Problem: show that the algebraic subsets of $\mathbb{A}^1(k)$ are just the finite subsets, together with $\mathbb{A}^1(k)$ itself.
My solution: I'm self-teaching from an AG textbook so please bear with me, here's my best shot. Obviously the whole set counts because we have the trivial ideal generated by $(0)$, so $V((0))$ is the whole set. It would seem to me that given any finite subset $\{a_i\}$ of elements in $k$, I can give you a polynomial $p = \Pi_i (x - a_i)$ that trivially has that given subset of zeros. But clearly $(p)$ is not the ideal generating that set, since by the definition of an ideal, there would be other zeros floating around in there. So I thought, okay, imagine the natural homomorphism of rings $\pi : k[X] \to k[X]/(p)$. I suppose ker $\phi$ would be an ideal such that $V(\ker \phi)$ is the required algebraic set. I suppose in this way I prove existence, but just not construction? (Obviously there can't be infinite such algebraic sets, since that would imply an infinite product of linear monomials, that part I feel like I get).
Am I on the right track?
TIA!
There is (in my opinion) a much simpler way of proving it:
Let $X \subseteq \mathbb{A}^1_k$ be an algebraic set; i.e, $X=V(\mathfrak{a})$ for some ideal $\mathfrak{a} \subseteq k[x]$. Since $k[x]$ is a PID, there exists some polynomial $F \in \mathfrak{a}$ such that $\mathfrak{a}=(F)$. We distinguish two cases:
i) If $F=0$, then $\mathfrak{a}=\{0\}$ and $X=V(0)=\mathbb{A}^1_k$.
ii) If $F \neq 0$, then $F$ has a finite number of roots (possibly none if the field $k$ is not algebraically closed). Hence, $$X=V(F)=\{p \in \mathbb{A}^1_k / F(p)=0\}=\{\text{roots of } F\}$$ is a finite set of $\mathbb{A}^1_k$.