# [ag/0] Notes from Hartshorne

Paris, Chennai

Modern algebraic geometry is built on the theory of schemes and cohomology. The subject is intricately linked with commutative algebra and homological algebra.

## Affine varieties

Let $k$ be an algebraically closed field. Then affine variety over an $n$-space, $A_k^n$, is defined as the set of $n$-tuples over the set $k$. $P = (a_1, \ldots, a_n)$ is called a point in $A^n$, and $a_i$s are called coordinates.

Let $A = k[x_1, \ldots, x_n]$ be a polynomial ring in $n$-space over $k$. Elements of $A$ are then functions from affine $n$-space to $k$, defined by $f(P) = f(a_1, \ldots, a_n)$, where $f \in A, P \in A^n$. The zero-set of $T \in A$ is defined as the set of simultaneous zeros of $f$:
$$Z(T) = \{P \in A^n \mid f(P) = 0 \quad \forall f \in T\}$$

Note that if $a$ is an ideal, then $Z(a) = Z(T)$.

The set $Y \in A^n$ is called an algebraic set if there exists $T \subseteq A$ such that $Y = Z(T)$. The union and intersection of any two algebraic sets is an algebraic set, as is the empty set, and the whole space.

Define the Zariski topology in $A^n$ by taking, as open subsets, the complement of algebraic sets. Considering the Zariski topology on the affine line $A^1$, we note that every ideal $A = k[x]$ is principal, so every algebraic set is the set of zeros of a single polynomial. Thus algebraic sets in $A^1$ are just the finite subsets and the whole space; in particular, this is not Hausdorff.

The nonempty subset $Y$ of a topological space $X$ is irreducible if it cannot be expressed as the union, $Y = Y_1 \cup Y_2$, of two proper closed sets. The empty set is not irreducible, but $A^1$ is.

An affine variety can then be expressed as an irreducible subset of $A^n$. An open subset of an affine variety is called quasi-affine.

An ideal $I(Y)$ for $Y \subseteq A^n$ is defined as:
$$I(Y) = \{f \in A \mid f(P) = 0 \quad \forall P \in Y\}$$

So, we have the functions $Z$ and $I$, $Z$ mapping subsets of $A$ to an algebraic set, and $I$ mapping the subsets of $A^n$ to ideals. Their properties can be summarized as follows:

1. If $T_1 \subseteq T_2$, then $Z(T_2) \subseteq Z(T_1)$.
2. If $Y_1 \subseteq Y_2$, then $I(Y_2) \subseteq I(Y_1)$.
3. $I(Y_1 \cup Y_2) = I(Y_1) \cap I(Y_2)$.
4. $I(Z(a)) = \sqrt{a}$, a direct consequence of the Nullstellansatz.
5. $Z(I(Y)) = \bar{Y}$, the closure of $Y$.

There is a one-to-one correspondence between the algebraic sets in $A^n$ and the radical ideals in $A$, given by $Y \mapsto I(Y)$, and $a \mapsto Z(a)$. Furthermore, the set is irreducible iff its ideal is a prime ideal. For example, $A^n$ is irreducible, since it corresponds to the zero ideal in $A$.

If $f$ is an irreducible polynomial in $k = A[x_1, \ldots, x_n]$, then the affine variety is defined as $Y = Z(f)$, an $n$-dimensional hypersurface.

Let $Y \subseteq A^n$ be an affine algebraic set. Then, the affine coordinate ring, $A(Y)$, is defined to be $A/I(Y)$. Furthermore, if $Y$ is an affine variety, then $A(Y)$ is an integral domain, and $A(Y)$ is also a finitely generated $k$-algebra.

A topological space is termed noetherian if it satisfies the descending chain condition. For instance, $A^n$ is noetherian. In this space, every subspace $Y$ can be expressed as a union of irreducible subspaces, $Y = Y_1 \cup \ldots \cup Y_n$. If we also require that $Y_i \not\subseteq Y_j$ for $i \neq j$, then $Y_i$ is uniquely determined; they are termed the irreducible components of $Y$.

Every algebraic set of $A^n$ can be expressed uniquely as a union of varieties, with no single variety containing the other.

The dimension of an affine or quasi-affine variety is equal to the dimension of the topological space, and this is exactly the supremum integer $n$ in $Z_0 \subset \ldots \subset Z_n$, the irreducible closed subsets of the topological space. As an example, dimension of $A^n$ is $n$.

