# Abstract Nonsense

## 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.

June 18, 2011 -

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