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Basic definition of Quadratic Fields

  1. Jan 20, 2012 #1
    I am reading Dummit and Foote Chapter 7.

    D&F use a quadratic field as an example of a ring. I am trying to get a good understanding of this ring.

    D&F define a quadratic field as follows:

    Let D be a rational number that is not a perfect square in and define

    [tex] \mathbb{Q} ( \sqrt D ) = \{ \ a + b \sqrt D \ | \ a,b \in \mathbb{Q} \ \}[/tex]

    as a subset of [itex] \mathbb{C} [/itex]

    In this example D&F write ... "... It is easy to show that the assumption that D is not a square implies that every element of [itex]\mathbb{Q} ( \sqrt D ) [/itex] may be written uniquely in the form [itex] a + b \sqrt D [/itex]."

    How do you show this? Further, I am not sure why this assumption is needed?

    Is it because we have both positive and negative roots of a square number like 4, but then only consider the principal root [itex] + \sqrt 3 [/itex] of 3? This seems slightly inconsistent!

    Also how does the above fit with the idea that D must be not only not a perfect square but squarefree? Is the squarefree condition on D necessary? If so why?

    Can someone please clarify this situation for me?

    PeterMarshall
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  3. Jan 20, 2012 #2

    Deveno

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    it's really pretty simple. if D is a square of a rational number, then Q(√D) = Q.

    if D isn't square-free, say D = p2m, for some prime p, and integer m, then we get the same extension of Q with Q(√m) as we do for Q(√D) (in other words, we want "the totally irrational part" of the square root, with any rational factors (hence squares) already factored out).

    for example, 12 isn't square free, so adjoining √12 = 2√3, gives us the same field as adjoining √3:

    a+b√12 <=> a + 2b√3
    c+d√3 <=> c + (d/2)√12

    on the other hand, if D IS square-free, we get a UNIQUE quadratic extension (different D's, different extensions).

    you can visualize such a ring (which is actually a field, since 1/√D = √D/D, and one can use "the conjugate trick" to find an inverse for a + b√D, as long as not both a and b are 0) as a lattice of rational points in the plane, that is, as a vector space it's isomorphic to Q x Q.
     
  4. Jan 20, 2012 #3
    Thanks!

    Thinking through this now!
     
  5. Jan 20, 2012 #4
    Thanks again

    I can see that if D is not a squarefree number then the field specified by [itex] \mathbb{Q} ( \sqrt D ) [/itex] is not unique in the sense that [itex] \mathbb{Q} ( \sqrt 12 ) [/itex] is the same field as[itex] \mathbb{Q} ( \sqrt 3 ) [/itex].

    But D&F's statement seemingly refers to consequences within the field if D is not squarefree.

    They write:

    "It is easy to show that the assumption that D is not a square implies that every element of [itex] \mathbb{Q} ( \sqrt D ) [/itex] may be written uniquely in the form [itex] a + b \sqrt D [/itex]."

    So they are saying that there is a lack of uniqueness within the field.

    Do you agree? Can you clarify?

    Another thing that bothers me in D&F's statement is that they do not use the term "squarefree" but go for a lesser condition that D is not a square. Can you clarify this also?
     
  6. Jan 20, 2012 #5

    Deveno

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    if D is not a square, then Q(√D) is bigger than D. argue by contradiction:

    suppose Q(√D) = Q. then √D is in Q, hence we have some rational number m (= √D) with m2 = D, so D is a square.

    the uniqueness of the extension is what D being square-free entails. even if D is NOT square-free, there is still only one way to write an element of Q(√D):

    suppose a+b√D = c+d√D, where √D is not in Q.

    then a-c = (d-b)√D.

    case 1) d≠ b:

    then (a-c)/(d-b) = √D, and thus √D is in Q, contradiction.

    case 2) d = b:

    then a-c = 0, so that a = c, which shows uniqueness.
     
  7. Jan 20, 2012 #6
    Thanks.

    OK so the sqarefree part is only to get a unique description of the field

    Yes, follow you arguments regarding [itex] Q \sqrt D [/itex] being bigger that Q.

    I now regard D&F's statement - that got me going on this - as rather misleading!
     
  8. Jan 20, 2012 #7

    morphism

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    I fail to see what is misleading.

    If D is not a square then [itex]\sqrt D[/itex] is irrational. So if [itex]a+b\sqrt D = a'+b' \sqrt D[/itex] with [itex]a,b,a',b' \in \mathbb Q[/itex], then from [itex]a-a' = (b'-b) \sqrt D[/itex] one easily sees that b'=b and hence a'=a (since otherwise [itex]\sqrt D = (a-a')/(b'-b)[/itex] would be rational). It follows that an element x of [itex]\mathbb Q(\sqrt D)[/itex] has a unique expression of the form [itex]x=a+b\sqrt D[/itex] where [itex]a,b \in \mathbb Q[/itex], in the sense that a and b are uniquely determined by x.

    If D is a square, say D=E^2 with E rational, then it's easy to see that this fails. For example, we have [itex]a+b\sqrt D = (a-|E|)+(b+1)\sqrt D[/itex] for any [itex]a,b \in \mathbb Q[/itex].
     
  9. Jan 21, 2012 #8
    Thanks for that help and guidance

    Your last two sentences clarified the situation for me!
     
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