The height of a prime ideal $p$ is defined as the supermum of $n$ in the chain of distinct prime ideals $p_0 \subset p_1 \subset \ldots \subset p_n = p$. The Krull dimension of $A$ is the supremum of heights of the prime ideals in $A$.

The dimension of an affine algebraic set, $\text{dim } Y$, is equal to the dimension of its affine coordinate ring $A(Y)$.

Let $k$ be a field, and $B$ be an integral domain which is a finitely-generated $k$-algebra. Then:

1. Dimension of $B$ is equal to the transcendence degree of the quotient field $K(B)$ of $B$ over $k$.
2. For any prime ideal $p \in B$, we have $\text{height } p + \text{dim}(B \mathbin{/} p) = \text{dim}(B)$.

If $Y$ is a quasi-affine variety, then $\text{dim}(Y) = \text{dim}(\bar{Y})$. Let $Z_0 \subset \ldots \subset Z_n$ be a sequence of distinct closed irreducible sets in $Y$, $\bar{Z_0} \subset \ldots \subset \bar{Z_n}$ be that of $\bar{Y}$. $Y$ is finite, so we can choose a maximal $n = \text{dim}(Y)$. In that case, $Z_0$ must be a point $P$, and the chain will also be maximal. Now, $P_0$ corresponds to the maximal ideal in affine coordinate ring $A(\bar{Y})$ of $\bar{Y}$. $Z_i$s correspond to prime ideals contained in $m$, so $\text{height}(m) = n$. Since $P$ is a point in affine space, $A(\bar{Y}) \mathbin{/} m \cong k$. Hence, $\text{dim}(A(\bar{Y})) = \text{dim}(\bar{Y}) = n$, and $\text{dim}(Y) = \text{dim}(\bar{Y}) = n$.

What follows is a bunch of results:

1. Krull's Haputidealsatz: Let $A$ be a noetherian ring, and $f \in A$ be an element which is neither a zero-divisor nor unit. Then, every minimal prime ideal containing $f$ has height $1$.
2. Every noetherian integral domain $A$ is a UFD iff every prime ideal of height $1$ is principal.
3. A variety $Y \in A^n$ has dimension $n - 1$ iff it is the zero set of a single nonconstant irreducible polynomial in $A$.

## Projective varieties

We proceed as before, but in the projective space $P_k^n$, the set of equivalence classes of $(n + 1)$-tuples, denoted $(a_0, \ldots, a_n)$, nonzero elements of $k$ with the equivalence relation $(a_0, \ldots, a_n) \sim (\lambda a_0, \ldots, \lambda a_n)$, for all $\lambda \in k$. An element of $P^n$ is called a point, and the $(n + 1)$-tuple $(a_0, \ldots, a_n)$ in the equivalence class of $P$ is called set of homogenous coordinates for $P$.

A graded ring $S$ is a ring together with the decomposition $S = \bigoplus_{d \geq 0} S_d$, the direct sum of abelian groups, such that for $d, e \geq 0$, $S_d . S_e \subseteq S_{d + e}$. An element of $S_d$ is called a homogenous element of degree $d$. Ideal $a \subseteq S$ is a homogenous ideal if $a = \bigoplus_{d \geq 0} (a \cap S_d)$. An ideal is said to be homogenous if it is generated by homogenous elements. The sum, product, intersection, and radical of homogenous ideals yields a homogenous ideal.

A polynomial $f \in S$ cannot be used to define a polynomial in $P^n$ because of the nonuniqueness of homogenous coordinates. However, if $f$ is a homogenous polynomial of degree $d$, then $f(\lambda a_0, \ldots, \lambda a_n) = \lambda^d f(a_0, \ldots, a_n)$, so the property of $f$ being zero depends only on the equivalence class of $(a_0, \ldots, a_n)$. Thus, if $T \in S$ is a set of homogenous elements, the zero-set $Z$ can be defined as:
$$Z(T) = \{P \in P^n \mid f(P) = 0 \quad \forall f \in T\}$$

Since $S$ is noetherian, any set of homogenous elements $T$ has a finite subset $f_1, \ldots, f_r$ such that $Z(T) = Z(f_1, \ldots, f_r)$.

Subset $Y \subset P^n$ is an algebraic set if there exists a set of homogenous elements $T$ such that $Y = Z(T)$. The union of any two algebraic sets is an algebraic set, as is the intersection. The Zariski topology on $P^n$ is defined in the usual way.

