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