Let \(I\) be an ideal in the polynomial ring \(\mathbb{C}[x_1, \ldots, x_n]\) for some \(n \geq 1\). Prove that the following conditions on \(I\) are all equivalent:
\(\mathbb{C}[x_1, \ldots, x_n]/I\) is a finite-dimensional \(\mathbb{C}\)-vector space;
\(I \cap \mathbb{C}[x_i] \neq (0)\) for all \(1 \leq i \leq n\);
The set of common zeroes in affine \(n\)-space \(\mathbb{C}^n\) of all of the polynomials in \(I\) is a finite (or empty) set.
Solution
Write \[ R:=\mathbb C[x_1,\ldots,x_n] \] and let \[ V(I):=\{a\in\mathbb C^n:f(a)=0\text{ for all }f\in I\}. \]
1. Condition (i) implies condition (ii).
Assume \[ R/I \] is finite-dimensional over \(\mathbb C\). Fix \(i\) and let \(\bar x_i\) be the image of \(x_i\) in \(R/I\). The infinite sequence \[ 1,\bar x_i,\bar x_i^2,\ldots \] lies in a finite-dimensional vector space, so it is linearly dependent. Hence there is a nonzero polynomial \[ f_i(T)=c_0+c_1T+\cdots+c_dT^d\in\mathbb C[T] \] such that \[ f_i(\bar x_i)=0 \] in \(R/I\). Equivalently, \[ f_i(x_i)\in I. \] Since \(f_i(x_i)\neq0\) in the polynomial ring \(\mathbb C[x_i]\), \[ I\cap\mathbb C[x_i]\neq(0). \] This holds for every \(i\).
2. Condition (ii) implies condition (i).
Assume that for each \(i\) there is a nonzero polynomial \[ f_i(x_i)\in I\cap\mathbb C[x_i]. \] Multiplying by a nonzero scalar, take each \(f_i\) monic, and write \[ d_i:=\deg f_i\ge0. \] If some \(d_i=0\), then \(f_i\) is a nonzero constant, so \(1\in I\) and \(R/I=0\), which is finite-dimensional. Thus suppose all \(d_i\ge1\).
Modulo \(I\), the relation \(f_i(x_i)=0\) expresses \[ x_i^{d_i} \] as a \(\mathbb C\)-linear combination of lower powers of \(x_i\). Repeated reduction therefore expresses every monomial in \(R/I\) as a linear combination of monomials \[ x_1^{e_1}\cdots x_n^{e_n} \qquad (0\le e_i<d_i). \] There are only \[ d_1\cdots d_n \] such monomials. Hence \(R/I\) is spanned by finitely many vectors over \(\mathbb C\), so it is finite-dimensional.
3. Condition (ii) implies condition (iii).
For each \(i\), choose \[ 0\neq f_i(x_i)\in I\cap\mathbb C[x_i]. \] If \[ a=(a_1,\ldots,a_n)\in V(I), \] then \[ f_i(a_i)=0 \] for every \(i\). Thus \(a_i\) must lie in the finite set of roots of \(f_i\). Consequently \[ V(I) \] is contained in the finite Cartesian product of those root sets. Therefore \(V(I)\) is finite, possibly empty.
4. Condition (iii) implies condition (ii).
Assume first that \[ V(I)=\{a^{(1)},\ldots,a^{(r)}\} \] is nonempty and finite, where \[ a^{(j)}=(a_1^{(j)},\ldots,a_n^{(j)}). \] For each \(i\), define the nonzero one-variable polynomial \[ g_i(T):=\prod_{j=1}^r(T-a_i^{(j)})\in\mathbb C[T]. \] Then \[ g_i(x_i) \] vanishes at every point of \(V(I)\). By Hilbert’s Nullstellensatz, \[ I(V(I))=\sqrt I. \] Hence \[ g_i(x_i)\in\sqrt I. \] Therefore some positive power satisfies \[ g_i(x_i)^{N_i}\in I. \] This is a nonzero element of \(\mathbb C[x_i]\), so \[ I\cap\mathbb C[x_i]\neq(0). \]
If \(V(I)=\varnothing\), the Nullstellensatz gives \[ \sqrt I=R. \] Hence \(1\in\sqrt I\), so \(1\in I\) and \(I=R\). Then \[ 1\in I\cap\mathbb C[x_i] \] for every \(i\). Thus (ii) holds in the empty case as well.
5. Therefore (i), (ii), and (iii) are equivalent.
By 1 and 2, \[ \text{(i)}\Longleftrightarrow\text{(ii)}. \] By 3 and 4, \[ \text{(ii)}\Longleftrightarrow\text{(iii)}. \] Hence all three conditions are equivalent.