# Function field $K(C)$ isomorphic to $K(t)$ [duplicate]

Let $$C$$ be a curve in $$\mathbb{P}^2$$ over a field $$K$$ defined by an equation of the form $$f: y^2 - x^3 - ax^2$$ with $$a \in K$$.

I want to show that the function field of $$C$$ is isomorphic to $$K(t)$$.

The function field $$K(C)$$ is the fraction field of $$K[C] = K[x,y]/(f)$$.

Since we're quotienting by $$f$$, I think that every $$g \in K(C)$$ can be written as $$a(x) + b(x)y$$, where $$a,b \in K(x)$$. Therefore $$K(C) \cong K(x)[y]/(f)$$. (This coincides so far with a comment on a similar question here.)

However, I don't know how to get the last part. Is there an obvious isomorphism I'm overlooking, or do I have to use the equation of the curve somewhere explicitly? (I've shown that there exists a singular point on $$C$$, but I don't see any use of this fact.)

• Try $t=y/x$. From $f=0$ you can solve $t^2=x+a$. From there on it should be easy. Commented May 31, 2020 at 18:35
• Basically, the curve has a double point at the origin, and this is the usual blow-up resolving the singularity. Commented May 31, 2020 at 18:41
• A geometric interpretation of Jyrki Lahtonen's idea is that this gives a rational parameterization of your curve, which shows that your curve is birational to a line and thus has function field $K(t)$. Commented May 31, 2020 at 19:27