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vaia.com有大量「似乎正确」的解答，仔细看时，发现都是乱写的（批量生成的？），例如：
https://www.vaia.com/en-us/textb ... rnamedegd2-g-2-and/
Question: Show that a divisor D with degree \(2g - 2\) and a dimension of the Riemann-Roch space associated with the divisor equal to g is a canonical divisor.
Step by step solution
01 Apply the Riemann-Roch theorem to the given divisor D
We are given that the degree of D is \(2g - 2\), and \(l(D) = g\). We can now apply the Riemann-Roch theorem to the divisor D:\(l(D) - l(K_C - D) = \operatorname{deg}(D) + 1 - g\)Now, substitute the given information:\(g - l(K_C - D)= (2g - 2) + 1 - g\)
02 Simplify the equation
Simplify the equation to isolate \(l(K_C - D)\):\(l(K_C - D) = g - (2g - 2) + 1 - g = -g + 2 + 1\)
03 Evaluate the dimension of the Riemann-Roch space associated with K_C - D
We know that the dimension of a Riemann-Roch space is non-negative, so \(l(K_C - D) \geq 0\). From the previous step, we have found that \(l(K_C - D) = -g + 3\). Since \(l(K_C - D) \geq 0\), it must be the case that \(-g + 3 \geq 0\), which leads to:\(g \leq 3\)Now recall that \(l(D) = g\). Therefore, \(l(K_C - D) = g - 3\).
04 Apply Riemann-Roch theorem to K_C - D
Since we know \(l(K_C - D) = g - 3\), we can apply the Riemann-Roch theorem to the divisor \(K_C - D\):\(l(K_C - D) - l(K_C - (K_C - D)) = \operatorname{deg}(K_C - D) + 1 - g\)\(l(K_C - D) - l(D) = \operatorname{deg}(K_C - D) + 1 - g\)Substitute the values we found in the previous steps:\((g - 3) - g = \operatorname{deg}(K_C - D) + 1 - g\)
05 Simplify the equation and find the degree of K_C - D
Simplify the equation to find the degree of \(K_C - D\):\(-3 = \operatorname{deg}(K_C - D) + 1 - g\)\(\operatorname{deg}(K_C - D) = g -4\)
06 Show that D is a canonical divisor
Now recall that the degree of a canonical divisor is \(2g - 2\). By substituting our result for the degree of \(K_C - D\), we can show that D is a canonical divisor:\(\operatorname{deg}(D) = \operatorname{deg}(K_C) - (\operatorname{deg}(K_C - D))\)\(\operatorname{deg}(D) = (2g - 2) - (g - 4)\)\(\operatorname{deg}(D) = 2g - 2\)This shows that divisor D has the degree of a canonical divisor, and since \(l(D) = g\), D is a canonical divisor. This completes the proof, and the given properties characterize canonical divisors. |
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