I'm was browsing this question, where it is proven the quotient field of $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is isomorphic to the rational function field $\mathbb{Q}(t)$ under the isomorphism $$ (x,y) \mapsto \left( \frac{2t}{t^2+1}, \frac{t^2-1}{t^2+1} \right). $$ How can this isomorphism be used to show $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is normal in its quotient field?

I identified $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ with $\mathbb{Q}\left[\frac{2t}{t^2+1}, \frac{t^2-1}{t^2+1} \right]$, and was trying to show this is integrally closed in $\mathbb{Q}(t)$.

If $f/g\in\mathbb{Q}(t)$, with $f,g$ coprime, is integral over $\mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]$, then $$ (f/g)^n+h_{n-1}(f/g)^{n-1}+\cdots+h_1(f/g)+h_0=0,\qquad h_i\in \mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]. $$ Multiplying through by $g^n$, I found the relation $$ f^n=g(-h_{n-1}f^{n-1}-\cdots-h_0g^{n-1}) $$ so any prime dividing $g$ divides $f$, so $g$ is constant, and in fact $f/g\in\mathbb{Q}[t]$. I was hoping to see that is is in fact in $\mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]$, but I think I've made a mistake somewhere.


2 Answers 2


Let's prove that $R=\mathbb{Q}\left[\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right]$ is integrally closed.

We have $\mathbb{Q}\left[\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right]=\mathbb{Q}\left[\frac{1}{t^2+1},\frac{t}{t^2+1}\right]$. It's obvious that $\mathbb Q(t)$ is its field of fractions. Let $f(t)/g(t)\in\mathbb Q(t)$ integral over $R$ with $\gcd(f(t),g(t))=1$. For any element $h\in R$ there is $r\in\mathbb N$ such that $(t^2+1)^rh\in\mathbb Q[t]$. Consider an integral dependence relation $$(f/g)^n+h_{n-1}(f/g)^{n-1}+\cdots+h_1(f/g)+h_0=0$$ with $h_i\in R$ and multiply this by $(t^2+1)^{nr}$ with $r$ large enough in order to get an integral dependence relation for $(t^2+1)^rf(t)/g(t)$ over $\mathbb Q[t]$. Since $\mathbb Q[t]$ is integrally closed we get $(t^2+1)^rf(t)/g(t)\in\mathbb Q[t]$, and therefore $g(t)\mid(t^2+1)^r$. Since $t^2+1$ is irreducible over $\mathbb Q$ we have $g(t)=(t^2+1)^s$. Now it's clear that $f(t)/g(t)\in R$.

  • $\begingroup$ It might be a silly question. But how do you see that $\frac{1}{t^2+1} \in \mathbb{Q}[\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}]?$ $\endgroup$
    – Youngsu
    Commented Apr 9, 2014 at 17:13
  • $\begingroup$ @Youngsu: $\frac{t^2-1}{t^2+1} = 1 - 2\left(\frac{1}{t^2+1}\right)$ $\endgroup$
    – zcn
    Commented Apr 9, 2014 at 19:16

Due to both denominator's being the same that ring is isomorphic to the ring $\mathbf{Q}[2t, t^2-1]$. That is, the subring of $\mathbf{Q}[t]$ generated by $2t$ and $t^2-1$. But in the presence of $2t$ we do not need $t^2-1$. So this subring can be generated by just $2t$. So this is isomorphic to the PID $\mathbf{Q}[2t]$, hence normal.

  • 1
    $\begingroup$ $\mathbb Q[2t]=\mathbb Q[t]$ and $\mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]$ can't be isomorphic to $\mathbb Q[t]$ since it's not a UFD. $\endgroup$
    – user26857
    Commented Apr 9, 2014 at 10:13

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