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Showing that a ring is an integral domain

  1. Apr 12, 2017 #1
    1. The problem statement, all variables and given/known data
    Show that ##\mathbb{Z} [\sqrt{d}] = \{a+b \sqrt{d} \ | \ a,b,d \in \mathbb{Z} \}## is an integral domain

    2. Relevant equations


    3. The attempt at a solution
    Do I have to go through all of the axioms to do this? For example, do I have to show that it is an abelian group under addition, that multiplication is associative, and that the distributive property holds, on top of showing that multiplication is commutative and that there is a multiplicative identity and that there are no zero divisors?
    This would seem like a lot of unnecessary, but easy, work, so I was wondering if there is a faster way.
     
  2. jcsd
  3. Apr 13, 2017 #2

    fresh_42

    Staff: Mentor

    Well, most of these attributes are quite obvious by the arithmetic rules in this ring. The essential part here is to see, that you don't get zero divisors from such extensions. If you want to go the long way, then you have to do it, but only for terms with ##\sqrt{d}## in it, the rest is inherited from ##\mathbb{Z}##. Since ##d \in \mathbb{Z}##, the shortest answer is probably ##\mathbb{Z}[\sqrt{d}] \subseteq \mathbb{R}##. But it makes sense to convince yourself that ##\mathbb{Z}[\sqrt{d}]## hasn't zero divisors, which is short enough to do and shows which properties of ##\mathbb{Z}## are actually used, in case a ring isn't the integers.
     
  4. Apr 13, 2017 #3
    If I am trying to show that ##\mathbb{Z}[\sqrt{d}]## has no divisors, do I proceed by a contradiction argument, such as, assume that ##(a+b \sqrt{d})(c + e \sqrt{d}) = 0##, and show that ##a=b=c=e=0##?
     
  5. Apr 13, 2017 #4

    fresh_42

    Staff: Mentor

    Yes. And if you do it step by step, you see the properties of the integers which are needed, e.g. it isn't true for ##\mathbb{Z}_6[\sqrt{d}]##.
     
  6. Apr 13, 2017 #5
    Well if I expand it out and compare the two sides I get the two equations ac + bed = 0 and ae + bc = 0, but I don't see how this shows that a = b = c = e = 0
     
  7. Apr 13, 2017 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    This integral domain is actually a field. You should be able to show every non-zero element has a multiplicative inverse. That would show that it's an integral domain.
     
  8. Apr 13, 2017 #7
    I don't see how every element has an inverse. For example, what is the inverse of ##2+\sqrt{2}##? It can't be ##1-\frac{1}{2} \sqrt{2}##, because ##1/2## is not an integer.
     
  9. Apr 13, 2017 #8

    Dick

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    Science Advisor
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    Oh right, sorry. I forgot you were working over the integers, not the rationals. Nevertheless, if there are no zero divisors over the rationals, there won't be over the integers. Correct?
     
  10. Apr 13, 2017 #9
    So how can I show that there are no zero divisors?
     
  11. Apr 13, 2017 #10

    fresh_42

    Staff: Mentor

    You have ##ac + bed = 0## and ##ae + bc = 0##. What does this mean for ##ace##?
     
  12. Apr 13, 2017 #11

    Dick

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    Science Advisor
    Homework Helper

    Pick an arbitrary nonzero element of the ring and write down its inverse. Say why it's always well-defined.
     
    Last edited: Apr 13, 2017
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