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Hm. Well, yes, some transcendentals are at least a little bit special. For t=sqrt(pi), one does have the 4=t^4 -t^8/12+t^12/360 - ... so that's a monic polynomial, albeit an infinite one, with rational coefficients, that evaluates to an integer. What is the set of all such polynomials? What's the
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The discriminant of the quadratic field Q(√d) is d if d is congruent to 1 modulo 4, and otherwise 4d. For example, when d is −1 so that K is the field of so-called
Gaussian rationals, the discriminant is −4. The reason for this distinction relates to general algebraic number theory. The ring of
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I have several pages of notes and examples about quadratic number fields that I used for my advancement talk. There are many things that are special about the quadratic case (quadratic reciprocity law being just the most obvious). I'll get some of that up here, sometime (when I get the chance).
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I don't see a mismatch, in fact. We might as well take the square root of 6 instead of 2/3 - it's the same field in the end. That is, it is being claimed that rationals up to squares of rationals are represented uniquely by square free integers. Perhaps it helps to look at each prime
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I think it's a pretty short exercise to show that all quadratic extension of Q are of the form Q(sqrt(d)), for square-free d, so maybe this proof could just be spelled out (I think it's only a couple lines).
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integers of K is spanned by 1 and the square root of d only in the second case, and in the first case there are such integers that lie at half the 'lattice points'
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Point taken :-) I've changed Q to Z in the displayed equation, which I feel makes the fact that the next sentence begins with "That is" easier to swallow.
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Are there any results/statemens for irrational quadratic rings? I'm looking for something/anything related to or in the form of
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until a consensus is reached, and readers of this page are welcome to contribute to the discussion.
801:{\displaystyle \left(t-{\frac {a+b{\sqrt {d}}}{c}}\right)\left(t-{\frac {a-b{\sqrt {d}}}{c}}\right)}
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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The notation was inherited from an earlier version - probably needs further changes for clarity.
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a non-zero square free rational, that is a rational which is not the square of another rational.
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has integer coefficients. If you multiply this out and think it through a bit you'll see
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properties of this set? How do the properties of this set depend on the transcendental
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has to satisfy a certain congruence condition, specifically it must be 1 modulo 4.
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Is there a simple proof of this surprising fact, or even just a non-rigorous way to
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Knowledge:Redirects for discussion/Log/2022 February 8#Quadratic field extension
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but in my case, I clearly have a square-root of pi sitting next to an integer.
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all integers and their highest common factor being 1. This will be an
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An editor has identified a potential problem with the redirect
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that it is true? I'm finding it very hard to visualise.
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And its not just a ring, its a field, right? For any
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is a polynomial ring. Don't expect anything special.
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Basically, I'm pursuing a relationship analoguous to
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574:Hmm. OK. How about "for any
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