Let $K$ be a field and $X \subseteq \mathbb{A}^n(K)$ be a subset. Is it true that if $X$ is irreducible with respect to the Zariski topology, then $X$ does not contain any proper closed subset?
Irreducibility tells us that $X$ can not be written as the union of two proper closed subsets. But (how) can we exclude a case where for example
$$X= C_1 \cup Y,$$ where $C_1$ may be proper and closed but $Y$ may be open or neither closed nor open?
I ask that question because I would like to prove that the closure of $$C:=\{(n,n^2) \in \mathbb{C} : n\in \mathbb{N}\}$$ is the vanishing set of the polynomial $X^2-Y$. I know that the latter exercise has been solved in another post, but it uses Hilberts Nullstellensatz and I would like to conclude without this machinery.
What I have so far is that $C \subseteq \bar{C} \subseteq V_{X^2-Y}(\mathbb{C})$. $C$ is not closed so its closure is not $C$ itself. Now the question is whether $\bar{C} \subsetneq V_{X^2-Y}(\mathbb{C})$. If not, then we have found the closure.
Now $X^2-Y$ is irreducible, hence the vansihing ideal of $V_{X^2-Y}(\mathbb{C})$ is the ideal generated by the polynomial $X^2-Y$. Now, if I show that this vansihing ideal is a prime ideal, then $V_{X^2-Y}(\mathbb{C})$ is irreducible and with my argument from the question that I asked in the beginning I could conclude that $V_{X^2-Y}(\mathbb{C})$ has no proper closed subsets and thus $\bar{C} = V_{X^2-Y}(\mathbb{C})$.
Since my argument only works up to some assumptions, any other approach that does not use Hilberts Nullstellensatz, and that lets me conclude about the dimension of the closure would be nice!
Any singleton set is a closed set, hence any irreducible closed set with more than one point has nonempty proper closed subsets.
Another example . . .
Let $K$ be a field and $X \subset \mathbb{A}^3(K)$ be defined by $$X = \{(x,y,0)\mid x,y \in K\}$$ Then $X$ is an irreducible closed subset of $\mathbb{A}^3(K)$, but $X$ has many proper closed subsets other than singletons, such as $$C = \{(x,0,0)\mid x \in K\}$$ or $$C = \{(x,x^2,0)\mid x \in K\}$$
The exercise which prompted your question was this:
Let $k$ be a field extension of $\mathbb{Q}$ (e.g., $k = \mathbb{C}$).$\\[7pt]$
Let $C = \{(n,n^2) \in \mathbb{A}^2(k)\mid n \in \mathbb{N}\}$, and let $\bar{C}$ be the closure of $C$ in the Zariski topology on $\mathbb{A}^2(k)$.$\\[7pt]$
Prove that $\bar{C} = V(f)$, where $f \in k[x,y]$ is given by $f(x,y)=y-x^2$.
Here's a proof avoiding explicit use of the Nullstellensatz . . .
Let $I(C) = \{g \in k[x,y]\mid g(n,n^2) = 0,\;$for all$\;n \in \mathbb{N}\}$.
Then $\bar{C} = V(I(C))$, so to prove $\bar{C} = V(f)$, it suffices to prove $I(C) = (f)$.
First we show $(f) \subseteq I(C)$.
Let $(n,n^2) \in C$.
Then $f(n,n^2) = n^2 - n^2 = 0$, hence $f \in I(C)$.
It follows that $(f) \subseteq I(C)$.
For the reverse inclusion, we need a preliminary lemma . . .
Lemma (The Generalized Factor Theorem)$\,$:
Let $R$ be a commutative ring, and let $p \in R[x]$.$\\[7pt]$
If $c \in R$ is such that $p(c) = 0$, then $p \in (x-c)$.
Proof:
Let $p \in R[x]$, and suppose $p(c) = 0$, for some $c \in R$.
By elementary algebra, \begin{align*} x^0 - c^0 &= (x-c)(0)\\[4pt] x^1 - c^1 &= (x-c)(1)\\[4pt] x^2 - c^2 &= (x-c)(x + c)\\[4pt] x^3 - c^3 &= (x-c)(x^2 + cx + c^2)\\[4pt] x^4 - c^4 &= (x-c)(x^3 + cx^2 + c^2x + c^3)\\[3.5pt] &\;\,\vdots\\[-6pt] x^k - c^k &= (x - c)\left(\sum_{j=0}^{k-1} x^{(k-1-j)}\,c^j\right)\\[4pt] \end{align*} It follows that for all nonnegative integers $k$, we have $x^k - c^k \in (x-c)$.
Write $p(x) = {\displaystyle{\sum_{k=0}^n a_kx^k}}$. \begin{align*} \text{Then}\;\;p(x) &= p(x) - p(c)&&\text{[since $p(c)=0$]}\\[4pt] &=\sum_{k=0}^n a_kx^k - \sum_{k=0}^n a_kc^k\\[4pt] &=\sum_{k=0}^n a_k(x^k - c^k)\\[4pt] &\in (x-c)&&\text{[since each summand is in $(x-c)$]}\\[4pt] \end{align*} which completes the proof of the lemma.
Back to the main proof . . .
We've already have the inclusion $(f) \subseteq I(C)$.
For the reverse inclusion, let $g \in I(C)$. We want to show $g \in (f)$.
Since $g \in I(C)$, we have $g \in k[x,y]$ and $g(n,n^2) = 0$, for all $n \in \mathbb{N}$.
Let $h \in k[x]$ be defined by $h(x) = g(x,x^2)$. Then for all $n \in \mathbb{N}$, $$h(n) = g(n,n^2) = 0$$ Thus, $h$ has infinitely many roots in $k$, hence $h = 0$.
Let $R=k[x]$.
Regarding $k[x,y]$ as $R[y]$, we have $g(x,y) = p(y)$, for some $p \in R[y]$. \begin{align*} \text{Then}\;\;&h(x) = 0\\[4pt] \implies\;&g(x,x^2) = 0\\[4pt] \implies\;&p(x^2) = 0\\[4pt] \implies\;&p(y) \in (y-x^2)\;\text{in}\;R[y]\;\;\;\text{[by the lemma]}\\[4pt] \implies\;&g(x,y) \in (f)\;\text{in}\;k[x,y]\\[4pt] \end{align*} It follows that $I(C) \subseteq (f)$.
Thus, we have both inclusions, hence $I(C) = (f)$.
Therefore $\bar{C} = V(f)$, as was to be shown.