Module Equivalence: Understanding Ann(M)

In summary, Module "equivalence" refers to the concept of being able to view a module as both an R-module and an R/I-module. This is made possible by two well-behaved constructions, one using the quotient ring R/I and the other using the tensor product of a ring homomorphism. This allows for the category of R/I-modules to be studied using the simpler module theory of R, with the condition I \subseteq Ann(M) determining which R-modules can also be regarded as R/I-modules.
  • #1
Bleys
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Module "equivalence"

There is a problem in a book I'm not quite understanding.
Let M be an R-module and let I=Ann(M). Show that M can be regarder as an R/I-Module where scalar multiplication is given by the rule m(I+r)=mr

I don't understand what they mean by "regarded as". Am I suppose to show there is an isomorphism? (I don't know how to do that between modules over different rings), or just a bijection? Should I create a one-to-one mapping from the scalars R to R/I instead?

Any help is appreciated
 
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  • #2


Really, they probably just want you to show that the underlying additive group of M, together with that scalar multiplication, is an R/I-module.

Bonus points for explicitly showing that, if [itex]\pi[/itex] is the projection from R to R/I, that [itex]r \cdot m = \pi(r) \cdot m[/itex]. (where the two [itex]\cdot[/itex] symbols refer to the appropriate scalar products)



One main point is that, in the expression [itex]r \cdot m[/itex] where r is in R and m is in M, it doesn't matter if we mean this product as the scalar product on M, or if we are using r to name an element of R/I, and the product is the scalar product on the R/I-module defined in the problem; the result of the arithmetic expression is the same.
 
  • #3


Ok, thank you Hurkyl!
Just out of curiosity, what exactly does this result mean, the fact that the set M can be regarded as both an R-module and an R/I-module? Is this mostly a consequence of the properties of the quotient ring R/I (for example of, like you pointed out, the canonical homomorphism from R to R/I)?
I'm finding it hard to wrap my head around it because R and R/I are not isomorphic (possibly not even of the same order if finite) yet behave the same as scalars over M?
 
  • #4


The more opportunities you have to use this fact, the more it will make sense.

You already know that all of the elements of I behave exactly the same as scalar multipliers on M. So what do you get when you impose that they are all congruent? :smile:



Incidentally, there are two well-behaved constructions related to a ring homomorphism f:R --> S.
  • (The underlying group of) any S-module M can be given an R-module structure defined by [itex]r\cdot m = f(r)\cdot m[/itex].
  • By the above, S itself can be viewed as an R-module. For any R-module M, the (underlying group of) the tensor product [itex]S \otimes_R M[/itex] can be given an S-module structure defined by [itex]s \cdot (s' \otimes m) = (ss') \otimes m[/itex].
(These constructions extend to transform S-module homomorphisms into R-module ones and R-module into S-module ones respectively)


When S is the quotient ring R/I, the constructions are even more well-behaved. The underlying group of the second construction is just (isomorphic to) the quotient [itex]M / IM[/itex]. Furthermore, the two constructions essentially become one-sided inverses -- applying the first construction and then the second construction gives you a module naturally isomorphic to the one you started with.

Roughly speaking, this means that you can view the entire category of R/I-modules as if it was just a subcategory of R-modules. Another way of saying that is that it's unusually simple to use the module theory of R to study the module theory of R/I. How do you identify which R modules are also R/I modules under this correspondence? They are the ones with [itex]I \subseteq Ann(M)[/itex].
 
  • #5


I would approach this problem by looking at the definition of module equivalence. Two modules are considered equivalent if there exists a bijective homomorphism between them. In this case, we have an R-module M and an R/I-module, where I is the annihilator of M. So, to show that M can be regarded as an R/I-module, we need to find a bijective homomorphism between them.

To do this, we can define a mapping from M to M' (the R/I-module) as m → m(I+r), where m is an element of M and r is an element of R. This mapping is well-defined and bijective, as for any two elements m and m' in M, if m(I+r) = m'(I+r), then m = m' since I is the annihilator of M. Additionally, this mapping preserves the module structure, as (m+m')(I+r) = m(I+r) + m'(I+r) and (rm)(I+r) = (rm)(I+r) for any r in R.

Therefore, we have shown that there exists a bijective homomorphism between M and M', thus proving that M can be regarded as an R/I-module. This means that M and M' are equivalent modules, and we can use the scalar multiplication rule given in the problem to define the module structure of M'.
 

What is module equivalence?

Module equivalence refers to the concept of two modules being essentially the same, despite having different names or structures. It means that the two modules have the same functionality and can be used interchangeably.

Why is understanding Ann(M) important?

Ann(M), or the annihilator of a module M, is important because it helps us understand the structure of M. It is the set of elements in the module's ring that "kill" or annihilate the elements of M. This information can be useful in determining the properties and behavior of the module.

How do you determine if two modules are equivalent?

To determine if two modules are equivalent, we need to check if their Ann(M) sets are the same. If the two modules have the same Ann(M) set, then they are equivalent. This means that they have the same structure and can be used interchangeably.

What is the significance of module equivalence in algebraic structures?

Module equivalence is significant in algebraic structures because it allows us to simplify and generalize concepts. By understanding Ann(M), we can determine if two seemingly different modules are essentially the same, which can lead to a deeper understanding of the structure and properties of the modules.

Can module equivalence be extended to other mathematical structures?

Yes, module equivalence can be extended to other mathematical structures, such as vector spaces and algebras. In these structures, the concept of an annihilator may be different, but the idea of two structures being equivalent remains the same - they have the same properties and can be used interchangeably.

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