# The Unapologetic Mathematician

## Tensor products of modules

The notion of a tensor product also extends to modules, but the generalization is not quite as straightforward as it was for direct sums.

We start with a ring $R$ and consider a right $R$-module $A_R$ and a left $R$-modules ${}_RB$. We consider functions $f:A\times B\rightarrow X$ which take an element of each module and return an element of some abelian group $X$. Such a function is called “middle linear” if

• $f(a_1+a_2,b)=f(a_1,b)+f(a_2,b)$
• $f(a,b_1+b_2)=f(a,b_1)+f(a,b_2)$
• $f(a\cdot r,b)=f(a,r\cdot b)$

That is, if it is bilinear as a function of abelian groups, and if we can pull the action of $R$ from the first argument to the second and back without changing the value of the function. This condition may seem a little artificial, but I’ll motivate it a bit more later.

The tensor product $A\otimes_RB$ is the abelian group for the universal middle linear function, just as the tensor product for abelian groups was the abelian group for the universal bilinear function. We show that such an object exists by a similar construction. Take the free abelian group generated by all elements of $A\times B$ and write a generator as $a\otimes b$. Then impose the relations $(a_1+a_2)\otimes b$, $a\otimes(b_1+b_2)$, and $(a\cdot r)\otimes b=a\otimes(r\cdot B)$. Then given any middle linear map $f:A\times B\rightarrow X$ there is a unique homomorphism of abelian groups $\bar{f}:A\otimes B\rightarrow X$ so that $f(a,b)=\bar{f}(a\otimes b)$.

In general, the tensor product over $R$ is just an abelian group — it eats $R$-module structures like $\hom$ does. That said, like $\hom$ plays well with extra module structures, so does tensor product. If ${}_SA_R$ is a right $R$-module and a left $S$-module, then $A\otimes_RB$ carries the structure of a left $S$-module. Indeed we can define $s\cdot(a\otimes b)=(s\cdot a)\otimes b$ and check that this action respects the relations we imposed. Similarly, if $A$ has an additional right $S$-module structure commuting with the action of $R$ then $A\otimes_RB$ is a right $S$-module, and the same goes for extra modules structures on $B$. Unlike in the case of $\hom$, no structures get “flipped over” in this process.

If $R$ is a commutative ring, then every module is both a left and a right $R$-module. Thus, the same is true of $A\otimes_RB$ — the tensor product eats the left module structure on $A$ and the right module structure on $B$, but leaves the other two structures.

Now for an example that should motivate the idea of a middle-linear map. Let $A_R$, ${}_SB_R$, and $C_R$ be right $R$-modules with an extra left $S$-module structure on $B$. Then $\hom(A,B)$ is a left $S$-module and $\hom(B,C)$ is a right $S$-module. We consider $f\in\hom(A,B)$ and $g\in\hom(B,C)$ and calculate the two composites
$\left[g\circ (s\cdot f)\right](a)=g(\left[s\cdot f\right](a))=g(s\cdot(f(a)))$
$\left[(g\cdot s)\circ f\right](a)=\left[g\cdot s\right](f(a))=g(s\cdot(f(a)))$
so we can use the $S$ action on either factor. This means that composition of these homomorphisms is a middle-linear function, and so defines a linear function $\hom(B,C)\otimes_S\hom(A,B)\rightarrow\hom(A,C)$.

April 30, 2007 Posted by | Ring theory | 6 Comments

## Direct sums of modules

We’ve covered direct sums in the case of abelian groups — that is, $\mathbb{Z}$-modules — but the concept extends to modules over arbitrary rings. This gives me a good chance to go back and clean up my coverage.

We build the direct sum of two $R$-modules $A$ and $B$ on the product $A\times B$ of the underlying sets. We define an abelian group structure by $(a,b)+(a',b')=(a+a',b+b')$, and an action of $R$ by $r\cdot(a,b)=(r\cdot a,r\cdot b)$.

