## 305 -Rings, ideals, homomorphisms (2)

March 16, 2009

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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