A projective variety is the irreducible algebraic sets of $P^n$ along with the induced topology. An open subset of a projective variety is a quasi-projective variety. If $Y \subset P^n$ is a subset, we define the projective ideal, $I(Y)$ to be the ideal $\{f | f \text{ is homogenous}, f(P) = 0 \, \forall P \in Y\}$. If $Y$ is an algebraic set, we define the homogenous coordinate ring to be $S(Y) = S/I(Y)$.

We want to show that every projective $n$-space has a covering by the affine $n$-space, and hence, every projective (or quasi-projective) variety has a covering by an affine (or quasi-affine) variety. If $f \in S$ is a linear homogenous polynomial, then the zero set of $f$ is called a hyperplane; we denote the zero-set of $x_i$ to be $H_i$. Let $U_i = P^n - H_i$ be open subsets. Then, $P^n$ is covered by $U_i$, since if $P = (a_0, \ldots, a_n)$ is a point, then atleast one $a_i$ is nonzero, and hence $P \in U_i$. We define mapping $\varphi_i : U_i \rightarrow A^n$, such that if $P \in U_i$ and $\varphi_i(P) = Q$, where $Q$ is the point defined by:
$$Q = \left(\frac{a_0}{a_i}, \ldots, \frac{a_n}{a_i}\right)$$

with $a_i/a_i$ omitted. Note that $\varphi_i$ is well-defined since $a_j/a_i$ is independent of the choice of homogenous coordinates.

$\varphi_i$ is a homeomorphism of $U_i$ with its induced topology to $A^n$ its Zariski topology. $\varphi_i$ is clearly bijective, so it will be sufficient to show that closed sets of $U_i$ are identified with the closed sets of $A^n$ by $\varphi_i$. We can assume that $i = 0$ and write $U$ for $U_0$, $\varphi$ for $\varphi_0$; then we have $\varphi : U \rightarrow A^n$. Let $A = k[y_1, \ldots, y_n]$, $\alpha$ be a map $S^h$, the set of homogenous elements of $S$ to $A$, $\beta$ be a map from $A$ to $S^h$. Set $\alpha(f) = [1, y_1, \ldots, y_n]$ and $\beta(g) = x_0^e g(x_1/x_0, \ldots, x_n/x_0)$. Let $Y \subseteq U$, and let $\bar{Y}$ be its closure in $P^n$. This is an algebraic set so $\bar{Y} = Z(T)$, for some subset $T \subseteq S^h$. Let $T' = \alpha(T)$, so $\varphi(Y) = Z(T')$. Checking the converse as well yields the result that both $\varphi$ and $\varphi^{-1}$ are closed maps, so $\varphi$ is a homeomorphism.

## Morphisms

A function $f: Y \rightarrow k$ is regular at point $P \in Y$, if $P \in U \subseteq Y$ for open neighborhood $U$, and $f, g \in A = k[x_1, \ldots, x_n]$, $h$ is nowhere zero on $U$, and $f = g \mathbin{/} h$. $f$ is regular on $Y$ if it is regular on every point of $Y$.

A regular function is continuous when $k$ is identified with $A_k^1$ in its Zariski topology. It suffices to show that $f^{-1}$ of a closed set is closed. A closed set of $A_k^1$ is a finite set of points, so it's enough to show that $f^{-1}(a) = \{P \in Y \mid f(P) = a\}$, for all $a \in k$. Subset $Z$ of a topological space $Y$ is closed iff it can be covered by open subset $U$, such that $Z \cap U$ is closed in $U$ for every $U$. Let $U$ be an open set on which $f$ can be expressed as $f = g \mathbin{/} h$, $g, h \in A$, $h$ nowhere zero in $U$. So, $f^{-1}(a) \cap U = \{P \in U \mid g(P) \mathbin{/} h(P) = a\}$. But $g(P) \mathbin{/} h(P) = a$ iff $(g - ah)(P) = 0$. Hence, $f^{-1}(a) \cap U = Z(g - ah) \cap U$, and $f^{-1}(a)$ is closed in $Y$. A similar argument holds for a quasi-projective variety.

If $f, g$ are regular functions in variety $X$, and $f = g$ in some subset $U \subseteq X$, then $f = g$ everywhere.

If $X, Y$ are two varieties (affine, projective, quasi-affine, or quasi-projective), and $\varphi : X \rightarrow Y$ a morphism, such that for each open subset $V \subseteq Y$, and every regular function $f : V \rightarrow k$, the function $f \circ \varphi : \varphi^{-1}(V) \rightarrow k$ is regular. This defines the isomorphism $\varphi : X \rightarrow Y$ and its inverse $\psi : Y \rightarrow X$ in the category of varieties. Isomorphisms are necessarily bijective and bicontinuous.

