Prove that \[ \mathbb Q[x,y]/(x^2+y^2-1) \] is an integral domain and that its field of fractions is isomorphic to \[ \mathbb Q(t). \]
Solution
Let \[ q(x,y)\coloneqq x^2+y^2-1. \]
1. The polynomial \(q\) is irreducible in \(\QQ[x,y]\).
Regard \(q\) as a monic quadratic polynomial in \(y\) over the UFD \(\QQ[x]\): \[ q=y^2+(x^2-1). \] If it were reducible, its monicity would give a factorization \[ q=(y+a(x))(y+b(x)) \] with \(a,b\in\QQ[x]\). Comparing the coefficient of \(y\) gives \[ b=-a, \] and comparing constant terms gives \[ a(x)^2=1-x^2. \] This is impossible. Indeed, \(a\) cannot be constant because the right side has degree \(2\). Thus \(\deg a=1\), but then the leading coefficient of \(a(x)^2\) is a square in \(\QQ\), whereas the leading coefficient of \(1-x^2\) is \(-1\), which is not a square in \(\QQ\).
Therefore no such factorization exists, so \(q\) is irreducible.
2. The quotient\[ R\coloneqq \QQ[x,y]/(q) \]is an integral domain.
The ring \(\QQ[x,y]\) is a UFD. In a UFD, every irreducible element is prime. By step 1, \(q\) is irreducible and therefore prime, so the principal ideal \((q)\) is prime. Hence the quotient by \((q)\) is an integral domain.
3. Let \(X,Y\in R\) be the residue classes of \(x,y\). Then \(X\) istranscendental over \(\QQ\).
The natural map \[ \QQ[x]\longrightarrow R \] is injective. Indeed, if a polynomial \(p(x)\in\QQ[x]\) belonged to \((q)\), then \[ p(x)=q(x,y)h(x,y) \] for some \(h\in\QQ[x,y]\). If \(h\neq0\), the right side has degree in \(y\) at least \(2\), whereas the left side has degree in \(y\) equal to \(0\), a contradiction. Thus \(p=0\).
Therefore no nonzero polynomial over \(\QQ\) vanishes at \(X\), so \(X\) is transcendental.
4. In the fraction field\[ K\coloneqq\operatorname{Frac}(R), \]define\[ t\coloneqq\frac{Y}{1+X}. \]Then\[ t^2=\frac{1-X}{1+X}. \]
Step 3 gives \(1+X\neq0\), so \(t\) is defined in \(K\). The relation \[ X^2+Y^2=1 \] holds in \(R\), hence in \(K\). Therefore \[ \begin{aligned} t^2 &= \frac{Y^2}{(1+X)^2}\\ &= \frac{1-X^2}{(1+X)^2}\\ &= \frac{1-X}{1+X}. \end{aligned} \]
5. One has\[ X=\frac{1-t^2}{1+t^2} \qquad\text{and}\qquad Y=\frac{2t}{1+t^2}. \]
Step 4 gives \[ t^2(1+X)=1-X. \] Thus \[ X(1+t^2)=1-t^2. \] Also, \[ 1+t^2 = 1+\frac{1-X}{1+X} = \frac{2}{1+X}, \] so \(1+t^2\neq0\). Hence \[ X=\frac{1-t^2}{1+t^2}. \] Finally, \[ Y=t(1+X) = t\left( 1+\frac{1-t^2}{1+t^2} \right) = \frac{2t}{1+t^2}. \]
6. The fraction field \(K\) is equal to \(\QQ(t)\).
By definition, \[ t=\frac{Y}{1+X}\in K, \] so \[ \QQ(t)\subseteq K. \] Conversely, step 5 shows \[ X,Y\in\QQ(t). \] Since \(K\) is generated as a field by the residue classes \(X\) and \(Y\), \[ K=\QQ(X,Y)\subseteq\QQ(t). \] Thus \(K=\QQ(t)\).
7. The element \(t\) is transcendental over \(\QQ\).
3. Hence \(t\) is transcendental.
8. Therefore\[ \boxed{ \operatorname{Frac}\left( \QQ[x,y]/(x^2+y^2-1) \right) \cong \QQ(t) }. \]
Step 6 identifies the fraction field with the field generated by \(t\), and step 7 shows that \(t\) is transcendental. Therefore that field is a rational function field in one variable over \(\QQ\).
9. Q.E.D.
Step 2 proves that the quotient is an integral domain, and step 8 identifies its field of fractions.