打字猴:1.701048418e+09
1701048418 ∀f∈H*,r∈R,则rf∈H*,规定为
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1701048420 (rf)(h)=rf(h), ∀h∈H.
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1701048422 不难验证,在这两种运算下H*为实线性空间.
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1701048429 设φ∶H1→H2是同态,则规定为不难验证φ*是线性映射.容易看出,若φ是同构,则φ*也是同构.
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1701048431 例如,Z*=R(1维实线性空间);设Q为有理数加群,Q*=R;若H的每个元素都是有限阶的,则H*=0(习题6).
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1701048434 命题A.5 (1)
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1701048437 (2)若H0是H的子群,并且H/H0的每个元素都是有限阶的,则
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1701048443 证明 (1)规定如下:令ζ(f1,f2)∈(H1H2)*为
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1701048446  ζ(f1,f2)(h1,h2)=f1(h1)+f2(h2),∀(h1,h2)∈H1H2,则ζ是线性映射.若ζ(f1,f2)=0,则∀h1∈H1,
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1701048448 f1(h1)=ζ(f1,f2)(h1,0)=0,
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1701048453 因此f1=0.同理f2=0,从而(f1,f2)=0,这说明ζ是单的.设g∈(H1H2)*,由f1(h1)=g(h1,0)规定由f2(h2)=g(0,h2)规定则ζ(f1,f2)=g.从而ζ是满的.于是ζ是同构.
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1701048456 (2)记i∶H0→H是包含映射.下面证是同构.
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1701048458 若i*(f)=0,即∀h0∈H0,f(h0)=0.由H/H0的元素是有限阶的,知C(H0)=H,即∀h∈H,存在r∈N,使得rh∈H0.于是rf(h)=f(rh)=0,从而f(h)=0.由h的任意性得出f=0.证明了i*是单的.
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1701048464 设令rh=min{r∈N|rh∈H0},规定f∶H→R为如果rh∈H0,则rh|r,从而由此事实不难验证f∈H*.显然i*(f)=g.i*是满的. ▎
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1701048466 定义A.3 当H是有限生成交换群时,称线性空间H*的维数为交换群H的秩,记作rankH.
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