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

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.

Solution 1:

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.