Field of fractions of $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$

Question

This problem goes as follows:

Prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$ is an integral domain and that its field of fractions is isomorphic to the ring of rational functions $\mathbb{Q}(t)$.

The first part is straightforward, $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$, and hence a prime ideal, as $\mathbb[Q][x,y]$ is a euclidean and hence a UFD. The quotient by a prime ideal is an integral domain.

How do I do the second part? I guess, I have to set a isomorphism by making a substitution such that only one variable remains. But I don't see how $x^2+y^2-1$ is going to help me make a choice.

${\bf Q}[x,y]$ is a UFD, but it isn't Euclidean. It's not even a PID. Think about the ideal generated by $x$ and $y$. — Gerry Myerson, Apr 26 '13 at 13:44
I have a feeling the map $x\mapsto2t/(1+t^2)$, $y\mapsto(1-t^2)/(1+t^2)$ is going to be involved somehow. — Gerry Myerson, Apr 26 '13 at 13:51
In any integral domain, prime elements generate prime ideals. — Martin Brandenburg, Apr 26 '13 at 13:54

Martin Brandenburg · Accepted Answer · 2013-04-26T14:27:48.003

6

This follows from the classification of Pythagorean triples.

A little bit more detailed: The rational solutions of $x^2+y^2=1$ are parametrized $x=\frac{2t}{1+t^2}$ and $y=\frac{1-t^2}{1+t^2}$. But actually the equation $\left(\frac{2t}{1+t^2}\right)^2 + \left(\frac{1-t^2}{1+t^2}\right)^2=1$ holds formally in $\mathbb{Q}(t)$. Thus, there is a homomorphism $R=\mathbb{Q}[x,y]/(x^2+y^2-1) \to \mathbb{Q}(t)$ given by $x \mapsto \frac{2t}{1+t^2}$ and $y \mapsto \frac{1-t^2}{1+t^2}$. Verify that it is injective, hence extends to a homomorphism $\mathrm{Quot}(R) \to \mathbb{Q}(t)$. Check that $t$ lies in the image. Hence, this homomorphism is an isomorphism.

edited Apr 26 '13 at 14:27

answered Apr 26 '13 at 13:52

Martin Brandenburg

163,620

3

This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. – Alexander Gruber Apr 26 '13 at 14:12
1

It does lead to an answer of the question when you think a little bit about it. But of course math.SE is not suited for answers which only give the main idea ... everything should be worked out, sigh – Martin Brandenburg Apr 26 '13 at 14:16
2

@MartinBrandenburg: To me, there is a very large gap between providing a full proof and a cryptic (to those who don't already know the link) remark about Pythagorean triples. Doesn't your original hint basically amount to "Go look up how Pythagorean triples are parametrized and then use that?" – Cam McLeman Apr 26 '13 at 14:51
See also http://en.wikipedia.org/wiki/Weierstrass_substitution#The_substitution – Andreas Caranti Apr 26 '13 at 15:05
@Andreas We think too much alike. I was adding a remark on such at the same time here. See esp. the tangent half-angle formula article – Math Gems Apr 26 '13 at 15:29
@MathGems, yours is a better reference, I was hurrying and couldn't find it myself. – Andreas Caranti Apr 26 '13 at 15:51
2

So where's the problem with a hint?! – Martin Brandenburg Apr 26 '13 at 16:04
The part I missed when I commented on the original question was how to show that $t$ is in the image. Now that I see how to do that, I don't know how I missed it! – Gerry Myerson Apr 27 '13 at 04:33
@GerryMyerson, how to see $t$ is in the image ? – MAS May 04 '23 at 14:19
1

@MAS, $(1-y)/x=t$. – Gerry Myerson May 04 '23 at 23:28

Field of fractions of $\mathbb{Q}[x,y]/\langle x^2+y^2-1\rangle$

1 Answers1

Linked

Related