Show that the Zariski topology over polynomials over $Q[x_1,...,x_n]$ in $\Bbb R^n$ is not $T_1$

Question

Can a polynomial in $\Bbb R^n$ with rational coefficients have a root that either

a) has at least one irrational component , ex.: in $\Bbb R^3 (1,1, e)$

b) if not a), then all components irrational, ex.: in $\Bbb R^3 (e,e,e)$

I realize that this is true for $\Bbb R^1$ (looking at discussions of transcendental numbers). But what about $\Bbb R^n$?

Edit: Answering my own question : for a) $f(x,y,z) = x-y$ works for the point I give, b) $f(x,y,z)=2x-y-z$ works for the point I give.

This arises in my attempt at showing that the Zariski Topology with rational coefficient polynomials on $\Bbb R^n$ is not a $T_1$ space, i.e. that it has a single point set that is not closed, or in other words a point that cannot be the root of a polynomial on $Q[x_1,...x_n$]. Maybe there is a simpler way of proving this. Hints appreciated.

Answer 1

How about $n=1$ and $P=(\sqrt2)$? Any rational polynomial vanishing on $P$ has a factor $X^2-2$, and so also vanishes at $-\sqrt{2}$. So $\{P\}$ is not closed.