# Proving that the fraction field of a $k[x,y]/(f)$ is isomorphic to $k(t)$

Let $f \in k[x,y]$ be an irreducible polynomial of degree $2$, $k$ is a field, and there exist $a,b\in k$ such that $f(a,b) = 0$. I'm trying to prove that $\mathsf{frac}(k[x,y]/(f))$ is isomorphic to $k(t)$, the field of rational functions with single variable.

What I know:

$f$ is irreducible and so its prime, so the ring $k[x,y]/(f)$ is an integral domain, so the fraction field is just the quotient of its elements. But I am stuck at this point. The only thing I can think of is finding some embedding from $k[x,y]/(f)$ into $k(t)$, then $\mathsf{frac}(R)$ would be contained in $k(t)$ and go from there.

Any hints would be appreciated...

• If $k=\mathbb R$ and $f=x^2+y^2$ then the fraction field doesn't seem isomorphic to $\mathbb R(t)$. Commented Jan 16, 2018 at 20:59
• That's the problem statement as given to me by my professor. All of us including our TA are stuck. What even do the elements in this ring look like? Commented Jan 17, 2018 at 3:42
• The fraction field of $\mathbb R[x,y]/(x^2+y^2)$ is $\mathbb R(y)[x]/(x^2+y^2)$, and therefore the elements of it are of the form $a(y)+b(y)x$ with $a(y),b(y)\in\mathbb R(y)$. This is a field, but not a purely transcendental extension of $\mathbb R$ as $\mathbb R(t)$ is. Commented Jan 17, 2018 at 7:11
• I received a hint from professor. apparently it involves mapping x/y to t from the quotient ring to k(t). Commented Jan 18, 2018 at 8:18
• If one follows the hint then in my example $t=x/y$ is algebraic over $\mathbb R$ since $t^2+1=0$. (I still support the opinion that your claim is wrong.) Commented Jan 18, 2018 at 10:10

The fraction field of $$R=\mathbb R[x,y]/(x^2+y^2)$$ is $$K=\mathbb R(y)[x]/(x^2+y^2)$$, that is, $$K=\mathbb C(y)$$, and this is not a purely transcendental extension of $$\mathbb R$$ (why?). It follows that $$K\not\simeq\mathbb R(t)$$.