# automorphism

• May 18th 2009, 07:56 PM
TheEmptySet
automorphism
let k be a field.

a) show that the mapping $\varphi : k[t] \to k[t]$ definded by $\varphi(f(t))=f(at+b)$ for fixed $a,b \in k,a \ne 0$ is an automorphism of k[t] which is the identity on k.

b) conversely, let $\varphi$ be an automorphism of k[t] which is the identity on k. Prove that thre exist $a,b \in k$ with $a \ne 0$ such that $\varphi(f(t)=f(at+b)$ 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 $a=1$ and $b=0$ 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
• May 18th 2009, 08:07 PM
NonCommAlg
Quote:

Originally Posted by TheEmptySet
let k be a field.

a) show that the mapping $\varphi : k[t] \to k[t]$ definded by $\varphi(f(t))=f(at+b)$ for fixed $a,b \in k,a \ne 0$ is an automorphism of k[t] which is the identity on k.

b) conversely, let $\varphi$ be an automorphism of k[t] which is the identity on k. Prove that thre exist $a,b \in k$ with $a \ne 0$ such that $\varphi(f(t)=f(at+b)$ 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 $a=1$ and $b=0$ 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

you only need to show that $\varphi(t)=at + b,$ for some $0 \neq a, b \in k,$ because $\varphi(a_0 + a_1t + \cdots + a_nt^n)=a_0 + a_1\varphi(t) + \cdots + a_n (\varphi(t))^n.$ so suppose $\varphi(t)=g(t).$

show that if $\deg g(t) \neq 1,$ then $\varphi$ wouldn't be onto. (look at $\varphi^{-1}(t)$)
• May 18th 2009, 08:40 PM
TheEmptySet
Quote:

Originally Posted by NonCommAlg
you only need to show that $\varphi(t)=at + b,$ for some $0 \neq a, b \in k,$ because $\varphi(a_0 + a_1t + \cdots + a_nt^n)=a_0 + a_1\varphi(t) + \cdots + a_n (\varphi(t))^n.$ so suppose $\varphi(t)=g(t).$

show that if $\deg g(t) \neq 1,$ then $\varphi$ wouldn't be onto. (look at $\varphi^{-1}(t)$)

ahhh. Thank you very much

so if $\deg(g)=0$ I get the identity map.(Clapping)

so Now if $\deg(g) > 1$ any linear polynomial does not have a preimmage.

let $f(x)=\sum_{i=0}^{n}a_{n-i}x^{n-i}$ then

$\varphi (f)=\sum_{i=0}^{n}a_{n-i}[g(t)]^{n-i}$

but since the degree of $g(t) > 1$ there will be no linear terms. In fact the lowest degree of any nonconstant polynomial would be $\deg(g)$ so there cannot be any isomporphisms except when g is linear.
• May 18th 2009, 08:51 PM
NonCommAlg
Quote:

Originally Posted by TheEmptySet
ahhh. Thank you very much

so if $\deg(g)=0$ I get the identity map.(Clapping)

i wouldn't call it the identitiy map though! if $g = c$ is constant, then the image of $\varphi$ would be $k$ instead of $k[t].$

Quote:

so Now if $\deg(g) > 1$ any linear polynomial does not have a preimmage.

let $f(x)=\sum_{i=0}^{n}a_{n-i}x^{n-i}$ then

$\varphi (f)=\sum_{i=0}^{n}a_{n-i}[g(t)]^{n-i}$

but since the degree of $g(t) > 1$ there will be no linear terms. In fact the lowest degree of any nonconstant polynomial would be $\deg(g)$ so there cannot be any isomporphisms except when g is linear.
that's right! you can also argue that in this case $\varphi^{-1}(t)$ wouldn't exist, because $\deg \varphi(f(t))=0,$ if $\deg f = 0,$ and $\deg \varphi(f(t)) \geq 2,$ if $\deg f \geq 1.$ so we'll never have $\varphi(f(t)) = t.$