator for . Note that
n 2 x = m 2 x and n 1 x = xn + b(n 2 x)
as before (after some calculation). However, we have
c(n) = n + flx(n 2 x),
after some calculation, for some fl 2 k. |___|
We must now come to grips with the associativity isomorphism
a: C, B,A -! ,A C,B.
If is a free bimodule generated by m, then both the domain and range of a
are isomorphic to R 4 as left R-modules, with summands generated by m m,
m 2 x m, m m 2 x, and m 2 x m 2 x. If has dimension 3, on the
other hand, both the domain and range of a are isomorphic to R k R, with
summands generated by m m, m 2 x m and m m 2 x, respectively. For
example, in the domain we have
m xm 2 x = m 2 x m 2 x,
but in the range we have
m xm 2 x = m 1 x m 2 x = xm m 2 x + (1 + b1x)m 2 x m 2 x.
With respect to this basis, write
a(m m) = (e1, e2, e3, e4) and a(m 2 x m) = (f1, f2, f3, f4).
If the dimension of is 3, then we take e4 = f4 = 0. In addition, in that case
f03= 0, since x(m m 2 x) = 0.
Lemma 3.11. With the above definitions, if a is a map of 3-fold bimodules,
then the left and right unit coherence diagram commutes if and only if e1 = 1,
e2 = 0, f1 = 0, and f2 = 1.
Proof. Apply the coherence diagram to m m 1 and to m 2 x m 1. |___|
We still have to determine what conditions are necessary for a to be a map
of 3-fold bimodules.
Lemma 3.12. In order for the map a defined above to be a map of 3-fold
bimodules making the left and right unit coherence diagram commute, must
be the free bimodule on m, and
e1 = 1, e2 = 0, e03= 0, f1 = 0, f2 = 1, f3 = 1, f4 = b1 + e3.
18
Proof. Of course, we know already that e1 = 1, e2 = 0, f1 = 0, f2 = 1, with
f03= 0 if has dimension 3. We have implicitly assumed that a is a map of left
R-modules by defining it only in terms of generators. To ensure that a preserves
the right module structure represented by A in
a: C, B,A -! ,A C,B,
we must have
a(m m 2 x) = (1, 0, e3, e4) A x = (0, 1, 0, e3)
and
a(m 2 x m 2 x) = (0, 0, 0, f3).
We now turn to the right module structure represented by B. Here we must
have
a(m m 1 x) = a(m m) B x.
Calculation shows that this forces f3 = 1, and this rules out the case when
the dimension of is 3 (since f03= 0 if the dimension of is 3). The same
calculation shows that f4 = b1 + e3. Further calculation shows that this is
enough to ensure that a preserves the right module structure represented by B.
We now must ensure that a preserves the right module structure represented by
C. More calculation of the relation
a(m 1 x m) = a(m m) C x
gives e3x = 0, so e03= 0. Further calculation implies that this enough to make_
a preserve the right module structure represented by C. |__|
We turn finally to the associativity pentagon.
Lemma 3.13. Given that a satisfies the conditions of Lemma 3.12, a makes
the associativity pentagon commute if and only if
e3 = 0, f4 = b1, e04= 0 and eithere14= 0 orb1 = 0.
Proof. If we apply the associativity pentagon to m 2x m m, we eventually
find
e3 = 0 and e04= 0.
and so f4 = b1 + e3 = b1. Further computation with the associativity pentagon
applied to m m m eventually yields
e14b1 = 0, so e14= 0 orb1 = 0.
These conditions then make the associativity pentagon commute. |___|
19
We are now left with determining which of these different define additively
equivalent symmetric monoidal structures. Recall that for such an equivalence,
we need an element X of the bimodule Picard group, an isomorphism j :X
k -!k, and an isomorphism
X X -!X
making various diagrams commute.
Lemma 3.14. Let R = k[x]=x2. Up to isomorphism, the only invertible R-
bimodules are the Ru, where u is a unit in k, Ru = R as a left module, and
1 x = ux.
This follows immediately from [Yek99, Lemma 3.3]. With all these lemmas
in hand, we can now complete the proof of Theorem 3.6.
Proof of Theorem 3.6.Let X = Ru, and suppose we have an isomorphism
j :k -!X k of left modules and an isomorphism
q : X X -!X
of 2-fold bimodules making the compatibility diagrams of Theorem 2.4 commute.
The map j is determined by j(1) = 1 ae for some nonzero ae in k, and the map
q is determined by q(m 1 1) = oe(1 m) + o(1 m 2 x) with oe, o 2 R,
where we have used m for a bimodule generator in both and satisfying the
condition on c(m) as in Lemma 3.10. Since q is a map of bimodules, it follows
that
q(m 2 x 1 1) = u-1q(m 1 . x 1) = u-1oe(1 m 2 x).
