305 -Rings, ideals, homomorphisms (2)

Let’s begin by verifying:

Theorem 1 If {R,S} are rings and {h:R\rightarrow S} is a homomorphism, then {h^{-1}(0)=\{a\in R: h(a)=0\}} is an ideal of {R.}

 

Proof: Clearly {0\in h^{-1}(0),} so this set is nonempty. If {a,b\in h^{-1}(0),} then {h(a-b)=h(a)+h(-b)=h(a)-h(b)=0,} so {a-b\in h^{-1}(0).} Finally, if {a\in h^{-1}(0)} and {b\in R} then {h(ab)=h(a)h(b)=0h(b)=0} and {h(ba)=h(b)h(a)=h(b)0=0} so both {ab} and {ba} are in {h^{-1}(0).} \Box

In a sense, this is the only source of examples of ideals. This is shown by means of an abstract construction.

Theorem 2 If {I} is an ideal of a ring {R} then there is a ring {S} and a homomorphism {h:R\rightarrow S} such that {I=h^{-1}(0).}

 

Proof: The proof resembles what we did to define the rings {{\mathbb Z}_n:} Begin by defining a relation {\sim} on {R} by ({a\sim b} iff {a-b\in I}). Check that {\sim} is an equivalence relation. We can then define {S=R/{\sim} =\{[x]:x\in R\}} where {[x]=[x]_\sim} is the equivalence class of {x,} i.e., {[x]=\{y:x\sim y\}.}

We turn {S} into a ring by defining {[x]+[y]=[x+y],} {[x]\times[y]=[xy]} and {0=[0].} Check that these operations are well defined. Then check that {(S,+,\times,0)} satisfies the axioms of rings.

Finally, let {h:R\rightarrow S} be the quotient map, {h(x)=[x].} Check that this is a homomorphism and that {h^{-1}(0)=I.} \Box

Definition 3 An isomorphism is a bijective homomorphism. If {h:R\rightarrow R} is an isomorphism, we say that it is an automorphism.

 

Proposition 4 Suppose that {{\mathbb F}} is a field and {I} is an ideal of {{\mathbb F}.} Then either {I=\{0\}} or {I={\mathbb F}.} {\Box}

 

It will be very important for us to understand the automorphisms of the field extensions {{\mathbb F}^{p(x)}} where {p(x)\in{\mathbb F}[x].} For this, we will need some tools of linear algebra, so it will be useful to review Chapter 12 and Appendix C of the book.

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