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Math Help - automorphism

  1. #1
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    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
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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 \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))
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    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.

    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.
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    ahhh. Thank you very much

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


    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.
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