I have an algebraic set $X \subseteq \mathbb{A}_k^n$. I want to show that its dimension is the same as the dimension of its projective closure. I am trying to do double inequality, and for $\dim(X)\geq \dim(\bar{X})$ the exercise suggests to use induction on the dimension of the closure.
I am having troubles with the induction step. Let's say that $\bar{X}$ has dimension $k$, then one can find projective varieties such that $Z_0 \subsetneq Z_1 \subsetneq ... \subsetneq Z_{k-1}\subsetneq Z_k \subseteq \bar{X}$. By induction hypothesis, since $Z_{k-1}$ has dimension $k-1$ , $Z_{k-1}\cap\mathbb{A}_k^n$ has greater or equal dimension, so it has more than $k-1$ varieties under it, so $\dim(X)$ is greater and I am done.
The problem is, can I find a chain of projective varieties that are not strictly contained in the hyperplane of infinity? So that intersection is not empty. I saw this post but I was wondering if there was a simpler way, by contradiction assuming that every $Z_{k-1}$ (nonempty, since it has more varieties under it) projective variety contained in $\bar{X}$ is indeed in that hyperplane. I guess it has something to do with $X$ being an affine set. I tried passing to the affine space but the correspondence between algebraic and projective varieties only works when the latter have non empty affine part, and that's exactly the opposite of what I am assuming...
If someone has a simple way to do that by the way I'm suggesting, I would be grateful.
I've came up with an elegant solution which does not require induction:
Actually, one can see the dimension of an algebraic set as the maximum of the dimensions of its irreducible components. So, we can suppose without loss of generality that $X$ is indeed an affine variety.
$X$ is a quasi-projective variety, being contained in its closure (Shafarevich $1$) and it is known that if $Z$ is a variety and $U \subseteq Z$ is an open subset, then their dimensions coincide so... it is done!