Talk:Quadratic field

Square-free

The equation

 $d\in \mathbb {Q} ^{*}/\mathbb {Q} ^{2}$

and the statement

 That is, we may take d to be a square free integer

don't match up; the equation makes d a non-zero square free rational, that is a rational which is not the square of another rational.

Which one is correct? For example, is $\mathbb {Q} ({\sqrt {2/3}})$ a quadratic field, and is it expressible as $\mathbb {Q} ({\sqrt {d}})$ for d a square-free integer?

Lupin 13:31, 7 Feb 2004 (UTC)

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 p separately, with the exponent then any integer r. What is being said is that rmod 2 can be taken as 0 or 1.

Charles Matthews 13:38, 7 Feb 2004 (UTC)

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.

Lupin 13:59, 7 Feb 2004 (UTC)

The notation was inherited from an earlier version - probably needs further changes for clarity.

Charles Matthews 14:03, 7 Feb 2004 (UTC)

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). Revolver 17:03, 9 Feb 2004 (UTC)

---

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). Revolver 17:01, 9 Feb 2004 (UTC)

"irrational" quadratic "field"?

Are there any results/statemens for irrational quadratic rings? I'm looking for something/anything related to or in the form of

$K=\mathbb {Q} [{\sqrt {\pi n}}]$

Basically, I'm pursuing a relationship analoguous to $\exp(\pi {\sqrt {163}})$ but in my case, I clearly have a square-root of pi sitting next to an integer. linas 18:27, 20 January 2006 (UTC)

If t is transcendental, Q[t] is a polynomial ring. Don't expect anything special. Charles Matthews 21:01, 20 January 2006 (UTC)

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 properties of this set? How do the properties of this set depend on the transcendental t? My apologies, I'm thinking aloud, rather than trying to present a coherent train of thought, excuse me if the quesitons just sound weird/wrong. linas 22:25, 20 January 2006 (UTC)

And its not just a ring, its a field, right? For any x in Q[t], I can always find a y in Q[t] such that xy=1. Whatever. Rhetorical quesiton. linas 22:50, 20 January 2006 (UTC)

Wrong. Charles Matthews 23:23, 20 January 2006 (UTC)

Hmm. OK. How about "for any x in Q[t] s.t. |x|< infty" ? Its not obvious how to construct a counterexample, but maybe it'll hit me while I'm walking to the kitchen... linas 01:42, 21 January 2006 (UTC)

Half-integers

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 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' Is there a simple proof of this surprising fact, or even just a non-rigorous way to see that it is true? I'm finding it very hard to visualise. 91.105.2.149 (talk) 14:04, 13 January 2009 (UTC)

If you consider a general number in the field Q(√d), it will look like

$\alpha ={\frac {a+b{\sqrt {d}}}{c}},$

with a, b, and c all integers and their highest common factor being 1. This will be an algebraic integer if and only if its minimal polynomial is monic, i.e. if and only if

$\left(t-{\frac {a+b{\sqrt {d}}}{c}}\right)\left(t-{\frac {a-b{\sqrt {d}}}{c}}\right)$

has integer coefficients. If you multiply this out and think it through a bit you'll see c has to be 1 or 2, and if it's 2 then d has to satisfy a certain congruence condition, specifically it must be 1 modulo 4. Chenxlee (talk) 12:14, 21 March 2009 (UTC)

"Quadratic field extension" listed at Redirects for discussion

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