Let \(V\) be an \(n\)-dimensional vector space over a field of characteristic \(0\). Prove that \(V\) is not the union of finitely many subspaces of dimension \(n-1\).
Solution
Let \(F\) be the base field of \(V\). The case \(n=0\) is immediate, so assume \(n\ge1\).
1. The field \(F\) is infinite.
Since \(F\) has characteristic \(0\), the elements \[ 0,1,1+1,1+1+1,\ldots \] are pairwise distinct.
2. If \(H\subset V\) has dimension \(n-1\), then there is a nonzero linear functional\[ \lambda_H:V\to F \]such that\[ H=\ker\lambda_H. \]
Choose a basis \(e_1,\ldots,e_{n-1}\) of \(H\) and extend it to a basis \[ e_1,\ldots,e_{n-1},e_n \] of \(V\). Define \(\lambda_H(e_i)=0\) for \(i<n\) and \(\lambda_H(e_n)=1\). Then \(\lambda_H\) is nonzero and its kernel is exactly \(H\).
3. For any finite collection of hyperplanes \(H_1,\ldots,H_m\subset V\), the function\[ P\coloneqq\prod_{j=1}^m\lambda_{H_j} \]is a nonzero polynomial function on \(V\) and vanishes on \(H_1\cup\cdots\cup H_m\).
Choose coordinates \(V\cong F^n\). Each \(\lambda_{H_j}\) is then a nonzero linear polynomial in \(F[x_1,\ldots,x_n]\). Since this polynomial ring is an integral domain, their product \(P\) is nonzero. If \(v\in H_j\), then \(\lambda_{H_j}(v)=0\), so \(P(v)=0\).
4. A nonzero polynomial in \(F[x_1,\ldots,x_n]\) does not vanish at every point of \(F^n\).
1. For \(n=1\), a nonzero polynomial has only finitely many roots. For \(n>1\), write a nonzero polynomial \(Q\) as\[ Q(x_1,\ldots,x_n) = \sum_{k=0}^d Q_k(x_1,\ldots,x_{n-1})x_n^k \]with some \(Q_k\ne0\). By induction there is\[ a\in F^{n-1} \]for which that coefficient \(Q_k(a)\) is nonzero. Hence \(Q(a,x_n)\) is a nonzero one-variable polynomial, so it is nonzero for some \(x_n\in F\).
5. The space \(V\) is not a finite union of \((n-1)\)-dimensional subspaces.
4. Therefore\[ \boxed{V\ne H_1\cup\cdots\cup H_m}. \]
6. Q.E.D.
Step 5 proves the assertion for every finite family of hyperplanes.