Now, here’s the diagram:

The direct sum $A\oplus B$ of two left $R$-modules $A$ and $B$ comes equipped with four module homomorphisms. For $A$ we have the pair $\pi_A:A\oplus B\rightarrow A$ and $\iota_A:A\rightarrow A\oplus B$. These are defined as follows:

• $\pi_A(a,b)=a$
• $\iota_A(a)=(a,0)$

There is a similar pair with a similar definition for $B$. These homomorphisms satisfy the identities

• $\pi_A\circ\iota_A=1_A$
• $\pi_B\circ\iota_B=1_B$
• $\pi_A\circ\iota_B=0$
• $\pi_B\circ\iota_A=0$
• $\iota_A\circ\pi_A+\iota_B\circ\pi_B=1_{A\oplus B}$

where $1_M$ is the identity homomorphism on the module $M$, and ${}0$ is the homomorphism between two modules sending every element of the domain to the element ${}0$ in the codomain.

Now if we have any two homomorphisms $f_A$ and $f_B$ from a module $X$ to $A$ and $B$ respectively, then there is a unique homomorphism $f_A\oplus f_B:X\rightarrow A\oplus B$ making the two triangles on the top commute. That is, $\pi_A\circ f_A\oplus f_B=f_A$, and similarly for $f_B$. In fact, we can define $f_A\oplus f_B=\iota_A\circ f_A+\iota_B\circ f_B$, for then
$\pi_A\circ(\iota_A\circ f_A+\iota_B\circ f_B)=\pi_A\circ\iota_A\circ f_A+\pi_A\circ\iota_B\circ f_B=f_A$
and so on. Similarly, given two homomorphisms $g_A$ and $g_B$ from $A$ and $B$ to a module $Y$, then the homomorphism $g_A\circ\pi_A+g_B\circ\pi_B$ is the unique homomorphism making the lower two triangles commute.

The upshot of all this is that the direct sum $A\oplus B$ behaves like both the direct product and the free product of two groups, since it satisfies both universal properties. For any finite number of modules $A_i$ we can build the direct sum $\bigoplus\limits_{i=1}^nA_i$ and it also satisfies the analogous universal properties, and comes equipped with analogous injections and projections satisfying analogous relations to those above.

The upshot of all this is that the direct sum of a finite collection of $R$-modules behaves like both the direct product and the free product of groups. In fact, we can take that as the definition, derive the relations between the injections and projections, and use the above construction to show such a thing actually exists. On the other hand, we can take the relations between injections and projections as the definitions, use the construction to show existence, and derive the universal property from the relations as above.

For an infinite index set $\mathcal{I}$ the situation is a bit more complicated. Here we use the definition from the injections and projections with the specified relations. Then the underlying set of the infinite direct sum $\bigoplus\limits_{i\in\mathcal{I}}A_i$ is not the infinite cartesian product of the underlying sets. It’s actually the list of all such “$\mathcal{I}$-tuples” where all but a finite number of the entries are the ${}0$ elements of the respective modules. This only satisfies the universal property for the $\iota_i$ — the bottom of the diagram above. For the top we really do need the infinite direct product of the modules, which uses the whole infinite cartesian product of the underlying sets. However, for most purposes the direct sum is all we need, and the relations between the injections and the projections are the most useful part of this definition.

April 29, 2007 Posted by | Ring theory | 8 Comments

## Divisibility

There is an interesting preorder we can put on the nonzero elements of any commutative ring with unit. If $r$ and $s$ are nonzero elements of a ring $R$, we say that $r$ divides $s$ — and write $r|s$ — if there is an $x\in R$ so that $rx=s$. The identity $1$ trivially divides every other nonzero element of $R$.

We can easily check that this defines a preorder. Any element divides itself, since $r1=r$. Further, if $r|s$ and $s|t$ then there exist $x$ and $y$ so that $rx=s$ and $sy=t$, so $r(xy)=t$ and we have $r|t$.