Let $Y$ be a variety, $\mathcal{O}(Y)$ the ring of regular functions on $Y$. If $P$ is a point of $Y$, we define local ring of $P$ on $Y$, $\mathcal{O}_P$, as the ring of germs of regular functions of $Y$ near $P$. An element of $\mathcal{O}_P$ is the pair $\langle U, f \rangle$ such that $U$ is an open subset of $Y$ near $P$, and $f$ is a regular function on $U$. We identify two such pairs $\langle U, f \rangle$ and $\langle V, g \rangle$ if $f = g$ in $U \cap V$.

$\mathcal{O}_P$ is indeed a local ring; its maximal ideal $m$ is the set of germs of regular functions that vanish at $P$. If $f(P) \neq 0$, then $1/f$ is regular on some neighborhood of $P$. The residue field $\mathcal{O}_P \mathbin{/} m \cong k$.

Elements of function field $K(Y)$ of variety $Y$ are the equivalence classes of the pair $\langle U, f \rangle$, where $U$ is an open neighborhood of $Y$, and $f$ is a regular function on $U$.

$K(Y)$ is a field that can be made into a ring by defining addition and multiplication. Set $V = U - (U \cap Z(f))$ so that $1/f$ is a regular function on $V$. Then, $\langle V, 1/f \rangle$ is the inverse of $\langle U, f \rangle$.

By restricting functions, we obtain the natual map $\mathcal{O}(Y) \rightarrow \mathcal{O}_P \rightarrow K(Y)$ which is injective. We normally treat $\mathcal{O}(Y), \mathcal{O}_P$ as subrings of $K(Y)$. If we replace $Y$ by an isomorphic variety, we obtain isomorphic rings, and say that $\mathcal{O}(Y), \mathcal{O}_P, K(Y)$ are invariants of the variety $Y$.

Let $A(Y)$ be the affine coordinate ring, and $S(Y)$ be the homogenous coordinate ring (as in projective spaces). Then, $A(Y) = \mathcal{O}(Y)$, but $S(Y)$ is not invariant; it depends on the embedding of $Y$ in the projective space.

Let $Y \subseteq A^n$ be an affine variety, and $A(Y)$ the affine coordinate ring. Then:

1. $\mathcal{O}(Y) \cong A(Y)$.
2. For each point $P \in Y$, let $m_P \subseteq A(Y)$ be the set of functions vanishing at $P$. Then, there is a one-to-one correspondence between the points of $Y$ and the maximal ideals of $A(Y)$, given by $P \mapsto m_P$.
3. For each $P$, $\mathcal{O}_P \cong A(Y)_{m_P}$, and $\text{dim } \mathcal{O}_P = \text{dim } Y$.
4. $K(Y)$ is isomorphic to the quotient field of $A(Y)$, and hence $K(Y)$ is a finitely generated extension field of $k$, with transcendence degree = $\text{dim } Y$.

To prove (a), notice that every polynomial $f$ is regular on $A^n$, and hence $Y$; thus, the kernel of the injective homomorphism $A \rightarrow \mathcal{O}(Y)$ is just $I(Y)$.

To prove (b), notice that there is a one-to-one correspondence between points of $Y$ and maximal ideals of $A(Y)$, passing to the quotient $I(Y)$. Identifying $A(Y)$ with regular functions on $Y$, the maximal ideal corresponding to $P$ is $m_P = \{f \in A(Y) \mid f(P) = 0\}$.

To prove (c), notice that, for every point $P$, there is a natural map $A(Y)_{m_P} \rightarrow \mathcal{O}_P$ which is injective; it is also surjective by definition of a regular function. Hence, $\mathcal{O}_P \cong A(Y)_{m_P}$. Now, $\text{dim } \mathcal{O}_P = \text{height } m_P$. Since $A(Y) \mathbin{/} m_P \cong k$, $\text{dim } \mathcal{O}_P = \text{dim } Y$.

From (c), it follows that the quotient field of $A(Y)$ is isomorphic to the quotient field of $\mathcal{O}_P$, and this is equal to $K(Y)$, because any rational function is in some $\mathcal{O}_P$. Since $A(Y)$ is a finitely generated $k$-algebra, $K(Y)$ is a finitely generated field extension of $k$. Furthermore, its transcendence degree equals $\text{dim } Y$. This proves (d).

