Results 1 to 3 of 3

Math Help - Ideal and homomorphism

  1. #1
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    469
    Thanks
    2
    Awards
    1

    Ideal and homomorphism

    1. Let f: R --> S be a ring homomorphism
    Let J be an ideal of S. Prove that  f^{-1}(J) is an ideal
    of R.
    I can see how to show this is true for an isomorphism but not for an homomorphism.
    I think there may be an error in the question. If the homomorphism is not bijective (thus is not an isomorphism) than it is possible that  f^{-1}(J) doesn't exist. In this case, the statement would not be true. Am I right?

    2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.
    Let  J = \{ j = f(i) : i \in I\}

    We know  0 \in I\ . Therefore, because f is an homomorphism,  f(0)=0 ,  0 \in I and  0 \in J

    We have  a,b \in I \implies a+b \in I and because f is an homomorphism  f(a+b) = f(a) + f(b) thus  f(a)+f(b) \in J

    We have  a \in I, r\in R \implies ar \in I and because f is an homomorphism  f(ar) = f(a)f(r) and that gives  f(a)f(r) \in J and conversely s  f(r)f(a) \in J

    Ok I miss some steps because it is tidious to write but you understand what I did. I don't understand where the fact that f is surjective intervene. Can you tell me where and why please?
    Last edited by vincisonfire; November 9th 2008 at 01:29 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by vincisonfire View Post
    1. Let f: R --> S be a ring homomorphism
    Let J be an ideal of S. Prove that  f^{-1}(J) is an ideal
    of R.
    I can see how to show this is true for an isomorphism but not for an homomorphism.
    I think there may be an error in the question. If the homomorphism is not bijective (thus is not an isomorphism) than it is possible that  f^{-1}(J) doesn't exist. In this case, the statement would not be true. Am I right?
    there's nothing wrong with the question. by definition: f^{-1}(J)=\{a \in R: \ f(a) \in J \}, which is defined for any function f. to show that f^{-1}(J) is an ideal of R, suppose that a,b \in f^{-1}(J) and

    r \in R. so f(a) \in J, \ f(b) \in J. thus: f(a+b)=f(a)+f(b) \in J, and: f(ra)=f(r)f(a) \in J, because J is an ideal of S. thus a+b \in f^{-1}(J) and ra \in f^{-1}(J).


    2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.
    Let  J = \{ j = f(i) : i \in I\}

    We know  0 \in I\ . Therefore, because f is an homomorphism,  f(0)=0 ,  0 \in I and  0 \in J

    We have  a,b \in I \implies a+b \in I and because f is an homomorphism  f(a+b) = f(a) + f(b) thus  f(a)+f(b) \in J

    We have  a \in I, r\in R \implies ar \in I and because f is an homomorphism  f(ar) = f(a)f(r) and that gives  f(a)f(r) \in J and conversely s  f(r)f(a) \in J

    Ok I miss some steps because it is tidious to write but you understand what I did. I don't understand where the fact that f is surjective intervene. Can you tell me where and why please?
    surjectivity of f is needed to prove that for any s \in S and a \in I we have: sf(a) \in f(I): since f is surjective, there exists r \in R such that s=f(r).

    since I is an ideal of R, we have: ra \in I. thus: sf(a)=f(r)f(a)=f(ra) \in f(I).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by vincisonfire View Post
    2. Prove that if f is surjective and I is an ideal of R then f(I) is an ideal.
    Let  J = \{ j = f(i) : i \in I\}
    Try this excercise. Show that the homomorphic image of an ideal is an ideal of the homomorphic image of the ring, i.e. in less fancy language, show that f(I) is an ideal of f(R).

    Now, if f is surjective then f(R) = R, from here it follows that f(I)\triangleleft R
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. ideal,nil,nilpotent ideal in prime ring
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 24th 2011, 07:57 AM
  2. prove N is a maximal ideal iff N is a prime ideal
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 20th 2011, 09:02 AM
  3. Ideal a is irreducible <--> a=p^n, p is prime ideal
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: July 3rd 2010, 10:54 PM
  4. Prime ideal or maximal ideal
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: January 21st 2010, 05:42 AM
  5. Prime ideal but not maximal ideal
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: November 14th 2007, 09:50 AM

Search Tags


/mathhelpforum @mathhelpforum