On the other hand, this preorder is almost never a partial order. In fact since $r(-1)=-r$ and $-r(-1)=r$ we see that $r|-r$ and $-r|r$, and most of the time $r\neq-r$. In general, when both $r|s$ and $r|s$ we say that $r$ and $s$ are associates. Any unit $u$ comes with an inverse $u^{-1}$, so we have $u|1$ and $1|u$. If $r=su$ for some unit $u$, then $r$ and $s$ are associates because $s=ru^{-1}$.

We can pull a partial order out of this preorder with a little trick that works for any preorder. Given a preorder $(P,\preceq)$ we write $a\sim b$ if both $a\preceq b$ and $b\preceq a$. Then we can check that $\sim$ defines an equivalence relation on $P$, so we can form the set $P/\sim$ of its equivalence classes. Then $\preceq$ descends to an honest partial order on $P/\sim$.

One place that divisibility shows up a lot is in the ring of integers. Clearly $n$ and $-n$ are associate. If $m$ and $n$ are positive integers with $m|n$, then there is another positive integer $x$ so that $mx=n$. If $x=1$ then $m=n$. Otherwise $m\lneq n$. Thus the only way two positive integers can be associate is if they are the same. The preorder of divisibility on $\mathbb{Z}^\times$ induces a partial order of divisibility on $\mathbb{N}^+$.

April 28, 2007 Posted by | Ring theory | 6 Comments

## Ordinal numbers

We use cardinal numbers to count how many elements are in a set. Another thing we think of numbers for is listing elements. That is, we put things in order: first, second, third, and so on.

We identified a cardinal number as an isomorphism class of sets. Ordinal numbers work much the same way, but we use sets equipped with well-orders. Now we don’t allow all the functions between two sets. We just consider the order-preserving functions. If $(X,\leq)$ and $(Y,\preceq)$ are two well-ordered sets, a function $f:X\rightarrow Y$ preserves the order if whenever $x_1\leq x_2$ then $f(x_1)\preceq f(x_2)$. We consider two well-ordered sets to be equivalent if there is an order-preserving bijection between them, and define an ordinal number to be an equivalence class of well-ordered sets under this relation.

If two well-ordered sets are equivalent, they must have the same cardinality. Indeed, we can just forget the order structure and we have a bijection between the two sets. This means that two sets representing the same ordinal number also represent the same cardinal number.

Now let’s just look at finite sets for a moment. If two finite well-ordered sets have the same number of elements, then it turns out they are order-equivalent too. It can be a little tricky to do this straight through, so let’s sort of come at it from the side. We’ll use finite ordinal numbers to give a model of the natural numbers. Since the finite cardinals also give such a model there must be an isomorphism (as models of $\mathbb{N}$ between finite ordinals and finite cardinals. We’ll see that the isomorphism required by the universal property sends each ordinal to its cardinality. If two ordinals had the same cardinality, then this couldn’t be an isomorphism, so distinct finite ordinals have distinct cardinalities. We’ll also blur the distinction between a well-ordered set and the ordinal number it represents.

So here’s the construction. We start with the empty set, which has exactly one order. It can seem a little weird, but if you just follow the definitions it makes sense: any relation from $\{\}$ to itself is a subset of $\{\}\times\{\}=\{\}$, and there’s only one of them. Reading the definitions carefully, it uses a lot of “for every”, but no “there exists”. Each time we say “for every” it’s trivially true, since there’s nothing that can make it false. Since we never require the existence of an element having a certain property, that’s not a problem. Anyhow, we call the empty set with this (trivial) well-ordering the ordinal ${}0$. Notice that it has (cardinal number) zero elements.

Now given an ordinal number $O$ we define $S(O)=O\cup\{O\}$. That is, each new number has the set of all the ordinals that came before it as elements. We need to put a well-ordering on this set, which is just the order in which the ordinals showed up. In fact, we can say this a bit more concisely: $O_1\leq O_2$ if $O_1\in O_2$. More explicitly, each ordinal number is an element of every one that comes after it. Also notice that each time we make a new ordinal out of the ones that came before it, we add one new element. The successor function here adds one to the cardinality, meaning it corresponds to the successor in the cardinal number model of $\mathbb{N}$. This gives a function from the finite ordinals onto the finite cardinals.

