3- For any R-module ,let M [x]denote the set of all polynomials in x with coefficient in M, that is tosay expression of the form m.+mix+...+m,x.(mi eM). Defining the product of an element of R[x]and an element of M[x] in the obvious way, show that M[x] is an R[x]-module. Show that M[x] =R[x] ®R M.

Answered: 3- For any R-module ,let M [x]denote…

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter5: Rings, Integral Domains, And Fields

Section5.2: Integral Domains And Fields

Problem 23E: [Type here] 23. Let be a Boolean ring with unity. Prove that every element ofexceptandis a zero...

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Since this section presents a method for constructing a field of quotients for an arbitrary integral domain D, we might ask what happens if D is already a field. As an example, consider the situation when D=5. a. With D=5, write out all the elements of S, sort these elements according to the relation , and then list all the distinct elements of Q. b. Exhibit an isomorphism from D to Q.
Let R be a ring, and let x,y, and z be arbitrary elements of R. Complete the proof of Theorem 5.11 by proving the following statements. a. x(y)=(xy) b. (x)(y)=xy c. x(yz)=xyxz d. (xy)z=xzyz Theorem 5.11 Additive Inverses and Products For arbitrary x,y, and z in a ring R, the following equalities hold: (x)y=(xy) b. x(y)=(xy) (x)(y)=xy d. x(yz)=xyxz (xy)z=xzyz
Let G=1,i,1,i under multiplication, and let G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. Find an isomorphism from G to G that is different from the one given in Example 5 of this section. Example 5 Consider G=1,i,1,i under multiplication and G=4=[ 0 ],[ 1 ],[ 2 ],[ 3 ] under addition. In order to define a mapping :G4 that is an isomorphism, one requirement is that must map the identity element 1 of G to the identity element [ 0 ] of 4 (part a of Theorem 3.30). Thus (1)=[ 0 ]. Another requirement is that inverses must map onto inverses (part b of Theorem 3.30). That is, if we take (i)=[ 1 ] then (i1)=((i))1=[ 1 ] Or (i)=[ 3 ] The remaining elements 1 in G and [ 2 ] in 4 are their own inverses, so we take (1)=[ 2 ]. Thus the mapping :G4 defined by (1)=[ 0 ], (i)=[ 1 ], (1)=[ 2 ], (i)=[ 3 ]
Prove that every ideal of n is a principal ideal. (Hint: See corollary 3.27.)

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