Let \(k\) be an algebraically closed field of characteristic zero. Classify the nonempty irreducible Zariski-closed subvarieties of \[ \AA^2_k. \] Show that they are exactly:
- \(\AA^2_k\) itself;
- irreducible plane curves \(V(f)\) for irreducible nonconstant \(f\in k[x,y]\);
- points.
Show moreover that every irreducible plane curve \(V(f)\) has infinitely many \(k\)-points.
Solution
Put \[ A=k[x,y]. \] Irreducible Zariski-closed subsets of \(\AA^2_k\) correspond to prime ideals of \(A\).
1. The prime ideal \((0)\) corresponds to the whole affine plane\[ V(0)=\AA^2_k. \]
The ring \(A=k[x,y]\) is a domain, so \((0)\) is prime. Its zero set is all of \(\AA^2_k\).
2. Every height-one prime ideal of \(A\) has the form\[ (f) \]for an irreducible nonconstant polynomial \(f\in k[x,y]\).
The polynomial ring \[ k[x,y] \] is a UFD. In a Noetherian UFD, every height-one prime ideal is principal and generated by an irreducible element PR-VARHT.
Thus every height-one prime is \[ \mathfrak p=(f) \] with \(f\) irreducible and nonconstant. Conversely, an irreducible polynomial is prime in the UFD \(A\), so \((f)\) is prime; Krull’s principal ideal theorem gives it height one.
3. The height-one primes correspond exactly to irreducible plane curves\[ V(f)\subseteq\AA^2_k. \]
If \(\mathfrak p=(f)\) is a height-one prime as in step 2, then \[ A/\mathfrak p = k[x,y]/(f) \] is a domain, so \(V(f)\) is irreducible. The affine dimension formula gives \[ \dim V(f) = \dim A-\height(f) = 2-1 = 1. \] Thus it is a curve.
Conversely, every irreducible closed curve in \(\AA^2_k\) has a prime vanishing ideal of height one, hence is of this form by step 2.
4. Every height-two prime ideal of \(A\) is the maximal ideal of a point:\[ \mathfrak m_{(a,b)} = (x-a,y-b). \]
The ring \(A\) has Krull dimension \(2\). A prime ideal of height \(2\) is therefore maximal. By the weak Nullstellensatz over the algebraically closed field \(k\), every maximal ideal of \(k[x,y]\) is \[ (x-a,y-b) \] for a unique point \[ (a,b)\in\AA^2_k. \] Its zero set is the singleton \[ \{(a,b)\}. \]
5. There are no other nonempty irreducible closed subvarieties of\(\AA^2_k\).
Every prime ideal of the two-dimensional domain \(A\) has height \[ 0,\ 1,\ \text{or }2. \] Step 1 classifies height zero, steps <1>2–<1>3 classify height one, and step 4 classifies height two. These exhaust all prime ideals and hence all nonempty irreducible closed subvarieties.
An irreducible topological space is nonempty by definition, and \[ \varnothing=V(1) \] corresponds to the unit ideal, which is not prime.
6. Every irreducible plane curve\[ C=V(f) \]has infinitely many \(k\)-points.
Write \(f\) as a polynomial in \(y\): \[ f(x,y) = a_r(x)y^r+\cdots+a_0(x), \qquad a_r\ne0. \]
If \(r=0\), then \(f\in k[x]\). Since \(k\) is algebraically closed and \(f\) is irreducible nonconstant, \(f\) is linear: \[ f=c(x-a). \] Then \[ C=\{a\}\cross\AA^1_k, \] which has infinitely many \(k\)-points because \(k\) is infinite.
Suppose \(r>0\). The nonzero polynomial \(a_r(x)\) has only finitely many roots. Hence there are infinitely many \[ c\in k \] with \[ a_r(c)\ne0. \] For each such \(c\), the specialization \[ f(c,y)\in k[y] \] has degree \(r>0\). Since \(k\) is algebraically closed, it has a root \[ b_c\in k. \] Thus \[ (c,b_c)\in C. \] Distinct values of \(c\) give distinct points, so \(C(k)\) is infinite.
7. Therefore the complete classification is\[ \boxed{ \AA^2_k,\quad V(f)\text{ for irreducible nonconstant }f,\quad \text{and points}. } \]
Step 5 gives the exhaustive classification, and step 6 proves the additional cardinality assertion for the curve case.
8. Q.E.D.
Step 7 is the required conclusion.