What’s left to check is the universal property. Here we can leverage the cardinal number model and this surjection of finite ordinals onto finite cardinals. I’ll leave the details to you, but if you draw out the natural numbers diagram it should be pretty clear how to how that the universal property is satisfied.

The upshot of all of this is that finite ordinals, like finite cardinals, give another model of the natural numbers, which is why natural numbers seem to show up when we list things.

April 26, 2007 Posted by | Fundamentals, Numbers, Orders | 2 Comments

## Some of my own stuff

I’m talking tomorrow in the Geometry, Symmetry, and Physics seminar here at Yale about the work that spun out of my realization of March 16. This ties into knot colorings, but goes far beyond that starting point. I won’t be able to say this with just the tools I’ve developed in the main line of my writings, so like the Atlas stuff it may not be comprehensible (yet!) to anyone beyond professionals.

So here’s the most general statement. Let $\mathcal{C}$ be an algebraic category and $X$ a co-$\mathcal{C}$ object in the category of pointed topological pairs up to homotopy. Write $P_n$ for the plane with $n$ marked points, and $C$ for the cube. Then every tangle $T:m\rightarrow n$ gives rise to a cospan in $\mathcal{C}$:

$\hom(X,P_m)\rightarrow\hom(X,(C,T))\leftarrow\hom(X,P_n)$

where the $\hom$-objects are taken in the category of pointed pairs up to homotopy. This then gives rise to an anafunctor from the comma category $(\hom(X,P_m),\mathcal{C})$ to the comma category $(\hom(X,P_n),\mathcal{C})$. This assignment is a monoidal functor from the category of tangles to the category of (categories, anafunctors). When $\mathcal{C}$ is the category of quandles, this functor categorifies the extension to tangles of the coloring number invariant of knots and links.

## Applicant-opaque departments

I’ve complained before about the opacity of the application procedure. I’ve gotten to the point now that I’m asking all outstanding schools what my status is. Following is a list — which I will keep updating — of departments that have closed their application procedures without bothering to inform applicants that they have been rejected. Petty? Maybe. But I’m the one who’s gotten screwed here, so the least they can do is be publicly known as not acting in good faith towards their applicants.

• George Mason University
• Kansas State University
• Northwestern University
• Purdue University
• Stony Brook University
• University of Arizona
• University of California, Davis
• University of California, Irvine
• University of California, Los Angeles
• University of Michigan
• University of Regina
• University of Tennessee
• University of Toronto

Note: I am giving departments the benefit of the doubt if they say they did send word. Still I am shocked — shocked — at how many letters the post office has lost.

April 25, 2007 Posted by | rants | 6 Comments

## Character Tables for Exceptional Real Reductive Groups

I’ve posted my notes for Zuckerman’s third and final talk. This should get us up to “what exactly does the Atlas Project do?”

Maybe we have enough pull around here to get Jeff Adams to come up and answer that question in a talk before I’m banished from the academy, but even if not I may be able to get time to sit down with him one-on-one once I head back to the Maryland area. Anyhow, my explanations are behind the jump, as usual.

April 25, 2007 Posted by | Atlas of Lie Groups | 2 Comments

Many times we’ll be interested in an abelian group on which more than one ring has an action, or on which a ring has two different actions. Most interesting are the cases when these different actions commute with each other. That is, if $M$ carries an $R$-module structure with action $r\cdot m$ and an $S$-module structure with action $s*m$, then it’s really nice if $r\cdot(s*m)=s*(r\cdot m)$. We’ll keep track of this sort of thing by hanging subscripts off of the module’s name to keep track of mutually commuting actions. In this case we’ll say ${}_{RS}M$ to denote two commuting left actions, one by $R$ and one by $S$.

