If $K$ is an algebraically closed field and $F \in K[X, Y]$ is irreducible, then $\dim K[X, Y]/(F) = 1$ $\newcommand{\trianglelefteqslant}{\leqslant \hskip{-7.8pt} \raise{0.9pt}\vartriangleleft}$
Let $K$ be an algebraically closed field. I will now state some basic definitions as I don't know how standard they are.


*

*$K^n$ is equipped with the Zariski topology. Any closed or open subset of $K^n$ will be understood with respect to this topology;

*The dimension of a closed set $V \subseteq K^n$ is
$$\dim V = \sup \{ n \in \mathbb{N} : (\exists V_0 \subsetneq V_1 \subsetneq \ldots \subsetneq V_n \subseteq V) \, V_k \text{'s are closed, irreducible} \};$$

*The (Krull) dimension of a ring $R$ is
$$\dim R = \sup \{ n \in \mathbb{N} : (\exists I_0 \supsetneq I_1 \supsetneq \ldots \supsetneq I_n ) \, I_k \text{'s are prime ideals in } R \};$$

*The coordinate ring of $V \subseteq K^n$ is 
$$K[V] = \{ F \upharpoonright V : F \in K[\overline{X}] \} \cong K[\overline{X}] / I(V).$$


The problem is:

Suppose $F \in K[X, Y]$ is irreducible and $V = Z(F) \subseteq K^2$. Prove that $\dim V = 1$.

What I know:


*

*Hilbert's Nullstellensatz: $I(Z(I)) = \sqrt{I}$;

*If $V \subseteq K^n$ is an affine algebraic set, then Zariski-closed subsets $U \subseteq V$ correspond bijectively to radical ideals $I \trianglelefteqslant K[\overline{X}]$ containing $I(V)$ by $U \substack{\xleftarrow{Z(I)} \\ \xrightarrow[I(U)]{}} I$;

*These in turn correspond to radical ideals $I \trianglelefteqslant K[V]$;

*In the above, irreducible closed subsets correspond to prime ideals;

*$\dim V = \dim K[V]$;

*I was presented a theorem that if $R$ is a domain and a finitely generated $K$-algebra, then $\dim R = \operatorname{td}_K R_0$, but it was said that the proof is somewhat complicated (uses an external theory), so I don't want to use that theorem;

*I can see how Krull's principal ideal theorem can be applied to immediately solve the problem, but I wasn't taught that theorem either, so if possible, I would rather avoid using it. However, if someone's really convinced there is no easier way, I will accept that as an answer.


Now my recognition of the problem: obviously there is some $\overline{a} = (a_1, a_2) \in V$ and $(X-a_1, Y-a_2) \supsetneq (F)$ is a chain of length $n=1$ of prime ideals in $K[\overline{X}]$ containing $I(V)$. Also if there were a chain $I_0 \supsetneq I_1 \supsetneq I_2$, we can assume that $I_0 = (X-a_1, Y-a_2)$ for some $\overline{a} \in V$ and $I_2 = (F)$. As $K[X, Y]$ is Noetherian, $I_1 = (G_1, \ldots, G_n)$ for some $G_1, \ldots, G_n \in K[X, Y]$, which can be assumed to be irreducible since $I_1$ is prime, and pairwise unassociated. But that doesn't seem to give any easy contradiction.
From the geometric side, suppose we have $V_0 \subsetneq V_1 \subsetneq V_2 \subseteq V$ closed, irreducible, so again we can assume $V_0 = \{ \overline{a} \}$ and $V_2 = V$ (and $V_1 = Z(I_1)$ if we want a connection). Any finite non-singleton is reducible, so $V_1$ is infinite (in fact, if $F$ depends on $Y$, the other case being trivial, then the projection $\pi_X[V_1]$ is co-finite). But I don't see any contradiction here either.
Any hint would be appreciated.
 A: There is indeed an elementary solution for the two-dimensional case.
Claim: Let $K$ be a field, $F, G \in K[X, Y]$. If $F$ and $G$ are coprime (have no common factors), then $V(F) \cap V(G)$ is finite.
Proof: As $F$ and $G$ have no common factors in $K[X, Y]$, they also have no common factors in $K(X)[Y]$ (this result is known as Gauss' Lemma and is an elementary property of principal ideal domains; Wikipedia and Proofwiki have relatively short proofs of it). As $K(X)[Y]$ is a principal ideal domain as a polynomial ring over the field $K(X)$, this means that there exist some $A, B \in K[X, Y]$ and $D \in K[X] \setminus \lbrace 0 \rbrace$ with $AF + BG = D$. Let now $(a, b) \in V(F) \cap V(G)$. Then $D(a) = 0$. This can hold for only finitely many $a \in K$. Similarly (or by symmetry) there can be only finitely many $b \in K$ such that $(a, b) \in V(F) \cap V(G)$. $ \square $
We can now tackle your problem. Let $F \in K[X, Y]$ be irreducible. You have already noted that $\dim K[X, Y]/(F) \geq 1$. We can show equality if we show that all prime ideals of $K[X, Y]$ strictly containing $(F)$ are maximal. So take a prime ideal $\mathfrak{p}$ of $K[X, Y]$ strictly containing $(F)$; take $G \in \mathfrak{p} \setminus (F)$. Then $G$ is coprime to $F$, whereby $V(F, G)$ is a finite set containing $V(\mathfrak{p})$, such that also the latter is finite. 
Now we apply the Nullstellensatz to find that $I(V(\mathfrak{p})) = \text{rad}(\mathfrak{p}) = \mathfrak{p}$ (last equality as prime ideals are radical). As $\mathfrak{p}$ is prime, $V(\mathfrak{p})$ is irreducible, whereby $V(\mathfrak{p})$ is a singleton. But then $\mathfrak{p} = I(V(\mathfrak{p}))$ is a maximal ideal.
Disclaimer: If $K$ is not algebraically closed, then the Nullstellensatz and thus the last part of the proof no longer work (and in general, the correspondence between irreducible varieties and prime ideals breaks down). One can still show that $\dim K[X, Y]/(F) = 1$ if $K$ is any field and $F$ is an irreducible polynomial, but this requires more work. One way to approach this would be to embed $K$ into its algebraic closure $\tilde{K}$, then consider the embedding $K[X, Y] \to \tilde{K}[X, Y]$. The Going Up theorem (many proofs online) allows us to 'lift' a chain of prime ideals from $K[X, Y]$ to $\tilde{K}[X, Y]$, whereby the two rings have the same Krull dimension, but the argument I gave for algebraically closed fields essentially proved $\dim \tilde{K}[X, Y] = 2$, so we are done. Alternatively, using Krull's Hauptidealsatz, one can prove that $\dim K[X_1, \ldots, X_n] = n$ for any field $K$ and any $n \in \mathbb{N}$.
