oct14 - x zero) of k [ x ] . Prove that x 2 is irreducible...

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Math 5125 Friday, October 14 October 14, Ungraded Homework Let R be a ring with a 1 and let θ : Q R be a nonzero ring homomorphism. Suppose that R has no nonzero zerodivisors (i.e. a , b R \ 0 implies ab 6 = 0). If m , n Z and n 6 = 0, then θ ( m / n ) = mn - 1 (of course m R means the element m 1). Since θ is not the zero homomorphism, there exists x Z such that θ ( x ) 6 = 0. Now θ ( x ) = θ ( x ) θ ( 1 ) and R has no nonzero zerodivisors, hence θ ( 1 ) = 1. Finally θ ( n ) θ ( 1 / n ) = θ ( 1 ) = 1 for 0 6 = n Z , so θ ( 1 / n ) = n - 1 . The result follows. Let k be a field and let R denote the subring { a 0 + a 2 x 2 + ··· + a n x n | a i k } (that is poly- nomials with coefficient of
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Unformatted text preview: x zero) of k [ x ] . Prove that x 2 is irreducible but not prime in R . Suppose x 2 = fg where f , g ∈ R . Since deg ( fg ) = deg ( f )+ deg ( g ) and R has no element of degree 1, we see that one of f , g has degree zero, say deg ( f ) = 0. But this implies 0 6 = f ∈ k and hence f is a unit. This shows that x 2 is irreducible. However x 2 is not prime because x 2 ± ± x 3 x 3 , but x 2-x 3 ....
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This note was uploaded on 01/02/2012 for the course MATH 5125 taught by Professor Palinnell during the Fall '07 term at Virginia Tech.

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