Every module over a commutative ring has both a left and a right action, and these clearly commute with each other because the ring is commutative. That is, if $M$ is an $R$-module for a commutative ring $R$, we automatically get two commuting $R$ actions, written ${}_RM_R$.

Another example we’ll see in more generality later is the tensor powers of an abelian group $A$. The group itself is a module over ${\rm End}(A)$. Now the tensor power $A^{\otimes n}$ also is a module over ${\rm End}(A)$. We just define $f\cdot(a_1\otimes...\otimes a_n)=f(a_1)\otimes...\otimes f(a_n)$ and extend linearly. But we also have a homomorphism from the symmetric group $S_n$ to the monoid of endomorphisms of $A^{\otimes n}$. For a permutation $\sigma$ we define $\sigma\cdot(a_1\otimes...\otimes a_n)=a_{\sigma^{-1}(1)}\otimes...\otimes a_{\sigma^{-1}(n)}$, shuffling around the factors. This extends to a unique homomorphism $\mathbb{Z}[S_n]\rightarrow{\rm End}(A^{\otimes n})$, which gives another module structure on $A^{\otimes n}$. These two actions, one of $\mathbb{Z}[S_n]$ and one of ${\rm End}(A)$, commute with each other.

Now that I have some examples down, I want to consider how these extra module structures play with $\hom_{R-{\rm mod}}$ — which tells us the homomorphisms between left $R$-modules — and $\hom_{{\rm mod}-R}$ — which does the same for right $R$ modules. I’ll mostly treat the left module case because the other side is very similar.

The first thing to make clear is that $\hom_{R-{\rm mod}}$ eats up the $R$-module structure. If we take left modules ${}_RM$ and ${}_RN$ then $\hom_{R-{\rm mod}}({}_RM,{}_RN)$ is just an abelian group with no $R$-module structure at all. The interesting things happen when we’ve got extra module structures floating around.

If $N$ also has a right $S$-module structure for another ring $S$, then we get a right $S$-module structure on the homomorphisms. We can define $\left[f\cdot s\right](m)=f(m)\cdot s$. On the right side of the equation we’re using the given action of $S$ on $N$. It’s not too hard to verify that this defines a right $S$ action on $\hom_{R-{\rm mod}}({}_RM,{}_RN_S)$.

On the other hand, if $M$ has a right $S$-module structure, we get a left $S$ action on the homomorphisms. We define $\left[s\cdot f\right](m)=f(m\cdot s)$. Let’s verify this one carefully:
$\left[(s_1s_2)\cdot f\right](m)=f(m\cdot(s_1s_2))=f((m\cdot s_1)\cdot s_2)=$
$\left[s_2\cdot f\right](m\cdot s_1)=\left[s_1\cdot(s_2\cdot f)\right](m)$

There are a number of similar cases, which you should check through:

• $\hom_{R-{\rm mod}}({}_{RS}M,{}_RN)$ is a right $S$-module.
• $\hom_{R-{\rm mod}}({}_RM_S,{}_RN)$ is a left $S$-module.
• $\hom_{R-{\rm mod}}({}_RM,{}_{RS}N)$ is a left $S$-module.
• $\hom_{R-{\rm mod}}({}_RM,{}_RN_S)$ is a right $S$-module.
• $\hom_{{\rm mod}-R}({}_SM_R,N_R)$ is a right $S$-module.
• $\hom_{{\rm mod}-R}(M_{RS},N_R)$ is a left $S$-module.
• $\hom_{{\rm mod}-R}(M_R,{}_SN_R)$ is a left $S$-module.
• $\hom_{{\rm mod}-R}(M_R,N_{RS})$ is a right $S$-module.

In general, $\hom_{R-{\rm mod}}$ eats a left $R$-module structure from each module we stick in. Extra module structures in the second slot carry through, while extra module structures in the first slot get flipped over from right to left and back. The same goes for $\hom_{{\rm mod}-R}$, except it eats a right $R$-modules structure from each slot.

