305 -Rings, ideals, homomorphisms (2)

Let’s begin by verifying:

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


Proof: Clearly {0in h^{-1}(0),} so this set is nonempty. If {a,bin h^{-1}(0),} then {h(a-b)=h(a)+h(-b)=h(a)-h(b)=0,} so {a-bin h^{-1}(0).} Finally, if {ain h^{-1}(0)} and {bin 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:Rrightarrow 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 ({asim b} iff {a-bin I}). Check that {sim} is an equivalence relation. We can then define {S=R/{sim} ={[x]:xin R}} where {[x]=[x]_sim} is the equivalence class of {x,} i.e., {[x]={y:xsim 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:Rrightarrow 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:Rrightarrow 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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