Algebraic Geometry Spring 2022

We have some association \(A\) some ring \(\rightarrow \Spec(A)\) with some features:

1. The basis for the topology are the opens \(D(a)\)
2. If \(a = a'a''\) in \(A\) then
   1. \(D(a) \subset D(a')\)
   2. There is a canonical map \(A_a \leftarrow A_{a'}\) where \(\frac{(a'')^nx}{a^n} \mapsto \frac{x}{(a')^n}\)

Two remarks: the converse of 2.a does not hold; secondly, if \(M\) is a \(A\)-module, the map in 2.b extends to \(M_a \leftarrow M_{a'}\) and the maps are independent of the choice of \(a''\).

Lemma: The structure sheaf of \(\Spec(A)\): there exists a sheaf of rings \(\mathcal O_{\Spec(A)}\) such that

1. For \(a \in A\) there exists a canonical isomorphism \(c_a: A_a \rightarrow \mathcal O_{\Spec(A)}(D(a))\).
2. If \(a = a'a''\) then the following diagram commutes:

Pf: Use either stalks (Hartshorne) or the fact that \(D(a)\) (Stacks Project) are a basis for the topology and general facts about sheaves on bases.

Ex: Let \(X = (\Spec(A), \mathcal O_{\Spec(A)}\), and \(D(a_1) \cup D(a_2) = U \subset X\) open. Then we get the exact sequence

\begin{gather*} 0 \rightarrow \mathcal O_X(U) \rightarrow O_X(D(a_1)) \times O_X(D(a_2)) \rightarrow \mathcal(O)_X(D(a_1a_2)) \\ 0 \rightarrow kernel \rightarrow A_{a_1} \times A_{a_2} \rightarrow A_{a_1a_2} \\ 0 \rightarrow k[x,y] \rightarrow k[x,y,x^{-1}] \times k[c,y,y^{-1}] \rightarrow k[x,y,(xy)^{-1}] \end{gather*}

The far map on the right takes \(F(U_1) \times F(U_2) \rightarrow F(U_1) \cap F(U_2)\) where \(s_1 \times s_2 \mapsto s_1|_{U_1 \cap U_2} - s_2|_{U_1 \cap U_2}\); that is, we take the difference.

Key algebra fact: if \(a_1, \dots, a_n\) generates the unit ideal, then \[ 0 \rightarrow A \rightarrow \prod_{i=1}^n A_{a_i} \rightarrow \prod_{i,j=1}^n A_{a_ia_j} \] is exact.

We can make an observation: the maps \(c_a\) determine isomorphisms \(A_\mathfrak p \rightarrow \mathcal O_{\Spec(A), \mathfrak p}\), the stalk of \(\mathcal O_{\Spec(A)}\) at the point \(\mathfrak p\), which by definition is the filtered colimit \(\colim_{\mathfrak p \in U \in \Spec(A)} \mathcal O_{\Spec(A)}(U)\). Note that we can take the colimit over \(D(a)\) and thus we get that there is a mapping from \(\colim_{a \in \mathfrak p} A_a\).

Corollary: \(\Spec(A), \mathcal O_{\Spec(A)}\) is a locally ringed space.

Def: A ringed space \(X, \mathcal O_X\) is a locally ringed space if every stalk of \(\mathcal O_X\) is a local ring. Let \((f, f^\#) : (X, \mathcal O_X) \rightarrow (Y, \mathcal O_Y)\) of ringed spaces. If \(f: X \rightarrow Y\) is continuous, you can think of \(f^\#: O_Y \rightarrow f_*\mathcal O_X\) (the push-forward of a sheaf) or as a map \(f^\#: f^{-1}\mathcal O_Y \rightarrow \mathcal O_X\) where \(f^{-1}\) is the pull-back (really, think of the push-back as an adjoint functor to the puhs-forward).

In some sense, think of \(f^\#\) as a rule that to each open \(U \subset X\) open and \(V \subset Y\) open with \(f(U) \subset V\) associates a map \((f^\#)^U_V: \mathcal O_Y(V) \rightarrow O_X(U)\) compatible with all restriction maps.

Thus, for all \(x \in X\) we get an induced map \(f^\#_x: \mathcal O_{Y, f(x)} \rightarrow \mathcal O_{X, x}\), where an element of \(\mathcal O_{Y, f(x)}\) is a “function” \(h \in \mathcal O_Y(V)\) determined on some \(f(x) = V \subset Y\) open, and then gets mapped to \((f^\#)^U_V(h)\) where \(x \in U \subset X\) open such that \(f(U) \subset V\).

