湖南师范大学 2025年高等代数第2题

由于 $h(\mathscr{A}) = f(\mathscr{A})g(\mathscr{A}) = g(\mathscr{A})f(\mathscr{A})$，对任意 $x \in h(\mathscr{A})V$，存在 $w \in V$ 使得 $x = f(\mathscr{A})g(\mathscr{A})w$。则 $x = f(\mathscr{A})(g(\mathscr{A})w) \in f(\mathscr{A})V$，且 $x = g(\mathscr{A})(f(\mathscr{A})w) \in g(\mathscr{A})V$，所以 $h(\mathscr{A})V \subseteq f(\mathscr{A})V \cap g(\mathscr{A})V$。因此只需讨论反向包含成立的条件。

设 $d(t) = \gcd(f(t), g(t))$，则存在多项式 $u(t), v(t)$ 使得 $d(t) = u(t)f(t) + v(t)g(t)$。将 $\mathscr{A}$ 代入得 $d(\mathscr{A}) = u(\mathscr{A})f(\mathscr{A}) + v(\mathscr{A})g(\mathscr{A})$。

若 $\gcd(f,g)=1$，则 $d(t)=1$，存在 $u,v$ 使得 $u(t)f(t)+v(t)g(t)=1$。对任意 $x \in f(\mathscr{A})V \cap g(\mathscr{A})V$，存在 $y,z \in V$ 使得 $x = f(\mathscr{A})y = g(\mathscr{A})z$。则 $$x = (u(\mathscr{A})f(\mathscr{A})+v(\mathscr{A})g(\mathscr{A}))x = u(\mathscr{A})f(\mathscr{A})x + v(\mathscr{A})g(\mathscr{A})x$$ $$= u(\mathscr{A})f(\mathscr{A})g(\mathscr{A})z + v(\mathscr{A})g(\mathscr{A})f(\mathscr{A})y = f(\mathscr{A})g(\mathscr{A})(u(\mathscr{A})z + v(\mathscr{A})y) \in h(\mathscr{A})V.$$ 因此 $f(\mathscr{A})V \cap g(\mathscr{A})V \subseteq h(\mathscr{A})V$，结合第一步得等式成立。

反证法。假设 $\gcd(f,g) \neq 1$，则存在非常数多项式 $d(t)$ 整除 $f$ 和 $g$。设 $f = d f_1$，$g = d g_1$，则 $h = d^2 f_1 g_1$。考虑 $\mathscr{A}$ 的若尔当标准型，取 $\lambda$ 为 $d(t)$ 的根，则存在特征向量 $v \neq 0$ 使得 $\mathscr{A}v = \lambda v$。于是 $f(\mathscr{A})v = f(\lambda)v = 0$，$g(\mathscr{A})v = g(\lambda)v = 0$，故 $v \in \ker f(\mathscr{A}) \cap \ker g(\mathscr{A})$。但 $v$ 不一定属于 $h(\mathscr{A})V$，因为 $h(\mathscr{A})V$ 是像空间。例如，取 $\mathscr{A}$ 为二维幂零若尔当块 $J_2(0)$，$f(t)=t$，$g(t)=t$，则 $f(\mathscr{A})V = g(\mathscr{A})V = \langle e_1 \rangle$，而 $h(\mathscr{A})V = \mathscr{A}^2 V = 0$，等式不成立。因此 $f$ 与 $g$ 必须互素。

综上所述，$f(\mathscr{A})V \cap g(\mathscr{A})V = h(\mathscr{A})V$ 当且仅当 $f(t)$ 与 $g(t)$ 互素，即 $\gcd(f,g)=1$。