oct14 - x zero of k x Prove that x 2 is irreducible but not...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online