Unformatted text preview: g + ··· + g n x n ) m = x . Since the coefﬁcient of x nm is f m g m n 6 = 0, we see that the degree of the left hand side is mn . Furthermore the degree of the right hand side is 1, so we must have mn = 1 and hence m = n = 1. We deduce that f 1 6 = 0 and f + f 1 g + f 1 g 1 x = x . This is possible if and only if f + f 1 g = 0 and f 1 g 1 = 1. Since f 1 6 = 0, this is always possible, by setting g 1 = 1 / f 1 and then g =f / f 1 . We conclude that all automorphisms of Q [ x ] are of the form q 7→ q for q ∈ Q and x 7→ ax + b where a , b ∈ Q and a 6 = 0. Another way to state this using the notation above, all maps of the form θ f where f is a polynomial of degree 1....
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 Fall '07
 PALinnell
 Math, Algebra, Vector Space, Isomorphism, Homomorphism, Polynomial ring, gn xn

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