HW5 - Lenny T. Evans MATH 534 Homework 5 March 18, 2010...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Lenny T. Evans MATH 534 Homework 5 March 18, 2010 3.2.1 A field has no ideal other than the zero and unit ideals. Proof: The zero ideal is obviously an ideal in the field. Let I be an ideal in the field that is not the zero ideal. For any nonzero b I (which exists because we assumed it is not the zero ideal) there exists an element in the field b- 1 , by the definition of fields. Since ideals are closed under multiplication (fields are commutative under multiplication so left multiplication is equivalent to right multiplication) by members of the field, this means that bb- 1 = e or b- 1 b = e must be in the ideal. Then, to be closed under multiplication by any element in the field, a, ae = a or ea = a must also be in the ideal. Thus, a nonzero ideal contains all elements of the field, and is thus the field itself. So, the only possible ideals are the zero and unit ideals. 3.2.2 If R is a commutative ring, and M 1 and M 2 are two ideals such that M 1 + M 2 = R, then M 2 1 + M 2 2 = R. Proof: Since M 1 + M 2 = R, this means that for every element r R, there exists m M 1 and M 2 such that r = m + . Notably, since R is a ring, it contains the multiplicative identity, therefore, there exists m e M 1 and e M 2 such that e = m e + e . Since e is a multiplicative identity, we know that r = eer. Exploiting the sum rules given above for r and e as well as the the distributivity relation of rings, we get r = ( m e + e )( m e + e )( m + ) = ( m e + e )( m e m + m e + e m + e ) = m 2 e m + m 2 e + m e e m + m e e + e m e m + e m e + 2 e m + 2 e . Exploiting the fact that rings are associative and this ring is commutative, we can rearrange the terms as r = m e ( m e m ) + m e ( m e ) + m e ( e m ) + e ( m e ) + m e ( e m ) + e ( m e ) + e ( m e ) + e ( e ) . Since M 1 and M 2 are ideals, they are by definition closed by multiplication by any member of the ring. Thus, any element in M 1 or M 2 multiplying an element in M 1 is an element in M 1 and likewise, any element in M 1 or M 2 multiplying an element in M 2 is an element in M 2 . Thus, we see r = m e m 1 + m e m 2 + m e m 3 + m e m 3 + e 1 + e 1 + e 2 + e 3 , where m i M 1 correspond to one of or e multiplying m, or m e in the equation above, and likewise i M 2 correspond to one of m or m e multiplying or e in the equation above. The first four terms in this expansion of r is the sum of products of two elements from M 1 , so m e m 1 + m e m 2 + m e m 3 + m e m 3 M 2 1 . Likewise, the last four terms is the sum of products of two elements from M 2 , so e 1 + e 1 + e 2 + e 3 M 2 2 ....
View Full Document

This note was uploaded on 03/13/2011 for the course MATH 534 taught by Professor Kumar during the Spring '11 term at UNC.

Page1 / 5

HW5 - Lenny T. Evans MATH 534 Homework 5 March 18, 2010...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online