let k be a field.
a) show that the mapping definded by for fixed is an automorphism of k[t] which is the identity on k.
b) conversely, let be an automorphism of k[t] which is the identity on k. Prove that thre exist with such that as in part a)
So I have done part a) but I can't seem to get out of the gate for part b).
I keep reading the problem and only seeing the trivial solution.
The identitiy map is an automorphism so and satisfy the above, but I have a feeling that is not what Dummit and Foote had in mind.
A push in the right direction would be great.
Thanks
TES
ahhh. Thank you very much
so if I get the identity map.
so Now if any linear polynomial does not have a preimmage.
let then
but since the degree of there will be no linear terms. In fact the lowest degree of any nonconstant polynomial would be so there cannot be any isomporphisms except when g is linear.
i wouldn't call it the identitiy map though! if is constant, then the image of would be instead of
that's right! you can also argue that in this case wouldn't exist, because if and if so we'll never have
so Now if any linear polynomial does not have a preimmage.
let then
but since the degree of there will be no linear terms. In fact the lowest degree of any nonconstant polynomial would be so there cannot be any isomporphisms except when g is linear.