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!

Is there an injective ring homomorphism?

  1. Mar 22, 2017 #1
    1. The problem statement, all variables and given/known data
    From ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##

    2. Relevant equations


    3. The attempt at a solution
    I know how to do this if we assumed that the rings had to be unital. In that case, there can be no non-trivial homomorphism. However, in my book rings don't need unity, and so a homomorphism is defined only such that the operations are preserved. In this case, how do I show that there is no injective homomorphism?
     
    Last edited: Mar 23, 2017
  2. jcsd
  3. Mar 23, 2017 #2
    An image of a homomorphism is always a sub-structure. In this case, if [itex]f:\mathbb{Z}_3\to\mathbb{Z}_{15}[/itex] was a ring homomorphism, then [itex]f(\mathbb{Z}_3)[/itex] is a sub-ring. I can certainly think of an injective ring homomorphism [itex]\mathbb{Z}_3\to\mathbb{Z}_{15}[/itex], though. Would you believe if I wrote [itex]\mathbb{Z}_{15}\cong \mathbb{Z}_3\times\mathbb{Z}_5[/itex]?
     
    Last edited: Mar 23, 2017
  4. Mar 23, 2017 #3
    I think that ##\mu## such that ##\mu (0) = 0##, ##\mu (1) = 5##, and ##\mu (2) = 10## is an injective homomorphism. However, I found this by trial-and-error, and am not completely sure of a systematic way to find the possible homomorphisms from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##
     
  5. Mar 23, 2017 #4
    Riddled with mistakes
     
    Last edited: Mar 23, 2017
  6. Mar 23, 2017 #5
    So basically you noted that ##\mathbb{Z}_{15}\cong \mathbb{Z}_3\times \mathbb{Z}_5##, and used this to your advantage, because ##f(k) = (k,0)## is an obvious injective homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_3\times \mathbb{Z}_5##, and then found the corresponding elements. However, while ##f(k) = (k,0)## is an obvious homomorphism, how do you know that there aren't others?

    For example, wouldn't ##\varphi## such that ##\varphi (0) = 0##, ##\varphi (1) = 10##, and ##\varphi (2) = 5## also be an injective homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##?
     
  7. Mar 23, 2017 #6
    Riddled with mistakes
     
    Last edited: Mar 23, 2017
  8. Mar 23, 2017 #7
    Could you address the first part of my last question? That is, why is ##f(k) = (k,0)## the only homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_3 \times \mathbb{Z}_5##?
     
  9. Mar 23, 2017 #8
    My argument is wrong. [itex]f[/itex] need not be unique. The image of the injective homomorphism (aka monomorphism) is unique, but it doesn't imply the mapping is unique. You proposed [itex]g(0)=0, g(1) = (2,0), g(2) = (1,0)[/itex] and this is also a monomorphism.

    Pardon the confusion.
     
    Last edited: Mar 23, 2017
  10. Mar 23, 2017 #9
    Well in that case we have two homomorphisms. But how do we know that ##f(k) = (k,0)##, where k = 1,2,3, and the ##g## that you describe are the only two such homomorphisms? How do you knowfor sure that other's can't be constructed?
     
  11. Mar 23, 2017 #10
    Riddled with mistakes
     
    Last edited: Mar 23, 2017
  12. Mar 23, 2017 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Actually, defining the map ##\varphi## such that ##\varphi (0) = 0##, ##\varphi (1) = 5##, ##\varphi (2) = 10##, doesn't give a ring homomorphism. ##\varphi (1*1) = \varphi (1) \varphi (1)## doesn't work.
     
  13. Mar 23, 2017 #12
    that is not even well-defined :/
     
  14. Mar 23, 2017 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sorry, typo. I corrected it.
     
  15. Mar 23, 2017 #14
    We are using the definition of the ring such that there isn't necessarily a multiplicative inverse, and hence the definition of homomorphism doesn't include the condition that ##\mu (1_A) = 1_B##, where ##\mu : A \rightarrow B##
     
  16. Mar 23, 2017 #15
    Nono, Dick is right.
    ##f(1\cdot 1) = f(1)f(1) ## must hold, regardless. But it doesn't.
    ##5 = f(1) = f(1\cdot 1) \neq f(1)f(1) = 10 ##

    What an oversight. It turns out you cannot define ##f## this way. The earlier work is not all for naught, you would still have to determine the only possible 3 element subring to be the image of your monomorphism.
     
  17. Mar 23, 2017 #16
    So does this just mean that there is no homomorphism from ##\mathbb{Z}_3## to ##\mathbb{Z}_{15}##? Very confused
     
  18. Mar 23, 2017 #17
    Oh my lord. This is all my fault, I made a huge mistake somewhere in my argumentation. I will attempt to make amends.

    1. Image of ring homomorphism is a subring. Since ##Z_3 ## contains three elements, we need a three element subring of ##Z_{15} ##.
    2. Because ##(Z_{15},+) ## is cyclic, its three element sub-abelian group must also be cyclic. Since the subgroup contains three elements, it must contain an element of order ##3## modulo ##15##. The only candidate is ##5## and it generates the subgroup ##\{0,5,10\}##. Is it a subring? Yes, it is, verify it.

    3. Since a ring homomorphism preserves zero element there are only two candidate monomorphisms
    ##f(0)=0, f(1)=5, f(2)= 10 ## OR ##g(0)=0, g(1)=10, g(2) =5 ##. Will either of them work?
     
  19. Mar 23, 2017 #18
    I don't see why it's not a subring. It has the zero element, the difference between any two elements is in the set, and the multiplication of any two elements is in the set.

    So will neither of them work for the reason that Dick gave?
     
  20. Mar 23, 2017 #19
    I am so sorry Mr Davis 97, it's such an obvious mistake, too :/

    Ok, back to business. The method I described earlier in ##Z_3\times Z_5 ## yields a monomorphism and there does, indeed, exist exactly one such monomorphism. You can verify that the ##g ## in my above post is a monomorphism.

    you can verify it by checking all combinations, but that can be counter-productive when you are dealing with more complicated mappings. Let us try to do it algebraically instead. Note that ##f(x) = 10x ## (this you can verify)
    Let ##x,y\in Z_3 ## Then

    ##f(x+y) = 10(x+y) = 10x + 10y = f(x)+f(y) ## and ##f(xy) = 10xy = 10\cdot 10 xy = 10x \cdot 10y = f(x)f(y) ##
    where I am using the fact that ##10=10\cdot 10 ## in ##Z_{15} ##.

    By the same method you can also see why Dick's answer is right. If you defined ##f(0)=0, f(1)=5, f(2)=10 ## i.e ##f(x) = 5x ## then ##f(xy)=5xy\neq 5x 5y ##.
     
    Last edited: Mar 23, 2017
  21. Mar 23, 2017 #20

    fresh_42

    Staff: Mentor

    You can simply analyze ##f(1)##. Since ##1\cdot 1= 1## you get ##f(1)\cdot (f(1)-1)=0## and ##f(1) \in \{0,3x,5y\}##.
    Now rule out every of these cases.
     
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: Is there an injective ring homomorphism?
  1. Ring homomorphisms (Replies: 3)

  2. Ring homomorphism (Replies: 7)

  3. Ring homomorphism (Replies: 3)

  4. Ring homomorphism (Replies: 10)

  5. Ring homomorphism (Replies: 1)

Loading...