RiemannRoch Theorem Exercises and Solutions

1. The theorem in a nutshell

Let \(C\) be a smooth projective curve over an algebraically closed field \(k\). For a divisor \(D\) on \(C\) the RiemannRoch theorem states

(D) - (K_C - D) = \deg D + 1 - g,        

where

• \((D) = \dim_k H^0(C,\mathcal O_C(D))\) is the dimension of the space of global sections,
• \(K_C\) is a canonical divisor,
• \(g\) is the genus of the curve.

Two immediate corollaries are:

• If \(\deg D > 2g-2\) then \((D)=\deg D+1-g\).
• If \(\deg D <0\) then \((D)=0\).

2. Strategy for solving the exercises

The exercises in Hartshorne 4.1 are designed to make the reader comfortable with the language of divisors, linear equivalence, and the use of the RiemannRoch formula. A few general tips:

1. Identify the genus. For most examples the genus is 0 (rational curve) or 1 (elliptic curve). The degree of a canonical divisor is \(2g-2\).
2. Use exact sequences. The short exact sequence

   0  O_C(DP)  O_C(D)  k(P)  0

   gives the adding a point trick: \((D) (D-P)+1\).
3. Apply Serre duality. In the curve case it reduces to \(H^1(C,\mathcal O_C(D))\cong H^0(C,\mathcal O_C(K_C-D))^\vee\), which is exactly the term \(\ell(K_C-D)\) appearing in the theorem.
4. Work with explicit equations when possible. For plane curves \(C\subset\mathbb P^2\) of degree \(d\) the canonical divisor is \((d-3)H|_C\), where \(H\) is a line.

3. Exercise1 Rational curves

Problem. Let \(C\cong \mathbb P^1\) and let \(D\) be a divisor of degree \(n\). Compute \((D)\).

Solution. The genus of \(\mathbb P^1\) is 0, so \(g=0\) and \(K_{\mathbb P^1}\sim -2\). The RiemannRoch formula becomes

(D) - (-2 - D) = n + 1.        

If \(n\ge0\) then \(-2-D\) has negative degree, hence \((-2-D)=0\). Therefore \((D)=n+1\). If \(n<0\) then \((D)=0\) because a divisor of negative degree on \(\mathbb P^1\) has no global sections. Consequently

(D)=\max\{0, n+1\}.        

4. Exercise2 Elliptic curves

Problem. Let \(C\) be a smooth cubic in \(\mathbb P^2\) (genus1). For a point \(P\in C\) compute \((mP)\) for any integer \(m\).

Solution. Here \(g=1\) and \(K_C\sim 0\). RiemannRoch reduces to

(mP) - (-mP) = m.        

When \(m>0\) the divisor \(-mP\) has negative degree, so \((-mP)=0\). Hence \((mP)=m\). For \(m=0\) we have \((0)=1\) (the constants). When \(m<0\) we switch the roles and obtain \((mP)=0\). In summary

(mP)=\begin{cases}m &\text{if } m>0,\\1 &\text{if } m=0,\\0 &\text{if } m<0.\end{cases}        

Because \(C\) is an elliptic curve, the linear series \(|mP|\) gives the complete linear system of degree \(m\) and dimension \(m-1\) for \(m\ge1\).

5. Exercise3 Canonical divisor on a plane curve

Problem. Let \(C\subset\mathbb P^2\) be a smooth plane curve of degree \(d\). Show that a canonical divisor is \((d-3)H|_C\) and compute \((K_C)\).

Solution. The adjunction formula gives

K_C = (K_{\mathbb P^2}+C)\big|_C = (-3H + dH)\big|_C = (d-3)H\big|_C .        

The degree of \(K_C\) is \((d-3)d\). Using RiemannRoch with \(D=K_C\) we have

(K_C) - (0) = \deg K_C + 1 - g .        

Since \((0)=1\) (constants) and \(g = \frac{(d-1)(d-2)}{2}\) (the genus of a plane curve), we obtain

(K_C) = (d-3)d + 1 - \frac{(d-1)(d-2)}{2} + 1 .        

Simplifying,

(K_C) = \frac{(d-1)(d-2)}{2} = g .        

Thus the space of global differential forms on \(C\) has dimension equal to its genus, as expected.

6. Exercise4 Special divisors

Problem. On a curve of genus \(g\ge2\) a divisor \(D\) is called special if \((K_C-D)>0\). Prove that if \(\deg D \le g-1\) then \(D\) is special.

Solution. Apply RiemannRoch to \(D\):

(D) - (K_C-D) = \deg D + 1 - g .        

If \(\deg D \le g-1\) then the righthand side is \(\le 0\). Since \((D)\ge0\), the only way for the equality to hold is that \((K_C-D)\ge1\); i.e. \(K_C-D\) has a nontrivial global section, so \(D\) is special.

7. Exercise5 Cliffords theorem

Problem. Let \(C\) be a smooth curve of genus \(g\ge1\) and let \(D\) be a special divisor with \((D)\ge2\). Prove Cliffords inequality

(D) \le \frac{\deg D}{2}+1 .        

Solution. Because \(D\) is special we have \((K_C-D)>0\). By RiemannRoch,

(D) = \deg D + 1 - g + (K_C-D) .        

Since \((K_C-D) \le \deg(K_C-D)+1 = (2g-2-\deg D)+1\), we obtain

(D)  \deg D + 1 - g + 2g-1-\deg D = g .        

Now write \((D) = r+1\) with \(r\ge1\). Substituting \(\deg D = 2r\) (the extremal case) gives the inequality. A more precise argument uses the fact that both \(D\) and \(K_C-D\) contribute to the series, leading to the bound

r  \frac{\deg D}{2},        

8. A worked example Hyperelliptic curves

Suppose \(C\) is a hyperelliptic curve of genus \(g\) with a degree2 map \(\pi:C\to\mathbb P^1\). Let \(Q\) be a branch point and set \(D = (g-1)Q\). Then \(\deg D = g-1\). Because \(\pi^*\mathcal O_{\mathbb P^1}(1)=\mathcal O_C(P+Q)\) for any two points in a fiber, one checks that

(D)=1,(K_C-D)=g-1 .        

Thus \(D\) is special (as predicted by exercise4) and equality in Cliffords inequality fails, illustrating the sharpness of the bound only for nonhyperelliptic curves.

9. Further reading

Hartshornes exposition is complemented by the following references:

• Arbarello, Cornalba, Griffiths, Harris Geometry of Algebraic Curves, vol.1.
• R.M.Friedman Algebraic Surfaces and Holomorphic Vector Bundles, Chapter4.
• J.H.Silverman The Arithmetic of Elliptic Curves (for concrete calculations on genus1 curves).