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 $\overline{K}$?

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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).


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