Def: If \((X, \mathcal O_X)\) and \((Y, \mathcal O_Y\) are locally ringed spaces, then \(f, f^\#\) is a morphism of locally ringed spaces iff \(f^\#_X\) is a local homomorphism \(\forall x \in X\).

Lemma: Let \(\varphi: A \rightarrow B\) be a ring map. There is a morphism of locally ringed spaces \(\Spec(\varphi), \Spec(\varphi)^\#\): \((\Spec(B), \mathcal O_{\Spec(B)}) \rightarrow (\Spec(A), \mathcal O_{\Spec(A)})\) such that given \(a \in A\) the diagram

commutes. In fact every such morphism arises as such a mapping induced by a ring homomorphism.

Let \(X\) be a scheme, \(x \in X\). Then, there exists \(x \in U \subset X\) open such that \(U\) is an affine scheme; that is, \(x \subset U (\Spec(A), \mathcal O_{\Spec(A)})\); but we can consider the restriction \((\Spec(A_a), \mathcal O_{\Spec(A)})\).

Thus, we see that on a scheme the opens which are affine form a basis for the topology; (but note that intersections of affine opens need not be affine).

Ex: Projective space.

Consider an algebraically closed field \(k\) and a nonnegative integer \(n\). Then, the (classical) projective plane is simply \[ \P^n(k) = (k^{n+1} - \{0\})/k^* \] Alternatively, we can take \(\P^n(k)\) the union over subsets of points such that \(x_i \neq 0\), which yields a bijection from the affine \(n\)-space over \(k\).

Def: A closed subset \(E\) of \(\P^n(k)\) is a subset such that \(E \cap U_i\) is closed in the Zarisky topology of \(U_i - \A^n(k) = k^n\).

Ex: Say \(F \in k[T_0, \dots, T_n\) is homogenous of degree \(d\). Then \[ V_+(F) = E = \{(x_0:\dots:x_n) \in \P^(k)\} \text{ such that } F(x_0, \dots, x_n) = 0 \] is a closed subset: \[ V_+(F) \cap U_i = \{(a_1, \dots, a_n) \in k^n \mid F(x_0, \dots, 1, \dots, x_n) = 0\} \]

Equivalently, we may see that every closed \(E \subset \P^n(k)\) is of the form \[ E = V_+(F_1) \cap \dots \cap V_+(F(t)) \] for some \(F_1, \dots, F_t \in k[T_0, \dots, T_n]\). The idea here is to have some correspondence

Def: We define the projective space as a scheme over a ring \(R\) as \[ \P^n_R = \Proj(R[T_0, \dots, T_n]) = \{\mathfrak p \subset R[T_0, \dots, T_n] \text{ homogenous prime ideals with } \mathfrak p \not\supset (T_0, \dots, T_n)\} \]

Note that the classical point \((x_1:\dots:x_n)\) corresponds to a line \(p = (x_iT_j - x_jT_i)\).

Lemma: Fix \(0 \leq i \leq n\). There is a bijection \[ U_i = \{\mathfrak p \subset R[T_0, \dots, T_n] \text{ homogenous prime}, T_i \notin \mathfrak p\} \leftrightarrow \{\mathfrak P_o \subset R[\frac{T_0}{T_i}, \dots, \frac{T_n}{T_i}] \text{ prime}\} = \Spec R[\frac{T_0}{T_i}, \dots, \frac{T_n}{T_i}] \] where \[ \mathfrak p \mapsto \mathfrak p R[T_0, \dots, T_n, \frac{1}{T_i}] \cap R[\frac{T_0}{T_i}, \dots, \frac{T_n}{T_i}] \] such that we may think of \[ \P^n_R = \bigcup_{i=0}^nU_i = \bigcup_{i=1}^n \Spec R[T_0, \dots, T_n, \frac{1}{T_i}] \] and more generally, \(F \in R[T_0, T_n\) homogenous of poisitive degree can be thought of as \[ D_+(F) = \Spec(R[T_0, \dots, T_n]_{(F)}) \] where \(R[T_0, \dots, T_n]_{(F)}\) is the set of degree 0 elements of \(R[T_0, \dots, T_n, \frac{1}{F}]\), which is \(R[\frac{T_0}{T_i}, \dots, \frac{T^n}{T_i}]\).

