Abstract Nonsense

Crushing one theorem at a time

Ring Homomorphisms (Pt. I)


Point of Post: In this post we discuss the notion of ring  homomorphisms, kernels, images, etc.

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Motivation As is standard in math, especially in algebra after defining a structure and its subobjects we define the morphisms between the two objects. In particular, we would now like to define the morphisms between two rings R and R'. What do we want these morphisms to do? Well, we clearly want the morphisms to preserve the ring structure. In other words, we’d like it to be additive (be a group homomorphism for the group structure) and multiplicative (be a semigroup homomorphism for the semigroup structure). After this we define the obvious notions of kernel and image and show that they are, in fact, subrings ,etc.

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Definitions and Basics

Let R and R' be rings. A map f:R\to R' is a ring homomorphism (alternatively homomorphism or even just morphism when the context is clear) if for all a,b\in R one has that f(a+b)=f(a)+f(b) and f(ab)=f(a)f(b). We denote the set of all ring homomorphisms R\to R' by \text{Hom}\left(R,R'\right). Some thing An injective homomorphism is called a monomorphism and a surjective homomorphism is called an epimorphism and is often denoted f:R\twoheadrightarrow R'. If f:R\to R' is both a bijective homomorphism we call it an isomorphism. If there exists an isomorphism R\to R' we say that R and R‘ are isomorphic and write R\cong R'. Some things are obvious:

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\text{ } Theorem: Let R_1,R_2,R_3 be rings and f\in\text{Hom}(R_1,R_2), g\in\text{Hom}(R_2,R_3). Then

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\begin{aligned}&\mathbf{(1)}\quad f(0)=0\\ &\mathbf{(2)}\quad f(-a)=-f(a)\\ &\mathbf{(3)}\quad f(na)=nf(a)\;\;n\in\mathbb{N}\\ &\mathbf{(4)}\quad f(a^n)=f(a)^n\;\;n\in\mathbb{N}\\ &\mathbf{(5)}\quad g\circ f\in\text{Hom}(g_1,g_3)\end{aligned}

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Proof: The first three of these follow immediately since f is a group homomorphism from (R_1,+) to (R_2,+). The fourth is immediate by induction. To see the fifth claim we know from first principles that g\circ f is a group homomorphism and it’s clear it’s a semigroup homomorphism for the multiplicative structure since g(f(ab)=g(f(a)(f(b))=g(f(a))g(f(b)) for all a,b\in R_1. The conclusion follows. \blacksquare

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Corollary: Let f:R_1\to R_2 and g:R_2\to R_3 be monomorphisms, epimorphisms, or isomorphism then g\circ f:R_1\to R_3 is a monomorphism, epimorphism, and isomorphism respectively.

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Moreover, we have that

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Theorem: Let f:R\to R' is an isomorphism then f^{-1}:R'\to R is an isomorphism.

Proof: We know that f^{-1} is a bijection so it suffices to prove its a homomorphism. The fact that f^{-1} is additive follows from basic ring theory and for any f(a),f(b)\in R' we have that f^{-1}(f(a)f(b))=f^{-1}(f(ab))=ab=f^{-1}(f(a))f^{-1}(f(b)) we may conclude the theorem. \blacksquare

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Since for any ring R we evidently have that \text{id}:R\to R is an isomorphism we get the following (really quite expected) theorem:

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Theorem: The relation “\cong” is an equivalence relation.

So, what about elements of R^\times? If this were a group homomorphism we would say something like “oh, well f(u^{-1})=f(u)^{-1} but is the same true here? Indeed it is:

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Image and Kernel

Similar to the case for groups the notions of image and kernel of a ring homomorphism are important  in the overall theory of rings. So, for a homomorphism f:R\to R' we call f^{-1}(\{0'\}) (where 0' is the zero element of R') the kernel of f and denote it \ker f. We note that the kernel of f:(R,+,\cdot)\to (R',+',\cdot') is really the same as the kernel of the group homomorphism f:(R,+)\to (R',+'). We denote f(R) as \text{im} f and call it the image of f. We first note that both of these respect the structure in an obvious way, namely:

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Theorem: Let f:R\to R' be a homomorphism, then \text{im }f and \ker f are subrings of R' and R respectively.

Proof: Let f(a),f(b)\in\text{im }f then f(a)-f(b)=f(a-b)\in\text{im }f and f(a)f(b)=f(ab)\in\text{im }f and so \text{im }f is a subring as desired. Similarly, let a,b\in\ker f then f(a-b)=f(a)-f(b)=0-0=0 and f(ab)=f(a)f(b)=00=0 and so a-b,ab\in\ker f and so the kernel is a subring as desired. \blacksquare

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The fact that f is also a group homomorphism gives us for free facts such as

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Theorem: Let f:R\to R' be a homomorphism, then f is injective if and only if \ker f is trivial.

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They also preserve other structure such as commutativity:

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Theorem: Let f:R\to R' be a homomorphism with R commutative, then \text{im }f is commutative.

Proof: Let f(a),f(b)\in\text{im }f be arbitrary. Then, f(a)f(b)=f(ab)=f(ba)=f(b)f(a), and since f(a),f(b) were arbitrary the conclusion follows. \blacksquare

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References:

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

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

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June 18, 2011 - Posted by | Algebra, Ring Theory | , , , , , , ,

10 Comments »

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