This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 5125 Monday, October 24 Eighth Homework Solutions 1. (a) Let a , b ∈ S and suppose S is not multiplicative, i.e. there exists 0 6 = r ∈ R such that abr = 0. Since b is a nonzerodivisor, we see that br 6 = 0. But then a ( br ) = 0, which contradicts the hypothesis that a is a nonzerodivisor and it follows that S is multiplicative. The fact that S contains 1 but not 0 is obvious. (b) The general element of S 1 R is of the form r / s with r ∈ R and s ∈ S . If r is a zero divisor, then rx = 0 for some x ∈ R 0, hence ( r / s )( x / 1 ) = 0 and we deduce that r / s is a zero divisor (note that x / 1 6 = 0). On the other hand if r is not a zero divisor, then r ∈ S and we see that r / s has a multiplicative inverse, namely s / r . 2. If S 1 R = 0, then 1 / 1 = / 1, that is we can find s ∈ S such that s ( 11 10 ) = 0 which means that s = 0. Since S does not contain 0, we conclude that S 1 R is never 0....
View
Full Document
 Fall '07
 PALinnell
 Math, Algebra, Prime number, Greatest common divisor, Divisor, ij, Universal property, ker α

Click to edit the document details