One explicit example of this effect: over a commutative ring $R$, every left module is also a right module and vice-versa. There’s really no difference between $\hom_{R-{\rm mod}}$ and $\hom_{{\rm mod}-R}$ here, so we’ll just write $\hom_R$. Now we’re looking at $\hom_R({}_RM_R,{}_RN_R)$, so one structure (say the left one, for now) on each module gets eaten, leaving a right $R$-module structure on each slot. The second slot carries through and the first slot flips over, giving a left and a right action of $R$ on $\hom_R(M,N)$. This will come in very handy when we start considering modules over fields.

April 24, 2007 Posted by | Ring theory | 4 Comments

## Homomorphisms of modules

Again, once we have a new structure it’s important to understand what sort of functions connect different instancees of the structure. For modules we have module homomorphisms.

By now this should be pretty straightforward. A left $R$-module $M$ is an abelian group equipped with an action of $R$. A homomorphism of $R$-modules is a homomorphism of abelian groups $f:M\rightarrow N$ that also preserves the action of $R$. That is, it’s a linear function satisfying $f(r\cdot m)=r\cdot f(m)$ for all $m\in M$. We call such a function $R$-linear.

In terms of the tensor product picture, any linear function $f:M\rightarrow N$ gives us a linear function $1_R\otimes f:R\otimes M\rightarrow R\otimes N$. The condition of $R$-linearity is that the following diagram commute:

where the horizontal arrows are the actions of $R$ on $M$ and $N$, respectively.

Now if we pick two left $R$-modules $M$ and $N$ we have a bunch of different $R$-module homomorphisms from $M$ to $N$. We’ll call the set of them $\hom_{R{\rm -mod}}(M,N)$, sometimes shortened to $\hom_R(M,N)$, or even just $\hom(M,N)$. The important thing about this set is that it inherits the structure of an abelian group from the one on $N$.

There’s always a homomorphism $0:M\rightarrow N$ sending every element of $M$ to the zero element of $N$. Also, given homomorphisms $f$ and $g$ we can define $\left[f+g\right](m)=f(m)+g(m)$. It’s straightforward to check that this preserves the action of $R$ as long as $f$ and $g$ both do. Finally, there’s a homomorphism defined by $\left[-f\right](m)=-f(m)$. We can easily see that these operations on $\hom(M,N)$ satisfy the axioms of an abelian group.

In fact it gets even better. Not only are the homomorphism sets abelian groups, but composition is bilinear! Let’s consider four homomorphisms between three modules $f_1:A\rightarrow B$, $f_2:A\rightarrow B$, $g_1:B\rightarrow C$, and $g_2:B\rightarrow C$. Then we build up the composition $(g_1+g_2)\circ(f_1+f_2)$. Let’s see what it does to an element $a\in A$.

$\left[(g_1+g_2)\circ(f_1+f_2)\right](a)=\left[g_1+g_2\right](\left[f_1+f_2\right](a))=$
$\left[g_1+g_2\right](f_1(a)+f_2(a))=g_1(f_1(a)+f_2(a))+g_2(f_1(a)+f_2(a))=$
$g_1(f_1(a))+g_1(f_2(a))+g_2(f_1(a))+g_2(f_2(a))=$
$\left[g_1\circ f_1+g_1\circ f_2+g_2\circ f_1+g_2\circ f_2\right](a)$

This gives us a linear function $\hom(A,B)\otimes\hom(B,C)\rightarrow\hom(A,C)$ for every three modules $A$, $B$, and $C$.

What happens if we pick all three modules to be the same one? Each homomorphism set is $\hom_R(M,M)$, which we’ll call ${\rm End}_R(M)$. Then we get a linear function ${\rm End}_R(M)\otimes{\rm End}_R(M)\rightarrow{\rm End}_R(M)$. This is a ring structure! We call it the ring of $R$-endomorphisms of $M$. If the ring $R$ is the ring of integers and $A$ is an abelian group, then ${\rm End}_\mathbb(A)$ is just the endomorphism ring ${\rm End}(A)$ we considered earlier. This is an example of how the theory of modules naturally extends the theory of abelian groups.

