1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Commutative Ring question

  1. Dec 2, 2005 #1
    I have this question, just curious if what I did is correct.
    ----
    Let A and R be rings, and let f:A->R be a function satisfying:

    f(a+b) = f(a)+f(b) and f(ab) = f(a)f(b).

    Prove that if f is surjective, then f(1) = 1.
    -----

    Note: I will use * for multiplication for clarity where things would have just otherwise been next to each other.
    Also, note that A and R are commutative rings, both of which have a multiplicative identity 1.




    First, R is a ring, so 1 is in R. Also, because this function is surjective, there must be some a in A, that satisfies f(a) = 1. So, we can rewrite f(a) = 1 as: 1 = f(a*1) = f(a)f(1) = 1*f(1) [because f(a) = 1] = f(1) = 1.

    That looks sufficient to me, but I don't like it when I don't use something I could have (ie, the + part of the function). Does this look correct? Thanks.
     
  2. jcsd
  3. Dec 2, 2005 #2

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Yes, but you have 1 = .... = f(1) = 1. You shouldn't have that underlined part.
     
  4. Dec 2, 2005 #3
    Thanks. You are right, I cannot say that f(1) = 1 in that sense.
     
  5. Dec 2, 2005 #4

    AKG

    User Avatar
    Science Advisor
    Homework Helper

    Yes, but of course you don't need to. You have 1 = ... = f(1), and that's all you needed.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Commutative Ring question
  1. Commutative rings (Replies: 4)

Loading...