When is the Zariski topology $T_2$? In my general topology course, our instructor introduced the Zariski topology as a topology on $F^n$, where $F$ is a field, as the topology $\tau$ generated by the basis $\mathcal{B} = \{ f^{-1} (F \setminus \{ 0 \}) : f \in F[x_1 , \ldots, x_n] \}$, or equivalently generated by closed sets which are the pre-image of $\{ 0 \}$ under some $n$-variate polynomial in $F$.

For $F$ infinite, $n = 1$, we can note that $F$ is given the cofinite topology, which is not $T_2$.

Now for $\mathbb{R}^{n}, n \geq 2$, the Zariski topology must necessarily be $T_1$ and not $T_2$ (a prescribed exercise). The former can be seen via the equivalent definition of $T_1$ as leaving all singletons closed (consider merely a linear polynomial). The not Hausdorff part, however, comes from observing that the pre-image of $\{ 0 \}$ under an $n$-variate non-constant polynomial can have dimension at most $n - 1$, so the open sets have too great a codimension for the topology to be Hausdorff, where the dimension is taken in the sense of Hausdorff (though I think the implicit function theorem would let us instead say simply that it's a manifold of dimension $&lt; n$). Similar justifications hold for $\mathbb{C}$. However, in these cases we're appealing to properties of $\mathbb{R}$ and $\mathbb{C}$ beyond merely algebraic properties, but geometric properties of the two; in particular we're considering properties of $\mathbb{R} ^ n , \mathbb{C} ^ n$ as Banach spaces, an object that doesn't even make sense except in terms of subfields of $\mathbb{C}$.

Alternatively, for finite fields $F = \mathbb{F}_{p^{k}}^{n}$, we can again show the topology is discrete, again by counting methods.

So the problem of discerning when $F^{n}$ is $T_2$ is simple for $n = 1$, coming down to cardinality arguments about $F$. Moreover, the problem is simple for finite $F$ for all $n \in \mathbb{N}$. The problem is also resolvable for $F = \mathbb{R}$ or $F = \mathbb{C}$, but these cases must be addressed by geometric methods, referring to relations between $\mathbb{R}[x_1 , \ldots, x_n]$ (resp. $\mathbb{C} [x_1 , \ldots, x_n ]$) and the thoroughly studied geometric properties of $\mathbb{R}$ (resp. $\mathbb{C}$).

But when I broached to my professor the question of how me might address the Zariski topology on, say, $\mathbb{Q}^2$, he said he'd leave the question to me, as it was beyond his area of knowledge. So I took it to the MSE community. How would we go about discerning whether the Zariski topology on $F^n$ is $T_2$ for arbitrary fields $F$, where $n \geq 2, |F| = \infty$?

My attempts thus far

What I've figured out is that $X$ is $T_2$ iff for every $x_1 , x_2 \in X, x_1 \neq x_2$, there exist closed $K_1, K_2$ such that $x_1 \in K_1 \setminus K_2, x_2 \in K_2 \setminus K_1, K_1 \cup K_2 = X$. My argument is as follows:

$(\Rightarrow)$ If $X$ is $T_2$, then there exist open $U_1, U_2$ such that $x_1 \in U_1, x_2 \in U_2, U_1 \cap U_2 = \emptyset$. Let $K_1 = U_2^{\complement}, K_2 = U_1^{\complement}$. Then $x_1 \in K_1, x_2 \in K_2$. Moreover, $K_1 \cup K_2 = U_1^{\complement} \cup U_2^{\complement} = (U_1 \cap U_2)^{\complement} = X$.

$(\Leftarrow)$ Let $x_1 \in K_1 \setminus K_2, x_2 \in K_2 \setminus K_1$, and let $U_1 = K_2^{\complement}, U_2 = K_1^{\complement}$. Then $x_1 \in U_1, x_2 \in U_2$, and $U_1 \cap U_2 = \left( K_1^{\complement} \cup K_2^{\complement} \right) ^{\complement} = \emptyset$.

This completes the proof (hopefully).

This means that I wanna be able for arbitrary point $\mathbf{x} = (x_{1}, \ldots, x_{n}), \mathbf{y} = (y_1 , \ldots, y_n ) \in F^n, \mathbf{x} \neq \mathbf{y}$, find polynomials $f, g \in F[x_1, \ldots, x_n]$ such that $f(\mathbf{x}) = 0 \neq f( \mathbf{y}), g(\mathbf{y}) = 0 \neq g(\mathbf{x})$, and $f^{-1}(\{0\}) \cup g^{-1}(\{0\}) = F^n$. Then $fg \equiv 0$. I feel as if there is a short line to add here to conclude that no such $f, g$ exist, but I don't know what it is.

Thanks!
The Zariski topology on $F^n$ is never Hausdorff if $n&gt;0$ and $F$ is infinite.  Indeed, you can reduce to the case $n=1$ by noting that the subspace topology on $F\times \{0\}^{n-1}\subseteq F^n$ coincides with the Zariski topology on $F$ (identifying $F$ and $F\times \{0\}^{n-1}$ in the obvious way): given a polynomial $f(x_1,x_2,\dots,x_n)$, the subset of $F\times \{0\}^{n-1}$ on which it vanishes is just the subset of $F$ where the single-variable polynomial $f(x,0,\dots,0)$ vanishes.  The Zariski topology on $F$ is just the cofinite topology since any nonzero single-variable polynomial can only vanish at finitely many points, and in particular is not Hausdorff.  Since a subspace of a Hausdorff space is Hausdorff, this means $F^n$ cannot be Hausdorff.
$T_1$ implies $T_2$ for any Zariski topology on a commutative ring. Note if $Z(R)$ is $T_1$, then the Krull dimension of $R$ must be $0$, from which the proof follows.