# all3 - Math 5125 Friday November 4 November 4 Ungraded...

This preview shows pages 1–3. Sign up to view the full content.

Math 5125 Friday, November 4 November 4, Ungraded Homework Remarks If M is an R -module, then 0 m = 0 = r 0, and ( - r ) m = - rm for all m M and r R . We shall use these elementary properties without comment. Exercise 10.1.3 on page 343 Assume that rm = 0 for some r R and some m M with m = 0. Prove that r does not have a left inverse (i.e., there is no s R such that sr = 1). Suppose to the contrary that there is an s R such that sr = 1. Then 0 = m = 1 m = ( sr ) m = s ( rm ) = s 0 = 0 which is a contradiction and the result is proven. Exercise 10.1.9 on page 344 If N is a submodule of M , the annihilator of N in R is defined to be { r R | rn = 0 for all n N } . Prove that the annihilator of N in R is a 2-sided ideal of R . Let I denote the annihilator of N in R . First note that 0 I , because 0 n = 0 for all n N . If x , y I , then xn = yn = 0 for all n N , hence ( x + y ) n = xn + yn = 0 + 0 = 0 and ( - x ) n = - ( xn ) = 0, and we deduce that x + y I and - x I . We have so far shown that I is an abelian group under addition. Suppose r R . Then ( rx ) n = r ( xn ) = r 0 = 0 and ( xr ) n = x ( rn ) = 0, and we conclude that rx and xr I . This completes the proof that I is an ideal of R . Exercise 10.2.8 on page 350 Let φ : M N be an R -module homomorphism. Prove that φ ( Tor ( M )) Tor ( N ) (where Tor ( M ) indicates the torsion elements of M ). Let m Tor ( M ) . Then there exists 0 = r R such that rm = 0. Then 0 = θ ( rm ) = r ( θ m ) , which shows that θ m Tor ( N ) as required.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Math 5125 Monday, November 7 November 7, Ungraded Homework Exercise 10.2.7 on page 350 Let z be a fixed element of the center of R . Prove that the map m zm is an R -module homomorphism from M to itself. Show that for a commutative ring R the map from R to End R ( M ) given by r rI is a ring homomorphism (where I is the identity endomorphism). Denote by θ the map m zm . Then for m , n M and r R , we have θ ( m + n ) = z ( m + n ) = zm + zn = θ m + θ n , and because z is in the center of R , we have zr = rz and hence θ ( rm ) = zrm = rzm = r θ m . This proves that θ End R ( M ) . Now suppose that R is commutative. This means that rI End R ( M ) for all r R , because it is the map m rm for m M , so we could use the above. Let us denote by φ the map r rI . Then for r , s R , we have φ ( r + s )( m ) = ( r + s ) m = rm + sm = φ ( r ) m + φ ( s ) m = ( φ ( r )+ φ ( s ))( m ) for m M , which shows that φ ( r + s ) = φ ( r )+ φ ( s ) . Also φ ( rs )( m ) = ( rs ) m = r ( sm ) = φ ( r ) φ ( s ) m for all m M , which shows that φ ( rs ) = φ ( r ) φ ( s ) . This proves that φ is a ring homomor- phism as required. Exercise 10.3.6 on page 356 Let R be a ring with a 1. Prove that if M is a finitely generated R -module that is generated by n elements, then every quotient of M may be generated by n elements. Deduce that quotients of cyclic modules are cyclic. Let M be generated by the n elements x 1 ,..., x n , and let M / N be a quotient of M , where N is a submodule of M . Thus we have M = Rx 1 + ··· + Rx n and given m M , we can find r 1 ,..., r n R such that m = r 1 x 1 + ··· + r n x n . Now consider M / N ; the general element can be written in the form m + N , where m M . Write m = r 1 x 1 + ··· + r n x n . Then m + N = r 1 ( x 1 + N ) + ··· + r n ( x n + N ) , which shows that M / N is generated by x 1 + N ,..., x n + N .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern