Results 1 to 3 of 3

Math Help - Rings

  1. #1
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132

    Rings

    1. R = \{ a +bi : a,b \in Z \}

    is a subring of a ring C (complex)

    Find an element of R different from the identity element which has a multiplicative inverse in R.

    2. Let R,S be rings and suppose f:R \rightarrow S is a ring homomorphism. Suppose im(f) = S

    (a) Show that if R has an identity element then S has an identity element.
    (b) Show that if R is commutative then S is commutative
    (C) Is it true that if R is an integral domain then S is an integral domain? Justify your assertion.
    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    I'm not 100% sure on some of this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,908
    Thanks
    1759
    Awards
    1
    For #1. Is it true that \left\{ {i, - i} \right\} \subseteq R?

    For #2b. If \left\{ {p,q} \right\} \subseteq S then because the mapping is onto we get \left( {\exists r_p } \right)\left( {\exists r_q } \right)\left[ {f(r_p ) = p\,\& \,f(r_q ) = q} \right].
    This means because R is commutative, \left( {r_p  * r_q } \right) = \left( {r_q  * r_p } \right). Which means p \cdot q = f\left( {r_p } \right) \cdot f\left( {r_q } \right) = f\left( {r_p  * r_q } \right) = f\left( {r_q  * r_p } \right) = f\left( {r_q } \right) \cdot f\left( {r_p } \right) = q \cdot p
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Jason Bourne View Post
    2. Let R,S be rings and suppose f:R \rightarrow S is a ring homomorphism. Suppose im(f) = S
    (a) Show that if R has an identity element then S has an identity element.
    Suppose that there exists an element 1 such that 1x=x1=x for all x\in R. Then we claim that f(1) is the identity element for S. We need to show sf(1)=f(1)s=s for all s\in S. Since the function is onto it means s=f(r). That means sf(1) = f(r)f(1)=f(r1)=f(r)=s and similarly f(1)s=s.

    (C) Is it true that if R is an integral domain then S is an integral domain? Justify your assertion.
    No, it is not true. Let R = \mathbb{Z} and S = \mathbb{Z}_4. For x\in R let f(x) = [x]_4 to be the congruence class of x mod 4. Then f is a ring homomorphism and \text{Im}(f) = S but R is an integral domain and S is not.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ring theory, graded rings and noetherian rings
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: January 4th 2012, 12:46 PM
  2. a book on semigroup rings and group rings
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: October 2nd 2011, 05:35 AM
  3. Rings.
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 24th 2010, 12:54 AM
  4. rings
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 21st 2010, 08:33 AM
  5. Rings with 0 = 1.
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: November 26th 2009, 06:34 AM

Search Tags


/mathhelpforum @mathhelpforum