Fundamental knowledge of polynomial over a field and its topology

Question

I am junior in this field and am self-studying the algebraic geometry by the following lecture:
http://www.math.uwaterloo.ca/~moraru/764AlgebraicSets.pdf

I met some problems:

Consider the polynomial ring $k[x_1,\ldots,x_n]$

1. (On p.5's Remark) "A polynomial over a field can only have finitely many roots." Does this only hold when $k$ is a finite field with $n\in \mathbb{N}$?

2. (1.2.9 Example) "The Zariski topology on $\mathbb{A}^n$ is Hausdorff if and only if $k$ is finite, in which case it is identical to the discrete topology." My question is:
   a. $k$ is finite here means number of elements in $k$ are finitely many? Like $\mathbb{N}$ is a finite field and $\mathbb{R}$ is infinite field?

   b. Can anyone please show me a simple example to illustrate this? I am particularly confused about how to find the disjoint neighborhood?

3. (p.3 ~ p.4 Hilbert basis theorem) If $k$ is Notherian, does this imply $k$ is finite and Zariski topology on $\mathbb{A}^n$ is Hausdorff?

Answer 1

1. This certainly holds when $k$ is finite and $n\in N$, but when $k$ is infinite, it only holds for [nonzero] polynomials of one variable. If $k$ is infinite, this does not hold for any $n > 1$. Look at $p(x,y) = xy - 1$, this has infinitely many roots in $k^2$ if $k$ is infinite. Namely, $p(\alpha,\alpha^{-1}) = 0$ for all $\alpha\in k^\times$.
2. Yes, $k$ finite means that $k$ has finitely many elements; i.e., $k\cong\Bbb F_q$ for some prime power $q$. $\Bbb N$ is not a field (it isn't even a group under addition), and it is not finite.

To see that $\Bbb A^n$ is Hausdorff when $k$ is finite, note first that the discrete topology is always Hausdorff (every set is both open and closed, so you can take $\{x\}$ and $\{y\}$ as disjoint neighborhoods of $x$ and $y$ ($x\neq y$)). Now, $V((x_1 - a_1,\dots, x_n - a_n)) = \{(a_1,\dots, a_n)\}\subseteq\Bbb A^n$, so all points are closed. But now $\Bbb A^n$ itself is finite, so any subset $S\subseteq\Bbb A^n$ is finite. You have $S = \bigcup_{x\in S}\{x\}$, and this is a finite union of closed sets (hence closed), so any subset of $\Bbb A^n$ is closed (and thus any subset is open). A proof that $k$ infinite implies $\Bbb A^n$ is not Hausdorff can be found in Georges' answer here.

3. Every field is Noetherian, because every field has only two ideals: $(0)$ and $(1)$. Recall that a ring $R$ is Noetherian if every ideal $I\subseteq R$ is finitely generated, and $(0)$ and $(1)$ are finitely generated (in fact, they're principal). So, $\Bbb A^n$ need not be Hausdorff when $k$ is Noetherian (as $k$ noetherian doesn't imply $k$ finite). Further, a general Noetherian ring need not be finite: $\Bbb Z$ is Noetherian but not finite.