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| #import "../../template.typ": * | ||
| #import "@preview/commute:0.2.0": node, arr, commutative-diagram | ||
|
|
||
| #show: note.with( | ||
| title: "作业12", | ||
| author: "YHTQ", | ||
| date: none, | ||
| logo: none, | ||
| withOutlined : false, | ||
| withTitle :false, | ||
| withHeadingNumbering: false | ||
| ) | ||
| = ex 1.5 | ||
| == 1 | ||
| #let CC = $cone(C)$ | ||
| 注意到： | ||
| $ | ||
| d^CC_n vec(c_(n-1), c_n) = vec(-d_(n-1) (c_(n-1)), d_n (c_n) - c_(n-1)) | ||
| $ | ||
| 由于它确实是复形，验证正合只需任取 $vec(c_(n-1), c_n) in ker d^CC_n$，验证它在 $im d^CC_(n+1)$ 中即可。事实上： | ||
| $ | ||
| vec(-d_(n-1) (c_(n-1)), d_n (c_n) - c_(n-1)) = 0 <=> c_(n-1) in ker d_(n-1) sect im d_n, c_(n-1) = d_n (c_(n))\ | ||
| d^CC_(n+1) vec(-c_(n), 0) = vec(d_(n) (c_(n)), c_(n)) = vec(c_(n-1), c_(n) ) | ||
| $ | ||
| 上两式表明： | ||
| $ | ||
| forall vec(c_(n-1), c_n) in ker d^CC_n, vec(c_(n-1), c_(n) ) = d^CC_(n+1) vec(-c_(n), 0) in im d^CC_(n+1) | ||
| $ | ||
| 足以给出正合性，同时表明题设的 $s$ 满足： | ||
| $ | ||
| d s|_(ker d) = id \ | ||
| d s|_(ker d) d = d \ | ||
| d s|_(im d) d = d \ | ||
| d s d = d \ | ||
| $ | ||
| 表明确实是分裂映射 | ||
| == 2 | ||
| 零调的定义是存在 $s$ 满足 | ||
| $ | ||
| f = d_D s + s d_C | ||
| $<def-null> | ||
| 而题设映射是复形间态射等价于： | ||
| $ | ||
| (-s, f) mat(-d_C, 0;-id, d_C) vec(c', c) = d_D (-s, f) vec(c', c)\ | ||
| (s d_C - f, f d_C) vec(c', c) = d_D (-s, f) vec(c', c)\ | ||
| $ | ||
| 表明： | ||
| $ | ||
| cases( | ||
| s d_C - f = -d_D s,\ | ||
| f d_C = d_D f | ||
| ) | ||
| $ | ||
| 其中 $f d_C = d_D f$ 来源与复形同态的定义，当然有上式与@def-null 等价 | ||
|
|
||
| == 3 | ||
| $f tilde g$ 的定义是存在 $s$ 使得： | ||
| $ | ||
| f - g = d_D s + s d_C | ||
| $<def-hom> | ||
| 而题设映射是复形间同态等价于： | ||
| $ | ||
| (f, s, g)mat(d_C, id, 0;0, -d_C, 0;0, -id, d_C) = d_D (f, s, g) | ||
| $ | ||
| 也即： | ||
| $ | ||
| cases( | ||
| f d_C = d_D f, | ||
| g d_C = d_D g, | ||
| f - s d_C - g = d_D s | ||
| ) | ||
| $ | ||
| 其中前两者是复形同态的定义，当然与@def-hom 等价 | ||
| = ex 2.1 | ||
| == 2 | ||
| 只证明同调 $delta-$函子情形，余调是类似的\ | ||
| 任给 $delta-$函子 $S$ 和自然变换 $f_0 :S_0 -> T_0$，既然 $T_n = 0$，当然 $f_n : S_n -> T_n$ 在 $n >= 1$ 时取并且只能取零映射，这就给出了泛性质 | ||
| = ex 2.2 | ||
| == 1 | ||