In particular, for q to be an isomorphism, we need oe to be a unit in R. We also
need
q(m 1 x 1 1) = u-1q(m 1 1 . x) = u-1q(m 1 1) 1 x.
Further calculation with this last equation yields b1 = u-1b1.
This means that if b1 is nonzero, we can choose u so as to ensure b1 = 1.
Said another way, given with b1 nonzero, we can define = Ru-1 Ru Ru
for a suitable u and get an additively equivalent symmetric monoidal structure
with b1 = 1. Any isomorphism between a and a both with b1 = 1 must have
u = 1, so we can think of it as an automorphism of . It is still useful to use*
* for
the domain copy of , because the choice of generator m as in Lemma 3.10 could
be different in the two copies of . We can then work through the compatibility
diagrams of Theorem 2.4 in this case. The unit diagram forces oe = ae1, so
oe 2 k. In order for the commutativity diagram to commute, we need ox = 0,
so o0 = 0. Then one finds that the fi in the commutativity isomorphism in
both and must be the same. Since we are in the case where b1 = 1, the
associativity isomorphism is completely determined by th preceding lemmas.
Thus we find that if b1 is nonzero, then the monoidal structure determined by
20
is additively equivalent to the one given by the Hopf algebra H1. Since fi does
not change under these isomorphisms, it could be anything, so the different
symmetric monoidal structures on - ^H1 - are classified by fi.
Now suppose b1 = b1 = 0. As above, we must have oe = ae-1 in order to
ensure the unit compatibility diagram commutes. The commutativity compat-
ibility diagram again forces o0 = 0, but this time we have fi = u-2fi . Since
b1 = 0, the preceding lemmas allow for a nontrivial e14as well, so we must
check the associativity compatibility diagram too. Painful computation then
gives that the e14for is u-3 times the e14for . Hence the different monoidal
structures on - ^H0 - are classified by
fl 2 {0} [ kx =(kx )3.
If fl = 0, then we can use Ru to make the symmetric monoidal structure cor-
responding to fi isomorphic to the one corresponding to u-2fi. Hence the sym-
metric monoidal structures when fl = 0 correspond to
fi 2 {0} [ kx =(kx )2.
When fl 6= 0, if we fix fl we can only use the Ru with u3 = 1. If k has no prim*
*itive
cube root of 1, then, the symmetric monoidal structures are parametrized by
fi 2 k, but if k does have a primitive cube root of 1, the symmetric monoidal
structures are parametrized by orbits of Z=3 acting on k by multiplying by_the
primitive cube root of 1. |__|
We now consider closed symmetric monoidal structures on k[Z=2]-modules
when the characteristic of k is not 2. Here the answer is wildly different; the*
*re
are a proper class of such structures!
Theorem 3.15. Suppose k is a field whose characteristic is not 2, and let
R = k[Z=2] ~=k[x]=(x2 - 1). If - ^ - is a closed k-linear symmetric monoidal
structure on the category of R-modules, then its unit K is isomorphic to R, k+ ,
or k- . If the unit is isomorphic to R, then -^- ~=- - as k-linear symmetric
monoidal functors. If K ~=k- , then - ^ - is k-linearly symmetric monoidal
equivalent to a closed k-linear symmetric monoidal structure whose unit is k+ .
Given any R-module M, there is a k-linear symmetric monoidal structure for
which the unit is k+ and k- ^ k- = M.
It is easy to see that symmetric monoidal structures with nonisomorphic
values of M cannot be equivalent. However, we do not know if there is more
than one closed symmetric monoidal structure for a given M.
Proof. Let x denote the element [1] of k[Z=2], so R = k[Z=2] ~=k[x]=(x2 - 1).
Since the characteristic is not 2, this ring is semisimple. Any module M splits
as M+ M- , where M+ is the 1-eigenspace of x and M- is the -1-eigenspace
of x. In particular, R itself so splits, with the splitting given by the orthog*
*onal
idempotents e+ = (1=2)(1+x) and e- = (1=2)(1-x). This produces a splitting
R-mod ~=k-mod x k-mod ,
21
up to equivalence, of the entire category of R-modules. Thus every R-module
is a direct sum of copies of k+ and k- , and there are no maps from k+ to k- ,
or from k- to k+ .
Given a k-linear closed symmetric monoidal structure, the corresponding
bimodule splits into 8 different spaces a,b,c, where a, b, c are each + or -
(by simultaneously diagonalizing the 3 actions of x). Here the a stands for the
left action of R on , and b and c for the two right actions. Of course the
unit K = K+ K- as well. One can easily check that ka,b,c c kc ~=ka,b, and
ka,b,c c kd = 0 if d 6= c. Since R = k+,+ k-,-, we see that there must be
dimension one terms +,+,aand Ka and dimension one terms -,-,band Kb.
