Use the universal coefficient theorem to show that if \(H_*(X; \mathbb{Z})\) is finitely generated, so the Euler characteristic \(\chi(X) = \sum_n (-1)^n \operatorname{rank} H_n(X; \mathbb{Z})\) is defined, then for any coefficient field \(F\) we have \(\chi(X) = \sum_n (-1)^n \dim H_n(X; F)\).
Solution
Write each finitely generated integral homology group as \[ H_n(X;\mathbb Z)\cong \mathbb Z^{r_n}\oplus T_n, \] with \(T_n\) finite. We compare the alternating sum of the dimensions of homology over a field \(F\) with \(\sum(-1)^nr_n\).
If \(\operatorname{char}F=0\), then \(F\) is torsionfree as a \(\mathbb Z\)-module, so the homology universal coefficient theorem gives \[ H_n(X;F)\cong H_n(X;\mathbb Z)\otimes F\cong F^{r_n}. \] Thus \(\dim_FH_n(X;F)=r_n\) in every degree.
Now suppose \(\operatorname{char}F=p\). Since \(F\) is a vector space over \(\mathbb F_p\), it is enough to compute with \(\mathbb F_p\) and then extend scalars. Let \[ t_n=\dim_{\mathbb F_p}(T_n/pT_n). \] For a finite abelian group the \(p\)-torsion subgroup \(T_n[p]\) has the same \(\mathbb F_p\)-dimension \(t_n\). The universal coefficient short exact sequence gives \[ 0\to H_n(X;\mathbb Z)\otimes\mathbb F_p \to H_n(X;\mathbb F_p) \to \operatorname{Tor}(H_{n-1}(X;\mathbb Z),\mathbb F_p)\to0. \] Hence \[ \dim_{\mathbb F_p}H_n(X;\mathbb F_p)=r_n+t_n+t_{n-1}. \] Taking the alternating sum, the two torsion contributions cancel: \[ \sum_n(-1)^n(t_n+t_{n-1})=0. \] Therefore \[ \sum_n(-1)^n\dim_FH_n(X;F) =\sum_n(-1)^nr_n =\chi(X). \]