## FANDOM

36 Pages

Springer GTM 52.

"This book provides an introduction to abstract algebraic geometry using the methods of schemes and cohomology."

Exercise Solutions Available:

# Chapter I: Varieties Edit

## Section I.1: Affine Varieties Edit

• Height of a prime ideal: Height of a prime ideal is like codimension of a subvariety.

## Section I.5: Nonsingular Varieties Edit

• One can show easily that [the Jacobian] definition of nonsingularity is independent of the choice of generators of the ideal: The point is that what we're really calculating is the dimension of the space spanned by the vectors $((\partial f / \partial x_1)(P), \ldots, (\partial f / \partial x_n)(P))$ for each $f \in I$. Suppose that $I = (f_1, \ldots, f_m)$ and $f \in I$; then $f = \sum_i p_i f_i$ for some $p_i \in k[x_1, \ldots, x_n]$, and so we have $\frac{\partial f}{\partial x_i}(P) = \sum_i \frac{\partial f_i}{\partial x_i}(P) p_i(P) + f_i(P) \frac{\partial p_i}{\partial x_i}(P)$. Since $f_i \in I$ and $P \in V(I)$, we have $f_i(P) = 0$ by definition, so this reduces to $\frac{\partial f}{\partial x_i}(P) = \sum_i \frac{\partial f_i}{\partial x_i}(P) p_i(P)$. Clearly, then, the vector $((\partial f / \partial x_1)(P), \ldots, (\partial f / \partial x_n)(P))$ lies in the span of the vectors $((\partial f_j / \partial x_1)(P), \ldots, (\partial f_j / \partial x_n)(P))$ for $j = 1, \ldots, m$.
• Example I.5.6: We need an additional commutative algebra result not found in the chapter to make sense of this, namely that $(k[x_1, \ldots, x_n]/I)_{\mathfrak{m}} \cong k[[x_1, \ldots, x_n]]/Ik[[x_1, \ldots, x_n]]$, where $\mathfrak{m} = (x_1, \ldots, x_n)$. See Chapter 7 of Eisenbud - Commutative Algebra - with a View Toward Algebraic Geometry.

# Chapter II: Schemes Edit

## Section II.3: First Properties of Schemes Edit

• Proposition II.3.1: Then $\mathcal{O}(U_1 \cup U_2) = \mathcal{O}(U_1) \times \mathcal{O}(U_2)$ which is not an integral domain. A product of nontrivial rings can never be an integral domain: (a, 0) * (0, b) = (0, 0), so (a, 0) and (0, b) are zero divisors. (In some sense it seems like Proposition 3.1 is a partial converse to this.)

## Section II.6: Divisors Edit

• Page 129: For each line L in P^2, we consider $L \cap C$ which is a finite set of points on C. Note that the "points" mentioned here are intrinsic points of C, not points of P^2 on the embedded copy of C.
• Proof of Proposition II.6.2: It is well-known that a UFD is integrally closed. This is a consequence of the rational root theorem.

# Chapter IV: Curves Edit

## Section IV.1: Riemann-Roch TheoremEdit

• Proof of Theorem IV.1.3: we have an exact sequence $0 \to \mathcal{L}(-P) \to \mathcal{O}_X \to k(P) \to 0$

This is an issue that comes up all over the place. It makes sense to talk about k(P), a k-valued skyscraper sheaf supported at P, only when we're thinking about k(P) as a sheaf of rings, like when we're thinking of it as the structure sheaf of P. In the exact sequence above, though, we're regarding it not as a sheaf of rings but as a sheaf of $\mathcal{O}_X$-modules, and in this case writing k(P) isn't that helpful because it doesn't really specify the $\mathcal{O}_X$-module structure. In this case, $\mathcal{O}_X$ acts on $k(P)$ by $f \cdot \alpha = \alpha f(P)$. This clarifies why $\mathcal{O}_X \otimes k(P) = k(P)$: we can simply move functions on the left over to the right side of the tensor product.

• Proof of Theorem IV.1.3:$\chi(k(P)) = 1$: Recall that k(P) here really means $j_\ast \mathcal{O}_P$, where j denotes the inclusion $j : P \hookrightarrow X$. By Lemma III.2.10, $H^i(X, j_\ast \mathcal{O}_P) = H^i(P, \mathcal{O}_P)$, so it follows that $\chi(j_\ast \mathcal{O}_P) = \chi(\mathcal{O}_P)$. By Grothendieck vanishing (Theorem III.2.7), a sheaf on the zero-dimensional space P only has zeroth cohomology, so $\chi(\mathcal{O}_P) = h^0(P, \mathcal{O}_P) = 1$.

# Chapter V: Surfaces Edit

## Section V.1: Geometry on a SurfaceEdit

• Page 357: This implies, by the way, that C and D are each nonsingular at P:

Since the maximal ideal of $\mathcal{O}_{D, P}$ is generated by f, {f} is a regular system of parameters. Since its cardinality is equal to the Krull dimension $\dim \mathcal{O}_{D, P} = 1$, $\mathcal{O}_{D, P}$ is a regular local ring.