Let $U_i \subseteq P^n$ be an open set defined by the equations $x_i \neq 0$. Then, $\varphi : U_i \rightarrow A^n$ is an isomorphism of varieties.

Let us introduce some notation. If $S$ is a graded ring, and $p$ is a homogenous prime ideal in $S$, then $S_{(p)}$ denotes the subring of elements of degree $0$ in the localization of $S$ with respect to multiplicative subset $T$ consisting of homogenous elements of $S \backslash (p)$. Notice that $T$ has a natural grading given by $\text{deg }(f/g) = \text{deg }f - \text{deg }g$, where $f$ is homogenous in $S$, and $g \in T$. $S_{(p)}$ is a local ring given by $(p . T^{-1} S) \cap S_{(p)}$. In particular, if $S$ is a domain, then for $p = (0)$, we obtain a field $S_{((0))}$. Similarly, if $f \in S$ is homogenous, $S_{(f)}$ denotes a subring with elements of degree $0$ in the localized ring $S_{(f)}$.

Let $Y \subseteq P^n$ be a projective variety with homogenous coordinate ring $S(Y)$. Then:

1. $\mathcal{O}(Y) = k$.
2. For any point $P \in S(Y)$, let $m_P \in Y$ be the ideal generated by the set of homogenous $f \in S(Y)$ such that $f(P) = 0$. Then, $\mathcal{O}_P = S(Y)_{(m_P)}$.
3. $K(Y) \cong S(Y)_{((0))}$.

Let $U_i \subseteq P^n$ be the open set $x_i \neq 0$, and let $Y_i = Y \cap U_i$. $U_i$ is isomorphic to $A^n$, and $Y_i$ is an affine variety. There is a natural isomorphism $\varphi^*$ of the affine coordinate ring $A(Y_i)$ with the localization $S(Y)_{(x_i)}$ of the homogenous coordinate ring of $Y$. First, make an isomorphism from $f[y_1, \ldots, y_n]$ to $f[x_1/x_0, \ldots, x_n/x_0]$ leaving out $x_i/x_i$; this isomorphism sends $I(Y)$ to $I(Y)S_{(x_i)}$, so passing to the quotient, we obtain $A(Y) \cong S(Y)_{x_i}$.

To prove (b), let $P$ be any point, and choose $i$ so that $P \in Y_i$. Then, $\mathcal{O}_P \cong A(Y_i)_{m'_P}$. where $m'_P$ is the maximal ideal of $Y_i$ corresponding to $P$. Since $\varphi_i^*(m'_P) = m_P . S(Y)_{x_i}$, $x_i \notin m_P$, and localization is transitive, we obtain $\mathcal{O}_P = S(Y)_{(m_P)}$.

To prove (c), notice that $K(Y)$, which is equal to $K(Y_i)$, which is the quotient field of $A(Y)$. But by $\varphi_i^*$, this is isomorphic to $S(Y)_{((0))}$.

To prove (a), let $f \in \mathcal{O}(Y)$ be a global regular function. For each $i$, $f$ is regular on $Y_i$, so $f \in A(Y_i)$. But we know that $A(Y_i) \cong S(Y)_{(x_i)}$, so $f$ can be written as $g_i/x_i^{N_i}$, where $g_i \in S(Y)$ is homogenous of degree $N_i$. Thinking of $\mathcal{O}(Y), K(Y), S(Y)$ as subrings of the quotient field $L$ of $S(Y)$, we get $x_i^{N_i} f = S(Y)_{N_i}$. Now, choose $N \geq \sum N_i$. Then, $S(Y)_N$ is spanned as a $k$-vector space as monomials of degree $N$ in $x_0, \ldots, x_n$, and in any such mononomial atleast one $x_i$ occurs as with power $\geq N_i$. Thus, $S(Y)_N . f^q \subseteq S(Y)_N \, \forall q \gt 0$ and $x_0^N . f^q \subseteq S(Y) \, \forall q \gt 0$. This shows that a subring $S(Y)[f]$ of $L$ is contained in $x_0^{-N} S(Y)$, which is a finitely-generated $S(Y)$-module. Since $S(Y)$ is noetherian, $S(Y)[f]$ is a finitely-generated $S(Y)$-module, and therefore, $f$ is integral over $S(Y)$. This means that there are elements $a_0, \ldots, a_n \in S(Y)$ such that:
$$a_n f^n + \ldots + a_0 = 0$$

Now, $S(Y)_0 = k$, so $a_i \in k$ and $f$ is algebraic over $k$. But $f$ is algebraically closed, so $f \in k$, completing the proof.