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Let $B = k[T]$ with $k$ a field; a $k$-automorphism of $B$ is a ring homomorphism $φ: B \to B$ that is the identity on $k$ and is an automorphism of $B$;
(i) Describe the group $Aut(B)$ of $k$-automorphism of $B$;
(ii) now do the same for the field $K = k(T)$ of rational function.

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You seem to have used the same notation for the polynomial ring and the field of rational functions. Anyway, what have you tried? What could the image of $T$ under such a map be? – Tobias Kildetoft Apr 10 '14 at 9:33

Any $k$-automorphism $\phi$ of $B$ is determined completely by its effect on $T$, viz. the polynomial $\phi(T)$ associated to the degree 1 polynomial $T$. If $\phi(T)$ is of degree $k$ then the image will have constants and polynomials of degree a multiple of $k$. Now you should be able figure out what choices will make $\phi$ an automorphism.

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