If a 6= b, then K = R, and so Proposition 3.1 tells us that our closed
symmetric monoidal structure is equivalent to R . We can therefore assume
a = b. The commutativity isomorphism tells us x,y,zhas the same dimension
as x.z.y. Therefore, taking z 6= a, we find that z,a,zis nonzero. If Kz were
also nonzero, we would get a term kz,ain B,K K ~=RR . Since we cannot
have such a term, we conclude that K = ka, so the unit is one-dimensional over
k.
There is an obvious self-equivalence of the category of R-modules that per-
mutes the two copies of k-mod . That is, it sends k+ to k- , and vice versa. Up*
* to
symmetric monoidal equivalence, then, we can assume the unit of our symmetric
monoidal structure is k+ . Let M = k- ^ k- . Then, using the decomposition of
R = k+,+ k-,- as bimodules, we get
~=k+,+,+ k-,-,+ k-,+,- M-,-,
where M-,- is the 2-fold bimodule whose underlying left module is M, and
where x acts as -1 in both right module structures.
Now suppose we are given M. We want to construct a k-linear closed sym-
metric monoidal structure on R-modules with k- ^ k- = M. We simply define
k+ ^ N = N ^ k+ = N for any R-module N, and k- ^ k- = M. On morphisms,
we note that there are no nonzero morphisms from k+ to k- or vice versa, and
that every endomorphism of k+ or k- is given by multiplication by an element
of k. So we define the induced morphism to be multiplication by the same el-
ement of k. Since every R-module is a direct sum of copies of k+ and k- , this
defines - ^ - as a bifunctor. We define the left unit to be the identity. The
commutativity isomorphism
cxy -!kx ^ ky -!ky ^ kx
is the identity, where x and y denote signs. We then extend through direct
sums. The associativity isomorphism
ax,y,z:(kx ^ ky) ^ kz -!kx ^ (ky ^ kz)
is the identity as long as at least one of x, y, or z is +. The map
a-,-,-: M ^ k- -! k- ^ M
is the commutativity isomorphism. We leave to the reader the check that_the
coherence diagrams hold. |__|
22
4. Structural results
In the examples in the previous section, especially in Theorem 3.15, we
saw that the 2-fold bimodule can be very complicated. It does not have
to be finitely generated, or even countably generated. However, it cannot be
completely random either. Furthermore, in all the examples we have, the unit
K of an additive symmetric monoidal structure on the category of R-modules
is always a prinicipal R-module. It is tempting to wonder whether this always
holds, or whether there are other properties that K must have.
In this section, we show that K is always a finitely generated module with
commutative endomorphism ring, and that is faithful in a very strong sense.
We begin by noting that tensoring with reflects any property of morphisms
that the tensor product preserves.
Proposition 4.1. Suppose R is a ring and 1,2is a 2-fold R-bimodule that
determines a closed symmetric monoidal structure on the category of R-modules
with unit K. Let P be a replete class of morphisms of abelian groups with the
property that if f 2 P, then A f and f B 2 P for any left R-module A or
right R-module B. If f is a morphism of left R-modules, then f 2 P if and only
if f 2 P. Similarly, if g is a morphism of right R-modules, then g 2 P if
and only if g 2 P. This statement holds with either right module structure
on .
Recall that a class of morphisms is replete whenever f 2 P and f ~=g in the
category of morphisms, then g 2 P. Said another way, if we have a commutative
square
A ---f-! B
? ?
i?y ?yj
A0 ----!g B0
where i, j are isomorphisms, then f 2 P if and only if g 2 P.
Proof. Suppose g is a morphism of right R-modules. By definition, if g 2 P,
then g 2 P. Conversely, suppose g 2 P, where we use the leftmost right
module structure on . Then
g ~=RR g ~=( B,K K) g ~=( B,K g) K,
and this is in P since B,K g is so. We use the commutativity isomorphism __
to prove the same thing for the other right module structure on . |__|
Taking the class P to tbe the collection of zero morphisms, the collection
of isomorphisms, and the collection of surjective maps gives us the following
corollary.
23
Corollary 4.2. Suppose R is a ring and 1,2is a 2-fold R-bimodule that de-
termines a closed symmetric monoidal structure on the category of R-modules
with unit K. Let f denote a morphism of right R-modules and let g denote a
morphism of left R-modules.
1.f = 0 if and only if f = 0. With either right module structure,
g = 0 if and only if g = 0.
2.f is an isomorphism if and only if f is so. With either right module
structure, g is an isomorphism if and only if g is so.
3.f is a surjection if and only if f is so. With either right module
structure, g is a surjection if and only if g is so.
In particular, these imply that is faithful.
