# Transcendence degree/dimension of a curve

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).

1. My understanding is that in $\bar{k}(x)[Y]$, $f$ generates a quadratic extension of $\bar{k}(X)$. Is this correct?

2. 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 $g\not= 0$.?

3. 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.

• indeed, but where does the coordinate ring fit in all of this? Is there a map that commutes between $\bar{k}(V_f)$, the coordinate ring, and $\bar{k}(x,y)$, and something else above the coordinate ring? ...maybe $\bar{k}[X,Y]$? – drawnonward Apr 25 '14 at 8:37
• The coordinate ring $\bar k[V_f]$ is, by definition, $\bar k[X,Y]/I$. You are denoting the images of $X$ and $Y$ in this quotient ring by $x$ and $y$, so you can also write it as $\bar k[x,y]$. The field of fractions of $\bar k[V_f]$ is denoted by $\bar k(V_f)$, so is $\bar k(x,y)$. The map from $\bar k[V_f] = \bar k[x,y]$ to $\bar k(V_f) = \bar k(x,y)$ is the inclusion map. Finally, $\bar k[X,Y]$ is not "above" the coordinate ring, but there is ofcourse the projection map $\bar k[X,Y] \to \bar k(x,y) = \bar k(V_f)$. – Magdiragdag Apr 25 '14 at 8:57
• Thank you so much, many blessings being sent your way. I was struggling with this a couple of weeks ago, and I had to put it on the back-burner. I think Hungerford has this in his book Algebra, but piecing togehter the notations between his book and Knapp's put me through a loop for a little while....Thanks again. – drawnonward Apr 25 '14 at 9:07

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.