Show there is an automorphism?

[kathrynag]

Define h:Z[sqrt2]--->Z[sqrt2] by h(m+nsqrt2)=m-nsqrt2 for all m,n in Z. Show that h is an automorphism of Z[sqrt2}

So a ring automorphism is when we have an isomorphism from a commutative ring R onto itself.
I'm a little confused on starting this one out.

 

[snipez90]

In the other thread you seemed to know what an isomorphism was. Here it is pretty easy to check that you have an isomorphism. Basically, if the problem gives you a map and asks you to show that it's an automorphism, then just do it. On the other hand, it's not always easy to exhibit an isomorphism between two rings (or groups or whatever) or prove that an isomorphism cannot possibly exist from first principles, so you may want to think of some theorems that can help you decide.

 

[kathrynag]

So I would just check that h:Z[sqrt2]--->Z[sqrt2] by h(m+nsqrt2)=m-nsqrt2 for all m,n in Z is 1-1 and onto?

 

[Robert1986]

So I would just check that h:Z[sqrt2]--->Z[sqrt2] by h(m+nsqrt2)=m-nsqrt2 for all m,n in Z is 1-1 and onto?

Yes, an automorphism is a special case of an isomorphism so all you need to do is to show that your map is an isomorphism. However, since you are mapping one set into itself, do you really need to show that it is 1-1 AND onto? Or could you just show one and somehow directly infer the other?