Good day all, I am trying to understand why the dimension of $f(X,Y) = Y^2 - X(X-1)(X+1) = 0$ is one via transcendence degree of $\bar{k}(V_f)/\bar{k}$ where $V_f$ is the variety associated to the ideal $I=(f(X,Y))\subset \bar{k}[X,Y]$ (This is an example from Knapp's Advanced Algebra).
My understanding is that in $\bar{k}(x)[Y]$, $f$ generates a quadratic extension of $\bar{k}(X)$. Is this correct?
The reason $x = X + I$ and $y= Y+I$ in $\bar{k}(V_f)$ are algebraically dependent is because we can parametrize $y$ in terms of a $r/g\in\bar{k}(x)$ for some $r$ and $g\not= 0$.?
Finally/vaguely, however you'd like to put it, is there a mapping diagram someone knows of that could help? I was thinking something along the lines of a tower of field extensions of $\bar{k}$ by the fields of quotients, with the transcendental extension of $\bar{k}$ and why not throw in the coordinate ring....as you can see, mass confusion here ???$\not8$(
Thanks.
1) Yes, but the formulation "$f$ generates a quadratic extension" is awkard. I would say "$\bar k(x,y)$ is a quadratic extension of $\bar k(x)$, because the minimum polynomial of $y$ over $\bar k(x)$ is $f(x,Y) \in \bar k(x)[Y]$".
2) Easier just to say that $f(x,y) = 0$, so $x$ and $y$ are algebraically dependent. But 1) also already says this.
3) You mean, for instance, $\bar k(x,y) = \bar k(V_f) \supseteq \bar k(x) \supseteq \bar k$? The first $\supseteq$ is algebraic of degree 2, the second transcendental of transcendence degree 1. Or the other way around, $\bar k(x,y) \supseteq \bar k(y) \supseteq \bar k$. Now the first $\supseteq$ is algebraic of degree 3, the second still transcendental of transcendence degree 1.