The Unapologetic Mathematician

Mathematics for the interested outsider

More about hom

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


  1. […] has deeper structure itself. For example, the set of homomorphisms between two abelian groups is itself an abelian group, because abelian groups are modules over the commutative ring . More generally, the set of […]

    Pingback by Enriched Categories « The Unapologetic Mathematician | August 13, 2007 | Reply

  2. Why is \sigma inverted in the definition of \mathbb{Z}[S_n]\rightarrow{\rm End}(A^{\otimes n})?

    Comment by MathOutsider | October 21, 2007 | Reply

  3. It’s so that the permutation (1\,2\,3) sends a_1\otimes a_2\otimes a_3 to a_3\otimes a_1\otimes a_2. That is, what was in slot 1 is now in slot 2, what was in slot 2 is now in slot 3, and what was in slot 3 is now in slot 1.

    Comment by John Armstrong | October 21, 2007 | Reply

  4. […] categories. In fact, since we’re working over a field (which is a commutative ring) the properties of -functors tell us that is enriched over […]

    Pingback by Linear Algebra « The Unapologetic Mathematician | May 19, 2008 | Reply

  5. This was really helpful. Tensor products seemed like a mess for me, because I could never remember where the extra module structures came from, and why it is there. Finally I have a good mnemonic: tensor products and hom are abelian groups. Taking the tensor products and taking hom eats the ring action. Extra actions on the modules yield extra actions on the tensor product and on hom. The first entry of hom turns a left to a right action and a right to a left action respectively. Thanks a lot!! 🙂

    Comment by Nico | May 5, 2021 | Reply

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