| - 首先，假设 $P$ 是 $C$ 中投射对象。显然它的每一项都是投射对象，由前面的命题只需证明存在 $s: P -> P$ 使得 $(-s, id): cone(P) -> P$ 是复形间的同态。 | ||
| 不难发现有交换图： | ||
| #align(center)[#commutative-diagram( | ||
| node((0, 0), $...$, 1), | ||
| node((0, 1), $P_(n-1) directSum P_n$, 2), | ||
| node((0, 2), $P_(n-2) directSum P_(n-1)$, 3), | ||
| node((1, 0), $...$, 4), | ||
| node((1, 1), $P_(n-1)$, 5), | ||
| node((1, 2), $P_(n-2)$, 6), | ||
| arr(1, 2, $$), | ||
| arr(2, 3, $mat(-d_C, 0;-id, d_C)$), | ||
| arr(4, 5, $$), | ||
| arr(5, 6, $-d_(n-1)$), | ||
| arr(1, 4, $$), | ||
| arr(2, 5, $id directSum 0$), | ||
| arr(3, 6, $id directSum 0$),)] | ||
| 换言之，$cone(P) -> P$ 间存在满同态，由投射对象的性质，将有交换图： | ||
| $ | ||
| #align(center)[#commutative-diagram( | ||
| node((0, 0), $cone(P)$, 1), | ||
| node((0, 1), $P$, 2), | ||
| node((1, 0), $P$, 3), | ||
| arr(1, 2, $id directSum 0$), | ||
| arr(3, 2, $id$), | ||
| arr(3, 1, $exists s'$),)] | ||
| $ | ||
| 设 $s' = vec(s'_(1), s'_(2))$，交换图给出： | ||
| $ | ||
| s'_(1) = id: P_(n-1) -> P_(n-1)\ | ||
| $ | ||
| 是复形间同态给出： | ||
| $ | ||
| mat(-d, 0;-id, d) vec(s'_(1), s'_2) = vec(s'_(1), s'_2)( -d) \ | ||
| -id + d s'_2 = -s'_2 d\ | ||
| d s'_2 + s'_2 d = id | ||
| $ | ||
| 由熟知的定理，这给出 $P$ 确实是分裂的正合列 | ||
| - 反之，假设 $P$ 是分裂的投射对象构成的正合列，任给 $X ->^f Y$ 是满射和 $P ->^g Y$，至少存在逐项的同态 $h: P_n -> X_n$ 使得： | ||
| $ | ||
| f h = g | ||
| $ | ||
| 注意到： | ||
| $ | ||
| id = d s + s d\ | ||
| $ | ||
| 令 $h' = d h s + h s d$，将有： | ||
| $ | ||
| f h' = f d h s + f h s d = d f h s + f h s d = d g s + g s d = g(d s + s d) = g\ | ||
| d h' = d h s d = h' d | ||
| $ | ||
| 表明 $h'$ 就是要找的复形间同态 | ||
| == 2 | ||
| 设有复形 $X_i$，并且 $X_i$ 分别成为投射对象 $P_i$ 的满射像。设 $pi: P_i -> X_i$ 是满射 | ||
| 断言有交换图表： | ||
| #align(center)[#commutative-diagram( | ||
| node((0, 0), $X_i$, 1), | ||
| node((0, 1), $X_(i-1)$, 2), | ||
| node((1, 0), $P_i directSum P_(i+1)$, 3), | ||
| node((1, 1), $P_(i-1) directSum P_i$, 4), | ||
| arr(1, 2, $d$), | ||
| arr(3, 1, $(pi, d pi)$), | ||
| arr(4, 2, $(pi, d pi)$), | ||
| arr(3, 4, $mat(0, 0;1, 0)$),)] | ||
| 不难验证下面一行是投射模构成的链复形，且： | ||
| $ | ||
| d(pi, d pi) = mat(-, 0;1, 0)(pi, d pi) | ||
| $ | ||
| 因此交换图表成立，这是链复形之间的满态射。此外，取： | ||
| $ | ||
