1
$\begingroup$

I am reading some introductory material on algebraic geometry and would like to understand the following statement:

If a variety $V \subseteq \mathbb{A}^n$ is given by a single polynomial equation $f(x_1, \ldots, x_n) = 0$ then $\dim(V) = n-1$.

The text that I'm using defines the dimension of a variety as the transcendence degree of the function field $\overline{K}(V)$ over $\overline{K}$; here $\overline{K}$ is the algebraic closure of some perfect field $K$.

I am new to these concepts so my biggest concern is having a natural/intuitive way to think about these objects, so that I can supplement the definitions with a "picture" in my head. The idea that I have comes from the analogous situation for the affine plane. To solve for $2X^2-Y^2-3=0$, I just need to know one variable, then the rest I can "solve for". I'm getting that the same idea applies here, so is $\{x_1, \ldots, x_{n-1} \}$ (or any other collection of $n-1$ variables) a choice for transcendence basis of $\overline{K}(V)$ over $\overline{K}$?

$\endgroup$
  • $\begingroup$ Kevin, if my answer has solved your question please consider accepting it by clicking the check-mark. This indicates to the wider community that you've found a solution and gives some reputation to both the answerer and yourself. There is, of course, no obligation to do this. $\endgroup$ – user379719 Oct 24 '16 at 0:26
4
$\begingroup$

Your intuition is somewhat correct. Let's just assume that $K = \bar K$ is algebraically closed. We define the coordinate ring of $V$ to be $K[V] := K[x_1,\dots,x_n]/(f)$, and we define the function field to be $K(V) := \mbox{frac}(K[V])$. So if $f$ is monic in one of the variables, say $x_n$, then your intuition is basically right; the image of $x_n$ in $K(V)$ will be algebraic over $K(x_1,\dots,x_{n-1}) \subseteq K(V)$, which is what you mean by being able to "solve for it", and $x_1,\dots,x_{n-1}$ form a transcendence basis for $K(V)$ -- but not every collection of $n-1$ of the variables necessarily does. In general, $f$ may not be monic in any of the variables, but you can do some clever change of variables to make it so (see a proof of Noether's normalization lemma or my answer to this math.SE post).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.