Abstract Nonsense

Crushing one theorem at a time

Module Homomorphisms

Point of Post: In this post we discuss the notion of module homomorphism.

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As always, when we define new structures our first job is to inspect the structure preserving homomorphisms associated with them. Fortunately for us, the morphisms for R-modules turn out to be precisely what we’d hope they’d be– “linear transformations“, or at least the analogue of them for modules (i.e. group homomorphisms that respect scalar multiplication).

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Module Homomorphisms

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Let M,N be two left R-modules. We say that a map f:M\to N is a Rhomomorphism (or just homomorphism, or R-morphsim for short) if f(rx+sy)=rf(x)+sf(y) for every x,y\in M and s,r\in R. Since f is, in particular, a group homomorphism between M\to N it enjoys all the pursuant benefits. We define, as per usual, the kernel \ker f of a R-morphism to be group-theoretic kernel of f (i.e. the preimage of 0_M) and the image to be the set theoretic image.

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Ok, so fine, these R-morphisms are nothing but linear transformations from linear algebra! Do they have the enjoy the same properties? Well, the answer is, for the most part yes. For exampe:

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Theorem: Let M,N be left R-modules and f:M\to N be a R-morphism. Then, for any submodules A\leqslant M and B\leqslant N one has that f^{-1}(B)\leqslant M and f(A)\leqslant N.

Proof: We know that f(A) and f^{-1}(B) are additive subgroups of their respective modules and so it just suffices to check the stability under R-multiplication. To see this we note that if f(a)\in f(A) then rf(a)=f(ra)\in f(A) since ra\in A. Moreover, if x\in f^{-1}(B) then since f(rx)=rf(x)\in B we have that rx\in f^{-1}(B). \blacksquare

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In particular, the above gives us that the kernel of a Rmorphism is a submodule of the domain space.

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We define a R-morphism f:N\to M to be a monomorphism if whenever g_1,g_2:L\to N are any R-morphisms (where L is some left R-modul)e such that f\circ g_1=f\circ g_2 then g_1=g_2. We can come up with a different, more manageable, characterization of monomorpisms for unital modules though:

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Theorem: Let M,N be unital left R-modules and f:M\to N a R-morphism. Then, f is a monomorphism if and only if f is injective.

Proof: Suppose first that f is a monomorphism, but there exists x\ne y\in M with f(x)= f(y). Define g_1:R\to M by g_1(r)=rx and g_2:R\to M with g_2(r)=ry. Clearly g_1 is a R-morphisms since g_1(r+r')=(r+r')x=rx+r'x=g_1(r)+g_1(r') and g_1(rr')=rr'x=r(r'x)=rg_1(r') and similarly for g_2. That said, note that f(g_1(r))=f(rx)=rf(x)=rf(y)=f(ry)=f(g_1(r)) so that f\circ g_1=f\circ g_2. But, this is a contradiction since f being a monomorphism should imply g_1=g_2 but this is untrue since g_1(1)=1x=x\ne y=1y=g_2(1). Thus, f is injective as claimed.

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Conversely, if f is injective then f is left cancellative for any set map L\to M (this is just basic set theory) and so the conclusion obviously follows for morphisms L\to M. \blacksquare

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We define an epimorphism f:M\to N to be a map such that given any two morphisms out of N, i.e. two morphisms g)1,g_2N\to L, such that g_1\circ f=g_2\circ f one can conclude that g_1=g_2. Just as for monomorphisms, for unital modules there is a nice way to characterize epimorphisms:

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Theorem: Let M,N be unital left R-modules and f:M\to N a morphism. Then, f is an epimorphism if and only if f is surjective.

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Sadly, we are not yet in a position to prove this, mostly because we haven’t developed enough definitions/machinery. So, until the point that we have and we can state the theorem fully we prove it for vector spaces–we do this because the proof is almost word-for-word the same, but it’s not clear we can do everything we are about to do for general left R-modules (e.g. the existence of quotient spaces):

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Theorem: Let V,W be two vector spaces over some field k. Then, a linear transformation T:V\to W is an epimorphism if and only if T is surjective.

Proof: Suppose first that T is an epimorphism. Consider then the quotient space W/T(V) and consider the two maps 0,\pi: W\to W/T(V) where 0 is the zero map and \pi is the canonical projection. Note then that \text{im }T\subseteq=\ker\pi and so T\circ\pi=T\circ 0=0. But, this implies by assumption that T is an epimorphism that 0=\pi and so W/T(V) is zero-dimensional, which implies that T(V)=W as desired.

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The converse follows because surjections are right-cancellative for any set map, let alone any morphism. \blacksquare

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Of course, after monomorphism and epimorphism comes the stronges morphism, the isomorphism. In particular if M,N are left R-modules we say that morphism f:M\to N is an isomorphism if there exists a morphism g:N\to M such that g\circ f=\text{id}_M and f\circ g=\text{id}_N. Considering the following:

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Theorem: Let M,N be left R-modules and f:M\to N a bijective morphism. Then, f^{-1} is also a morphism.

Proof: It’s basic group theory that f^{-1} is a group homomorphism, to see that it respects R-multiplication we merely note that f(rx)=rf(x) says rx=f^{-1}(rf(x)), or letting f(x)=y (which we can uniquely since f is bijective) rf^{-1}(y)=f^{-1}(ry). \blacksquare

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From this it easily follows, as it did quite nicely for epi and monomorphisms that a morphism is an isomorphism if and only if it’s bijective. If there exists an isomorphism between two modules we call them isomorphic and denote this by saying M\cong N

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Remark: The reason I am making such a big fuss over these seeming trivialities is that the category R-\bold{Mod} of left R-modules is going to serve as a prime example of a concrete category when I get around to discussing such categorical matters–and so the above gives a nice example of a concrete category where the epis and monos are just the surjections and injections respectively.

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Now, it’s handy to note that since every R-morphism is a group homomorphism we have that a R-morphism will be a monomorphism if and only if it has trivial kernel.

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Of course, it’s clear from the above that the composition of epi,mono,iso-morphisms are epi-mono,iso-morphisms. Moreover, if g\circ f is an epimorphism then g is an epimorphism and if g\circ f is a monomorphism then f is a monomorphism.

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[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.


October 28, 2011 - Posted by | Algebra, Module Theory, Ring Theory | , , , , , , , , , , ,


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