Abstract Nonsense

Module Homomorphisms

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

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Motivation

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 $R$homomorphism (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 $R$morphism 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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References:

[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 -

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