$\{ x \in \mathbb R^n : \forall f_i \in \mathbb R[X_1, X_2, \dots X_n], f_i(x) = 0 \}$

Open sets (the complement of closed sets) are of them form:
$\{ x \in \mathbb R^n : \exists f_i \in \mathbb R[X_1, X_2, \dots X_n], f_i(x) \neq 0 \} \in \tau$

The empty set is generated as $\{ x \in \mathbb R^n : 0 \neq 0 \}$ and the
full set is generated as $\{ x \in \mathbb R^n : 1 \neq 0 \}$.
`IN-SET`

.
if the point is not in $O$, $T_O$ infinite loops. Formally:
$\begin{aligned}
x &&\in O \iff \text{$T_O$ halts on input $x$} \\
x &&\not \in O \iff \text{$T_O$ does not halts on input $o$}
\end{aligned}$

Alternatively, for a closed set $C \in \tau$, we associate a a turing machine
$T_C$ which semi-decides exclusion. That is, if a point is ```
# return NOT-IN-SET of x is not a zero of the polynomial
def is_not_in_zeros(poly, x0):
precision = 0 # start with zero precision
while True:
if poly(x0[:precision]) != 0: return NOT-IN-SET
precision += 1 # up the precision
```

Since we can only evaluate a polynomial up to some finite precision, we
start with zero precsion and then gradually get more precise. If some
$x0$ is `poly(x0[:precision]) = 0`

for all
levels of precision.
`Spec(R)`

- To setup a topology for the prime spectrum of a ring, we define the topological space as $Spec(R)$, the set of all prime ideals of $R$.
- The closed sets of the topology are $\{ V(I) : I \text{is an ideal of } R \}$, where the function $V: \text{Ideals of } R \rightarrow 2^{\text{Prime ideals of } R}$each ideal to the set of prime ideals that contain it. Formally, $V(I) = \{ p \in Spec(R) : I \subseteq p \}$.
- We can think of this differently, by seeing that we can rewrite the condition as $V(I) = \{ P \in Spec(R) : I \xrightarrow{P} 0 \}$: On rewriting using the prime ideal $P$, we send the ideal $I$ to $0$.
- Thus, the closed sets of $Spec(R)$ are precisely the 'zeroes of polynomials' / 'zeroes of ideals'.
- To make the analogy precise, note that in the polynomial case, we imposed a topology on $\mathbb R$ by saying that the closed sets were $V(p_i) = \{ x \in \mathbb R : p(x) = 0 \}$for some polynomial $p \in \mathbb R[x]$.
- Here, we are saying that the closed sets are $V(I) = \{ x \in Spec(R) : I(x) = 0 \}$for some ideal $I \in R$. so we are looking at ideals as functions from the prime ideal to the reduction of the ideal. That is, $I(P) = I/P$.

$\begin{aligned}
V(I) = \{ x \in Spec(\mathbb Z) : I(x) = 0 \}
V((n)) = \{ (p) \in Spec(\mathbb Z) : (n)/(p) = 0 \}
V((n)) = \{ (p) \in Spec(\mathbb Z) : (n) \mod (p) = 0 \}
\end{aligned}$

So in our minds eye, we need to imagine a space of prime ideals (points), which
are testing with all ideals (polynomials). Given a set of prime ideals (a tentative locus, say a circle),
the set of prime ideals is closed if it occurs as the zero of some collection of ideas
(if the locus occurs as the zero of some collection of polynomials).
- Consider the functions $f(x) = (x - 1)$ and $g(x) = (x - 1)^2$. as functions on $\mathbb R$. They are indistinguishable based on Zariski, since their zero sets are the same ( $\{ 1 \}$).

- If we now move to the scheme setting, we get two different schemes: $R_g \equiv \mathbb R[X] / (x - 1) \simeq \mathbb R$, and $R_g \equiv R[X] / (x - 1)^2$.

- In the ring $R_g$, we have an element $(x-1)$such that $(x-1)^2 = 0$, which is a nilpotent. This "picks up" on the repeated root.

`math.se`

question: 'Geometric meaning of the nilradical'
- Another example is to consider (i) the pair of polynomials $y = 0$(the x-axis) and $y = \sqrt(x)$. The intersection is the zero set of the ideal $(y, y^2 - x) = (y, 0 - x) = (y, x)$.

- Then consider (ii) the pair of $y = 0$ (the x-axis) and $y = x^2$. Here, we have "more intersection" along the x-axis than in the previous case, as the parabola is "aligned" to the x-axis. The intersection is the zero set of the ideal $(y, y - x^2) = (y, 0 - x^2) = (y, x^2)$.

- So (i) is governed by $\mathbb R[X, Y] / (y, x)$, while (ii) is governed by $\mathbb R[X, Y] /(y, x^2)$which has a nilpotent $x^2$. This tells us that in (ii), there is an "infinitesimal thickening" along the $x$-axis of the intersection.