oct12 - g + + g n x n ) m = x . Since the coefcient of x nm...

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Math 5125 Wednesday, October 12 October 12, Ungraded Homework Determine all ring automorphisms of Q [ x ] . Let f Q [ x ] . Then by the universal property for polynomial rings, there exists a unique ring homomorphism θ f : Q [ x ] Q [ x ] such that θ ( q ) = q for all q Q and θ ( x ) = f , and in fact θ ( q 0 + q 1 x + ··· + q n x n ) = q 0 + q 1 f + ··· + q n f n . Next if θ : Q [ x ] Q [ x ] is a nonzero homomorphism, then ( θ 1 ) 2 = θ 1 6 = 0 and we deduce that θ 1 = 1. Therefore θ r = r for all r Z and we deduce that θ ( r / s ) = r / s for r , s Z with s 6 = 0. Thus θ q = q for all q Q . If θ x = f , then θ = θ f . We deduce that all nonzero ring homomorphisms Q [ x ] Q [ x ] are of the form θ f for some f Q [ x ] . All that remains is to decide which f yield a bijective map θ f . Since bijective maps are nonzero and θ x is the identity map, we see that θ f is bijective if and only if there exists g Q [ x ] such that θ g θ f ( x ) = x , equivalently θ g ( f ) = x . Write f = f 0 + ··· + f m x m and g = g 0 + ··· + g n x n , where f m , g n 6 = 0 (clearly we may assume that f , g 6 = 0). Then θ g ( f ) = x if and only if f 0 + f 1 ( g 0 + ··· + g n x n )+ ··· + f m (
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Unformatted text preview: g + + g n x n ) m = x . Since the coefcient 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 f where f is a polynomial of degree 1....
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