代数几何Prove that any finite set of points S⊂P² can
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代数几何
Prove that any finite set of points S⊂P² can be defined by two equations
Prove that any finite set of points S⊂P² can be defined by two equations
先考虑仿射情形.
对S = {(a[i],b[i]) | i = 1,2,...,n},若a[i]两两不等,则可取f(x) = ∏{1 ≤ i ≤ n} (x-a[i]),
g(x) = ∑{1 ≤ i ≤ n} b[i]·∏{j ≠ i} (x-a[j])/(a[i]-a[j]) (Lagrange插值多项式).
则f(x) = 0,f(x)+g(x) = y恰好给出S.
为处理a[i]相等情形,取适当的k使a[i]-k·b[i]两两不等(不适用有限域),进行坐标变换x' = x-ky即可.
尝试用上述方程的齐次化来处理射影情形.
注意到deg(f) = deg(g) = n (这是特意取f(x)+g(x)的用意).
当n > 1时,齐次化方程有一个无穷远点(0:1:0).
为此可取适当坐标变换,使S中恰有一个无穷远点(0:1:0) (不适用有限域),
然后对其余n-1个点用仿射情形结论(这要求n-1 > 1,即n > 2).
对n = 1,2可直接构造.
有限域情形没有细想,总觉得可能不成立.
对S = {(a[i],b[i]) | i = 1,2,...,n},若a[i]两两不等,则可取f(x) = ∏{1 ≤ i ≤ n} (x-a[i]),
g(x) = ∑{1 ≤ i ≤ n} b[i]·∏{j ≠ i} (x-a[j])/(a[i]-a[j]) (Lagrange插值多项式).
则f(x) = 0,f(x)+g(x) = y恰好给出S.
为处理a[i]相等情形,取适当的k使a[i]-k·b[i]两两不等(不适用有限域),进行坐标变换x' = x-ky即可.
尝试用上述方程的齐次化来处理射影情形.
注意到deg(f) = deg(g) = n (这是特意取f(x)+g(x)的用意).
当n > 1时,齐次化方程有一个无穷远点(0:1:0).
为此可取适当坐标变换,使S中恰有一个无穷远点(0:1:0) (不适用有限域),
然后对其余n-1个点用仿射情形结论(这要求n-1 > 1,即n > 2).
对n = 1,2可直接构造.
有限域情形没有细想,总觉得可能不成立.
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