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Let $A = \mathbb Q [x, y] / (x ^ 2 + y ^ 2 - 1)$ and note that $A$ is a domain. How to show that $\operatorname{Quot} (A)$ (or $\operatorname{Frac} (A)$, i.e. the "field of fractions") is isomorphic to $\mathbb Q (t)$?

I know that the parametrization of the classic circle is $$t\mapsto \left(\frac{1 - t ^ 2}{ 1 + t ^ 2}, \frac{2t}{1 + t ^ 2}\right)$$

Karl Kroningfeld
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jenny
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1 Answers1

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Define a $\mathbb{Q}$-algebra homomorphism $\mathbb{Q}[x,y]$ to $\mathbb{Q}(t)$ by sending $x$ to $\frac{1 - t ^ 2}{ 1 + t ^ 2}$ and $y$ to $\frac{2t}{ 1 + t ^ 2}$. The ideal generated by $x^2+y^2-1$ is clearly in the kernel, so the map descends to a map $A=\mathbb{Q}[x,y]/(x^2+y^2-1) \rightarrow \mathbb{Q}(t)$.

A map between integral domains induces a map between their fraction fields, so you obtain $\mathrm{Frac}(A) \rightarrow \mathbb{Q}(t)$ (since $\mathbb{Q}(t)$ is its own fraction field). This is a nonzero map between fields, so it is automatically injective. Thus, it suffices to show surjectivity. This can be done by constructing an explicit preimage of an arbitrary rational function $p(t)/q(t)$ (hint: start by finding a preimage of $t/1$).

user26857
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Cass
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    "A map between integral domains induces a map between their fraction fields": what about $\mathbb Z\to\mathbb Z/2\mathbb Z$? Conclusion: "An *injective* map between integral domains induces a map between their fraction fields". So you must prove that the map is injective which is not obvious. – user26857 Apr 05 '14 at 09:40
  • do you have a methode for that ? – jenny Apr 05 '14 at 19:33
  • Surjectivity : the above injective field homomorphism maps $\frac{y}{x+1}$ to $t$ –  Aug 22 '19 at 20:50
  • @user682705 The injectivity of the map $A=\mathbb{Q}[x,y]/(x^2+y^2-1) \rightarrow \mathbb{Q}(t)$ is equivalent to $\ker=(x^2+y^2-1)$. That is, this answer proves nothing. – user26857 Aug 23 '19 at 15:49