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  1. #1
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    automorphism

    let k be a field.

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

    b) conversely, let $\displaystyle \varphi$ be an automorphism of k[t] which is the identity on k. Prove that thre exist $\displaystyle a,b \in k $ with $\displaystyle a \ne 0$ such that $\displaystyle \varphi(f(t)=f(at+b)$ 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 $\displaystyle a=1$ and $\displaystyle b=0$ 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
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  2. #2
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    Quote Originally Posted by TheEmptySet View Post
    let k be a field.

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

    b) conversely, let $\displaystyle \varphi$ be an automorphism of k[t] which is the identity on k. Prove that thre exist $\displaystyle a,b \in k $ with $\displaystyle a \ne 0$ such that $\displaystyle \varphi(f(t)=f(at+b)$ 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 $\displaystyle a=1$ and $\displaystyle b=0$ 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 $\displaystyle \varphi(t)=at + b,$ for some $\displaystyle 0 \neq a, b \in k,$ because $\displaystyle \varphi(a_0 + a_1t + \cdots + a_nt^n)=a_0 + a_1\varphi(t) + \cdots + a_n (\varphi(t))^n.$ so suppose $\displaystyle \varphi(t)=g(t).$

    show that if $\displaystyle \deg g(t) \neq 1,$ then $\displaystyle \varphi$ wouldn't be onto. (look at $\displaystyle \varphi^{-1}(t)$)
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    you only need to show that $\displaystyle \varphi(t)=at + b,$ for some $\displaystyle 0 \neq a, b \in k,$ because $\displaystyle \varphi(a_0 + a_1t + \cdots + a_nt^n)=a_0 + a_1\varphi(t) + \cdots + a_n (\varphi(t))^n.$ so suppose $\displaystyle \varphi(t)=g(t).$

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

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

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

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

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

    but since the degree of $\displaystyle g(t) > 1$ there will be no linear terms. In fact the lowest degree of any nonconstant polynomial would be $\displaystyle \deg(g)$ so there cannot be any isomporphisms except when g is linear.
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    Quote Originally Posted by TheEmptySet View Post
    ahhh. Thank you very much

    so if $\displaystyle \deg(g)=0$ I get the identity map.
    i wouldn't call it the identitiy map though! if $\displaystyle g = c$ is constant, then the image of $\displaystyle \varphi$ would be $\displaystyle k$ instead of $\displaystyle k[t].$


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

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

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

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