| s = mat(0, 1;0, 0): P_(i-1) directSum P_i -> P_i directSum P_(i+1) | ||
| $ | ||
| 不难验证将有： | ||
| $ | ||
| d s d = d | ||
| $ | ||
| 因此下面一行构成分裂的正合列，上面结论表明它是投射对象，证毕 | ||
| = ex 2.3 | ||
| == 1 | ||
| #let xl = $overline(x)$ | ||
| #let yl = $overline(y)$ | ||
| #let ul = $overline(u)$ | ||
| #let vl = $overline(v)$ | ||
| #let dl = $overline(d)$ | ||
| #let pl = $overline(p)$ | ||
| 由 Baer 法则，只需验证任给 $R = ZZ quo m ZZ$ 的理想 $J$ 以及 $f: J -> R$ 都可以延拓到 $R -> R$\ | ||
| 事实上，注意到 $R$ 是主理想环，可设 $J = (xl)$ 以及 $f(xl) = yl$（此时 $f$ 被唯一确定）\ | ||
| 由于 $f$ 是加法循环群之间的同态，将有： | ||
| $ | ||
| yl^(ord(xl)) = 0 => ord(yl) | ord(xl) | ||
| $ | ||
| 根据循环群的结论，$RR$ 中阶为 $ord(xl)$ 子群 $(xl)$ 恰有一个，而 $(yl)$ 只能是其子群，进而可设： | ||
| $ | ||
| yl = ul xl | ||
| $ | ||
| 因此： | ||
| $ | ||
| f' := [t: R -> ul t] | ||
| $ | ||
| 就是 $f$ 的一个延拓，证毕 | ||
|
|
||
| 对于第二个结论，假设 $d|m, p | m, m /d$，则 $ ZZ quo d ZZ$ 是 $R$ 的一个理想，也是加法子群，进而可设为唯一的 $d$ 阶子群 $(xl)$，同时设 $m/d$ 阶子群为 $(yl)$，$p$ 阶子群为 $(pl)$，则有： | ||
| $ | ||
| (pl) subset (xl), (yl) | ||
| $ | ||
| 考虑群同态（也是模同态）： | ||
| $ | ||
| funcDef(f, (yl), (xl), yl, pl) | ||
| $ | ||
| 假设它是 $f': R -> (xl)$ 的一个限制，并设： | ||
| $ | ||
| f'(1) = ul xl | ||
| $ | ||
| 则 $f'(yl) = ul xl yl = pl$\ | ||
| 然而注意到： | ||
| $ | ||
| d = ord(xl) = m/(gcd(x, m))\ | ||
| m/d = ord(yl) = m/(gcd(y, m))\ | ||
| gcd(x, m) gcd(y, m) = m \ | ||
| m/(gcd(x, m)) = gcd(y, m) | y => m | y gcd(x, m) = gcd(x y, m) => gcd(x y, m) = m | ||
| $ | ||
| 意味着 $xl yl = 0 => pl = 0$，矛盾！ | ||
| == 2 | ||
| 设 $a in A, e_A (a) = 0$，由定义，这将意味着： | ||
| $ | ||
| forall f in Hom(A, QQ quo ZZ), f(a) = 0 | ||
| $ | ||
| 然而考虑映射： | ||
| $ | ||
| funcDef(h, (a), QQ quo ZZ, a, 1/2) | ||
| $ | ||
| 只要 $a != 0$ 上式就是群同态\ | ||
| 由内射对象的性质，有交换图： | ||
| #align(center)[#commutative-diagram( | ||
| node((0, 0), $(a)$, 1), | ||
| node((0, 1), $A$, 2), | ||
| node((1, 0), $QQ quo ZZ$, 3), | ||
| arr(1, 2, $id$, inj_str), | ||
| arr(2, 3, $exists h'$), | ||
| arr(1, 3, $h$),)] | ||
| 显然 $h' in Hom(A, QQ quo ZZ), h'(a) !=0$，这表明 $e_A (a) = 0$ 除非 $a = 0$，也即 $e_A$ 是单射 | ||
| == 3 | ||
| $A = 0$ 时显有 $Hom(A, QQ quo ZZ) = 0$，反之由于上面给出了 $A -> Hom(QQ quo ZZ)$ 的单射，当然后者为零意味着前者为零 |
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