April 23, 2007 Posted by | Ring theory | 1 Comment

## Modules

The analogue in ring theory for the idea of a group action is that of a module. Again we want every element of the ring to behave like a function on a set and for multiplication to correspond to composition of functions, but now we have the addition floating around and we’d like to include it in the structure as well. We’ll handle this by letting the ring act on an abelian group.

So let’s take a ring $R$ and an abelian group $M$. We say that $M$ has the structure of a left $R$-module if each element $r\in R$ acts as a linear function from $M$ to itself. We write $r\cdot m$ for the effect of this function on an element $m\in M$. Linearity here means that $r\cdot(m_1+m_2)=r\cdot m_1+r\cdot m_2$. We also require that the ring structure play its part

• $(r_1+r_2)\cdot m=r_1\cdot m+r_2\cdot m$
• $(r_1r_2)\cdot m=r_1\cdot(r_2\cdot m)$
• $1\cdot m=m$ if the ring has an identity element $1$

We can say all this another way. Since this action of $R$ on $M$ is linear in both the ring and the abelian group we get a linear function from the tensor product $\alpha:R\otimes M\rightarrow M$ just like we had one for the multiplication of the ring: $\mu:R\otimes R\rightarrow R$. Not every such function will work, though. We also need that the following diagram commutes:

Around the top of the diagram we use the module action twice, while around the bottom we first multiply in the ring and then use the action once. If the ring has an identity we also require that the following diagram commute

where on the top we use the unique linea function sending the integer $1$ to the identity in $R$, and the diagonal is the canonical isomorphism from $\mathbb{Z}\otimes M$ to $M$.

And there’s another way to say it. Remember that every abelian group $M$ comes equipped with the endomorphism ring ${\rm End}(M)$. An $R$-module structure on $M$ is a ring homomorphism $R\rightarrow{\rm End}(M)$.

All three ways of defining a module — from the raw axioms, as a linear function $R\otimes M\rightarrow M$, or as a homomorphism $R\rightarrow{\rm End}(M)$ — are useful in various situations, and it’s important to be able to slide back and forth between the different pictures.

Of course, the fact that I said left $R$-module above should immediately lead you to think about right $R$-modules. These are the same, except that the module action is written $m\cdot r$ and satisfies $(m\cdot r_1)\cdot r_2=m\cdot(r_1r_2)$ — the order in which the factors are applied is reversed. Over a commutative ring we often ignore this distinction, since we can always switch the order of any product.

Every ring $R$ immediately has the structure of a left and a right $R$-module, using multiplication as the action. This is most clearly seen from the tensor-product definition, where the commuting square just expresses the associativity of ring multiplication. On the other hand, every abelian group $M$ immediately has the structure of an ${\rm End}(M)$-module. This is trivial in the homomorphism picture: just pick the identity homomorphism.

Somewhat more importantly, every abelian group is naturally a $\mathbb{Z}$-module. We define $n\cdot m=m+...+m$ — adding up $n$ copies of the group element $m$. Everything we say for modules in general will apply to abelian groups, and in fact much of what we’ve already said about abelian groups will extend to other modules.

At this point I want to make a point about language. The terms “abelian group” and “$\mathbb{Z}$-module” are interchangeable, but there are cases in which I feel the context clearly indicates using one or the other. Later we’ll come to (and many reader have already seen) “homology groups”, which are more naturally modules than groups. Similarly, the one-dimensional circle, torus, and real projective space are all semantically very different, even though they happen to be equivalent in many situations. I think that this imprecision can lead to confusion on the part of a student, so I’ll try my best to use the most apropriate of equivalent terms.

April 21, 2007 Posted by | Ring theory | 10 Comments