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 rings and unity element proof

  1. Apr 21, 2014 #1
    So this is a review problem in our book I came across and i really want to understand it but I am just not having any luck, I did some research and found a guide on solving it but that's not really helping either. We didn't talk about unity elements in class and there aren't any examples in our book just this problem. Would anyone be willing to take the time to fully explain this to me and show me how to prove it?

    Question:
    You may assume that
    E=even integers is a Commutative Ring.
    Prove that E does NOT have a unity element

    Information I found online: It suffices to show that there is at least a single element, n, of E , for which no element of E acts as a unity element for that specific n.

    So specifically, consider the case n = 2 and let m = any element of
    E. Then m MUST have exactly 1 of 3 possible properties:

    so let CASE 1 cover the 1st possible property of m and show that m cannot be a “unity” for 2

    and let CASE 2 cover the 2nd possible property of m and show that m cannot be a “unity” for 2

    and let CASE 3 cover the 3rd possible property of m and show that m cannot be a “unity” for 2

    Conclude that if there is no unity element ‘e’ for 2 such that 2*e=e*2=2

    then, of course, there is no unity element for all of
    E …

    E does NOT have a unity element
     
  2. jcsd
  3. Apr 21, 2014 #2

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I don't understand your cases. What are these three properties of ##m## that you are mentioning?

    I think you are making this harder than it has to be. A unity element ##e## must satisfy ##er= r## for every ##r\in R##. As you noted, it suffices to show that ##er=r## is impossible when ##r=2##. Since any candidate ##e## is an even integer, you can write it as ##e=2k## for some integer ##k##. What can you conclude from ##er=r##?
     
  4. Apr 21, 2014 #3
    the cases were a guide I found to solving the same problem . i'm not entirely sure what they were talking about but it was about the exact same problem so i thought they might be important. What you are saying makes sense tho. I'm not sure where you are going with your question, are you saying I should consider 2e=2 ?
     
  5. Apr 21, 2014 #4

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Just substitute ##e=2k## and ##r=2## into ##er=r##, and see if you can find a contradiction.
     
  6. Apr 21, 2014 #5

    Zondrina

    User Avatar
    Homework Helper

    You should look for a contradiction somewhere. Use jbun's hint.
     
  7. Apr 21, 2014 #6
    i think I got it!

    Suppose;

    1Er=r
    ∀ r∈E. 1E2=2. since 1E can be written as 2k for k∈Z we have,


    (2k)(2)=2
    this implies that 2l=1, which is false for ∀l∈Z. Contradiction

    Therefore, E does not have a unity element

    Does this look sound?
     
  8. Apr 21, 2014 #7

    jbunniii

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, it looks fine. A slightly different way to say it is that ##(2k)(2) = 4k = 2## implies that ##2## is a multiple of ##4##, which is obviously false.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Commutative rings and unity element proof
  1. Rings with unity (Replies: 11)

  2. Commutative rings (Replies: 4)

Loading...