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  1. How can I prove that $\mathbb{Q}[x,y]/\langle x^2+y^2-1 \rangle$ is an integral domain?

  2. Also, I need to prove that its field of fractions is isomorphic to the field of rational functions $\mathbb{Q}(t)$?

(The question is taken from UC Berkeley Preliminary Exam, Fall 1995.)

Bill Dubuque
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Karatuğ Ozan Bircan
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    Do you want a method to show it or an actual solution? It is a theorem that the quotient is an integral domain if the ideal is prime. In this case you can check it is prime by checking that the polynomial is irreducible. – Matt Jan 04 '12 at 17:15
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    This is the same as proving that $\langle x^2+y^2-1 \rangle$ is a prime ideal in $\mathbb Q[x,y]$. – Thomas Andrews Jan 04 '12 at 17:16
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    Actually, I couldn't solve the second question. – Karatuğ Ozan Bircan Jan 04 '12 at 17:21

3 Answers3

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Hint $ $ for $ (1)\!:\ x^2\!+y^2\!-1\,$ is irreducible (so prime) by $\, 1\!-\!y^2\, $ is not a square in the UFD $\,\mathbb Q[y],\,$ indeed, prime $\,y\!-\!1\,$ occurs to odd power (equivalently apply Eisenstein using the prime $\, y\!-\!1)$.

For $(2),$ besides the Wikipedia entry on Pythagorean triples, be sure to see also the article on the tangent half-angle formula. This provides further intuition behind such rational parametrization. In particular it explains the Weierstrass substitution, which is widely used to reduce the problem of integration of rational functions $\,R(\sin(\theta),\cos(\theta))\,$ to the much simpler problem of integrating a rational function $\,r(t).$

Bill Dubuque
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For 1., just prove directly that $x^2 + y^2 -1$ is a prime element of $\mathbb Q[x,y]$. If you want to, you can use Gauss's lemma, which reduces you to showing that it is irreducible, which is particularly simple. (But you can also prove it is prime directly pretty easily --- you will more or less recover the proof of Gauss's lemma in this special case.)

For 2., you need to rationally parameterize the conic $x^2 + y^2 - 1$. This is a standard part of the theory of pythagorean triples. (See the wikipedia entry; note that the quantity called $m/n$ there is your $t$.) The same geometric argument shows that any conic can be rationally parameterized if it contains at least one point defined over the ground field.

Matt E
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    A silly argument for irreducibility is: if the polynomial were reducible, it would be a product of linear factors, and therefore its zero set in $\mathbb R^2$ would be unbounded, but it isn't. – Mariano Suárez-Álvarez Jan 04 '12 at 18:33
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    @MarianoSuárez-Alvarez I think this is equivalent to saying that the unit circle cannot be written as a union of two lines. – Karatuğ Ozan Bircan Jan 04 '12 at 22:50
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We show that $x^2+y^2-1$ is irreducible in $\Bbb{Q}[x, y]$. It is the Exercise 9.4.11 in Dummit and Foote's Abstract Algebra. We view $x^2+y^2-1=y^2+(x^2-1)$ as a polynomial in $(\Bbb{Q}[x])[y]$.

Case 1. $y^2+(x^2-1)=[f(x)y+g(x)][h(x)y+i(x)]$. \begin{eqnarray*} &\text{If}& y^2+(x^2-1)\\ &=& [f(x)y+g(x)][h(x)y+i(x)] \\ &=& f(x)h(x)y^2+[g(x)h(x)+f(x)i(x)]y+g(x)i(x)\\ &\Rightarrow& f(x)h(x)=1\\ &\Rightarrow& f(x)=r\text{ and }h(x)=1/r\\ &\stackrel{g(x)h(x)+f(x)i(x)=0}{\Rightarrow}& (1/r)g(x)+ri(x)=0\\ &\Rightarrow& g(x)+r^2 i(x)=0\\ &\Rightarrow& g(x)=-r^2 i(x)\\ &\stackrel{g(x)i(x)=x^2-1}{\Rightarrow}& -r^2 i(x)^2=x^2-1\\ &\Rightarrow& r^2 i(x)^2=1-x^2\\ &\stackrel{2=\deg{1-x^2}=2\deg{ri(x)}}{\Rightarrow}& \deg{r i(x)}=1\\ &\stackrel{ri(x)=ax+b}{\Rightarrow}& (ax+b)^2=1-x^2\\ &\Rightarrow& a^2=-1, \text{ which is impossible because }a\in \Bbb{Q}. \end{eqnarray*}

Case 2. $y^2+(x^2-1)=f(x)[g(x)y^2+h(x)y+i(x)]$.

If $y^2+(x^2-1)=f(x)[g(x)y^2+h(x)y+i(x)]$, then $f(x)g(x)=1$ and $f(x)$ is a unit.

Note. Case 2 is necessary because we DON'T have the implication: If $f(x)$ can't be written as a product of two polynomials $g(x)$ and $h(x)$ whose degree less than $\deg{f(x)}$, then $f(x)$ is irreducible. For instance, $f(x)=2(x+1)\in \Bbb{Z}[x]$.

bfhaha
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