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A≡b mod n true in ring of algebraic integers => true in ring of integers

  1. Apr 10, 2012 #1
    "a≡b mod n" true in ring of algebraic integers => true in ring of integers

    Hello,

    So I'm learning about number theory and somewhere it says that if [itex]a\equiv b \mod n[/itex] is true in [itex]\Omega[/itex], being the ring of the algebraic integers, then the modular equivalence (is that the right terminology?) it also true in [itex]\mathbb Z[/itex], at least if [itex]a,b,m \in \mathbb Z[/itex].

    Fair enough, but as a prove it states: "This is a direct consequence of theorem blabla, because [itex]\Omega[/itex] is a ring."

    Now theorem blabla, as I called it so eloquently, is basically the statement that if x is an algebraic integer and a rational, that it is also an integer.

    But isn't theorem blabla all we need then? Why is there "because [itex]\Omega[/itex] is a ring"? I don't see the relevance of [itex]\Omega[/itex] being a ring. After all, if [itex]a\equiv b \mod n[/itex] is true in [itex]\Omega[/itex], it means that [itex]\frac{a-b}{n} \in \Omega[/itex], and since a,b,n are integers, this fraction is a rational number and hence by theorem blabla it's an integer. Done(?)
     
  2. jcsd
  3. Apr 10, 2012 #2
    Re: "a≡b mod n" true in ring of algebraic integers => true in ring of integers


    We have that [itex]a\equiv b\mod n[/itex] actually means [itex]a-b=rn\,,\,r\in\Omega[/itex] , as the equivalence is in [itex]\Omega[/itex] , and from

    here, SINCE WE'RE IN A RING. we can conclude that [itex]\,\,\frac{a-b}{n}=r\in\Omega[/itex] , and since the LHS is a rational and

    the RHS is an alg. integer, we deduce that in fact [itex]\,\,r\in\mathbb Z[/itex] .

    DonAntonio
     
  4. Apr 10, 2012 #3
    Re: "a≡b mod n" true in ring of algebraic integers => true in ring of integers

    Thank you for your reply, but I'm not yet getting your point.

    Can you explain more about how you go from a-b=rn to [itex]\frac{a-b}{n}=r[/itex] using Omega is a ring?

    I would think (and I realize I can be wrong) that given a-b=rn, [itex]\frac{a-b}{n}[/itex] is by definition equal to r. That being said, it needs to be shown that this definition is consistent, and I suppose for that one needs to know that if a-b=sn, that r = s. In other words Omega needs to be a ring without zero divisor. This is of course true for Omega, but the argument simply mentions "because [itex]\Omega[/itex] is a ring" and not "because [itex]\Omega[/itex] is a ring without zero divisors".

    So it seems that due your post I've now gone from "I don't see how being a ring is relevant" to "I don't see how being a ring is enough" haha.
     
  5. Apr 10, 2012 #4
    Re: "a≡b mod n" true in ring of algebraic integers => true in ring of integers

    ....
     
  6. Apr 10, 2012 #5
    Re: "a≡b mod n" true in ring of algebraic integers => true in ring of integers

    Okay, but then it seems like your first post doesn't make sense: you said (shortening) "we have [itex]a-b = rn, r \in \Omega[/itex] and from here, SINCE WE'RE IN A RING, we conclude that [itex]\frac{a-b}{n} = r \in \Omega[/itex]", yet in your second post you state there is no difference between writing these two expressions, so why did you write when going from the former to the latter expression "since we're in a ring", even in capital letters?
     
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