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R/kZ under multiplication

  1. Nov 25, 2013 #1
    I'm trying to figure out if ##\mathbb{R}/k\mathbb{Z}## would form a field if we allowed for an "induced" multiplication from ##\mathbb{R}##, with ##r\in\mathbb{R}/k\mathbb{Z}## identified with ##r+k\in\mathbb{R}/k\mathbb{Z}##. So, if we wanted to multiply ##3\in\mathbb{R}/4\mathbb{Z}## by ##\frac{1}{2}\in\mathbb{R}/4\mathbb{Z}##, we'd get 3/2, but if we wanted to multiply ##3\in\mathbb{R}/4\mathbb{Z}## and ##2\in\mathbb{R}/4\mathbb{Z}##, we'd get 6-4=2. I'm not convinced that this is a field, despite what my professor told me in a really off-topic discussion from polar coordinates.

    Clearly, it forms an abelian group under addition. It also has a commutative, associative, and distributive multiplication law with an identity. Now I just need to show that it has an inverse for every nonzero element. I tried to prove this by proving that there isn't a nontrivial proper ideal in ##\mathbb{R}/k\mathbb{Z}##, but that isn't getting me anywhere.

    Rather than help me with just this specific problem, though, I'd also like to know if there is a general algorithm/procedure/etc. for showing that a commutative ring is a field.

    Any help is much appreciated. Thank you.
     
  2. jcsd
  3. Nov 25, 2013 #2

    jgens

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    Your multiplication has some problems here. Notice that 0+4Z = (0+4Z)(1/2+4Z) = (4+4Z)(1/2+4Z) = 2+4Z using your rule.
     
  4. Nov 26, 2013 #3
    Aha! There we go. Thank you. :biggrin:

    Do you have any idea how I could notice things like that in the future?
     
  5. Nov 26, 2013 #4

    Office_Shredder

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    Anytime you are quotienting out by a set (in this case kZ) the first question should be "are my operations still well-defined?" And until you prove that they are there's no point in proceeding. And by 'well-defined' we always mean what happens if I pick different representatives of the same co-set, so if you tried some examples you would quickly find the problem here.
     
  6. Nov 26, 2013 #5

    jgens

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    Sure. If you have a ring R and a subgroup I then generally R/I inherits multiplication from R only when I is an ideal. So basically whenever you quotient out by something that is not an ideal, you should not expect to end up with a ring.
     
  7. Nov 26, 2013 #6
    I probably should have thought of that. I just wasn't thinking of ##\mathbb{R}## alone in terms of rings. :redface:

    Thank you. You've been very helpful and patient.
     
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