Corollary 4.3. Suppose R is a ring and 1,2is a 2-fold R-bimodule that de-
termines a closed symmetric monoidal structure on the category of R-modules.
Then is faithful as a left or right R-module, with either right module struct*
*ure.
Proof. Choose r 6= 0 2 R. Then the map R -!R that is left multiplication by
r induces left multiplication by r on = R . Since r 6= 0, this map is also
nonzero by Corollary 4.2. Hence r does not annihilate , so is faithful. Use *
*__
right multiplication by r to see that is faithful as a right R-module. |*
*__|
They also imply that K is finitely generated.
Theorem 4.4. The unit K in an additive closed symmetric monoidal structure
on the category of R-modules is finitely generated.
Proof. In thePunit isomorphism B,K K ~= RR , write 1 as the image of a
finite sum ~i ki. Let K0 denote the submodule of K generated by the ki,
and j :K0 -!K denote the inclusion. Then j is surjective, so j must_also
be. |__|
We suspect that the unit K must in fact be a principal R-module, but we
do not know how to prove this.
Another essential property of K, or the unit of any symmetric monoidal
category, is that its endomorphisms commute with each other. Somewhat more
is true in our case.
Theorem 4.5. Suppose K is the unit of an additive closed symmetric monoidal
structure on the category of R-modules. Then End R(K) is a subring of the
center Z(R) of R.
Proof. Suppose f 2 End R(K). Then f is a bimodule endomorphism of
R, through the unit isomorphism. Any bimodule endomorphism of R must
be given by x 7! rx for some r 2 Z(R). This defines a ring homomorphism
End R(K) -!Z(R). If f is in the kernel of this homomorphism, then f =_0,
but then f = 0 by Corollary 4.2. |__|
24
We note that EndR (K) can be a proper submodule of Z(R), as for example
when R = k[x]=(x2) and the unit is k (of characteristic 2).
It is pretty rare for an R-module to have a commutative endomorphism ring.
Using the work of Vasconcelos [Vas70], for example, we can deduce the following
corollary.
Corollary 4.6. Suppose R is a commutative Noetherian ring with no nonzero
nilpotent elements, and K is the unit of an additive closed symmetric monoidal
structure on the category of R-modules. Then
K ~=a=b
for some radical ideal b and some ideal a b with the ideal quotient (b:a) = b.
Recall that the ideal quotient (b:a) is the set of all x such that xa b.
Proof. Let b = ann(K), the annihilator of K. We have an obvious monomor-
phism of rings R=b -! EndR(K) that takes r to multiplication by r, But
End R(K) is a subring of R by Theorem 4.5. Hence R=b is a subring of R,
and therefore has no nilpotents, so b is a radical ideal. Thus K is a finitely *
*gen-
erated (by Theorem 4.4), faithful R=b-module with commutative endomorphism
ring, and R=b is a commutative Noetherian ring with no nonzero nilpotent el-
ements. Vasconcelos [Vas70] proves in this situation that K is an ideal in R=b.
Hence K ~=a=b for some ideal a of R. The condition on the ideal quotient is_so_
that the annihilator of K will in fact be b. |__|
Note that, if M is a submodule of the unit K, then the image of M in
K ~=R will be a sub-bimodule of R, and hence a two-sided ideal.
Corollary 4.7. Suppose K is the unit of an additive closed symmetric monoidal
structure on the category of R-modules. Then every nonzero proper submodule
of K gives rise to a nonzero proper two-sided ideal of R. Hence, if R is a simp*
*le
ring, then K is a simple left module.
We note that a simple commutative ring is of course a field, but there are
many simple noncommutative rings that are not division rings. We would like to
be able to say that the map from nonzero proper submodules of K to two-sided
ideals of R is one-to-one, but we do not know if this is true.
Proof. Suppose L is a proper submodule of K. Then the maps L -! K and
K -! K=L are both nonzero, so they remain so after tensoring with B,K by
Corollary 4.2. Hence the image of B,L L is a nonzero proper subbimodule_
of B,K K = RR . |__|
As above, we do not know if this map from submodules of K to two-sided
ideals in R is one-to-one, but it is on direct summands of K.
25
Corollary 4.8. Suppose K is the unit of an additive closed symmetric monoidal
structure on the category of R-modules. There is a one-to-one map from iso-
morphism classes of direct summands of K to central idempotents in R. In
particular, if R is indecomposable as a ring, then K is indecomposable as an
R-module.
Proof. Suppose M is a direct summand of K, so that there is a retraction
f :K -! M. Tensoring with gives us a retraction of bimodules R -! M.
The composite
R -! M -! R
must be multiplication by a central idempotent e of R, with M = eR. The
bimodule M determines e [Lam01 , Exercise 22.2], and we can recover M_
from M, up to isomorphism, by tensoring witk K. |__|
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