the set of ring automorphisms is an abstract group under composition

catcherintherye

1. The problem statement, all variables and given/known data

Aut(R) denotes the set of ring automorphisms of a ring R Show formally that Aut(R) is a group under composition.

2. Relevant equations



3. The attempt at a solution

I Have a very similar question to which I have the solution viz


Aut(G) denotes the set of group automorphisms of a Group G, show that Aut(G) is a group under composition.


Proof Let a,b: G -> G be automorphisms

then [tex] a\circ b: G \rightarrow g is also an auto [/tex]

[tex] (a\circ b)(xy) = a(b(xy))=a(b(x)b(y))= a(b(x))a(b(y)) [/tex]
[tex] =(a\circ b)(x)(a\circ b)(y) [/tex]

so [tex] a\circb [/tex] is a homomorphism

it is also bijective since a,b are bijective

[tex] \circ: aut(G)\times aut(G) \rightarrow aut(G) [/tex]
is automatically associative (because comp of mappings is associative)

As identity in Aut(G) [tex] take Id_g:G \rightarrow G [/tex]
finally inverses

Let a: G --> G be an auto

then a^-1:G -->G is atleast a mapping and bijective

need only show

[tex] a^-1(xy) = a^-1(x)a^-1(y)[/tex]

let [tex] x,y \in G [/tex]

choose [tex] c,d \in G [/tex] : a(c)=x, a(d)=y

[tex] a^-1(xy) = a^-1(a(c)a(d))=a^-1(a(cd)) = cd= a^-1(x)a^-1(y) [/tex]

q.e.d



.......the proof for rings is essentially the same right? The only thing that concerns me is the last part(above) since we used the fact that every element has it's inverse in a group but we don't have that in a ring....

matt grime

Where have you used that fact?

catcherintherye

ok so i haven't and these two proofs are essentially the same?

HallsofIvy

You did use a^-1 where a is automorphism. You say you know
Are you sure of that definition of Aut(G)? Homomorphism, in general, do not have inverses.

catcherintherye

no i've made a mistake it's supposed to be the group of isomorphisms