Topological, if \(\Omega \subset R[T_0, \dots, T_n]\) is a subset of homogenous elemeents, then \[ V_+(\Omega) = \{\mathfrak p \in \P^n_R \mid \Omega \subset \mathfrak p\} \]

1. These form closed subsets of a topology
2. The basis for the topology are \(D_+(F) = \P^n_R - V_+(F)\)
3. \(U_i = D_+(T_i) \cong \A^n_R = \Spec(R[\frac{T_0}{T_i}, \dots, \frac{T^n}{T_i}])\)

To identify the sheaf of rings, we can start with \(\mathcal O_{\P_R^n}(D_+(F) = R[T_0, \dots, T_n]_(F)\) and then extend exactly as in the case of affine schemes.

Aside: Why is projective space interesting? We are sort of adding things at infinity!

1. \(P^0(k)\) is one point.
2. \(P^1(k) = D_+(T_0) \sqcup V+(T_0)\) is a line with a point at infinity.
3. \(P^2(k) = \A^2(l) \sqcup \P(k)\) is a plane with a line and a point at infinity.

Def: Let \(R\) be a ring. A projective scheme over \(R\) is a scheme \(X\) equipped with a morphism \(X \rightarrow \Spec(R)\) such that there exists an \(n > 0\) and a closed immersion \[ X \rightarrow \P^n_R \] of schemes over \(R\).

If \(e\) is a category and \(s \in \Ob(e)\), we can consider th category \(e/s\) of “objects over \(S\)”.

1. Objects are arrows \(X \rightarrow S\)

2. Morphisms \(X/S \rightarrow Y/S\) are morphisms \(f: X \rightarrow Y\) in \(e\) such that the following diagram commutes:

   

3. schemes over \(R\) are schemes over \(\Spec(R)\).

Fact: The category of schemes over \(R\) is what you get when you replace the definition of the category of schemes, replacing ring with \(R\)-algebra everywhere; that is, scheme over \(R\) are schemes \((X, \mathcal O_X\) such that \(\mathcal O_x\) is given the structure of a sheaf of \(R\)-algebras.

Def: A morphism \(f: X \rightarrow Y\) of schemes is said to be affine, resp. finite, integral, a closed immersion iff

1. for all \(V \subset Y\) affine open the inverse image \(U = f^{-1}(V) \subset X\) is affine
2. the morphism \(f|_U: U \rightarrow V\) of affine schemes corresponds to a ring map \(A \rightarrow B\) which is abitrary, resp. finite, integral, surjective

Lemma: Let \(f: X \rightarrow Y\) be a morphism of schemes; let \(Y = \bigcup V_j\) be an affine open cover. Then, \(f\) is affine, resp. finite, integral, a closed immersion iff the earlier definition holds for each \(V_j\).

Example: \(\Spec(B) \rightarrow \Spec(A)\) is a closed immersion \(\iff\) \(B = A/I\) for some ideal \(I \subset A\).

Example (Fermat hypersurface): As a scheme, \(X: T_0^1 + \dots + T_n^1 = 0\) in \(\P^n_R\) (topologically, \(V_+(T_0^1 + \dots + T_n^1\)). Now we have some \(X \rightarrow \P^n_R\) with \(X \cap D_+(T_i) = \) the closed subscheme of \(\Spec(R[\frac{T_0}{T_i}, \dots, \frac{T_n}{T_i}])\) by taking \((\frac{T_0}{T_i})^d + \dots + 1 + \dots + (\frac{T_n}{T_i})^d = 0\). Algebraically, we can take this to mean that \[ X \cap D_+(T_i) = \Spec(R[\frac{T_0}{T_i}, \dots, \frac{T_n}{T_i}]) / (\frac{T_0}{T_i})^d + \dots + (\frac{T_n}{T_i})^d) = \Spec(R[t_0, \dots, t_n]) / (t_0^d + \dots + 1 + \dots + t_n^d) \]

The upshot is that if \(I \subset R[T_0, \dots, T_n]\) is a graded ideal (homogenous ideal) then we can construct a closed subscheme \(V_+(I) \rightarrow \P_R^n\) associated to \(I\).

We have a couple of facts:

1. \(V_+(I_ = \Proj(R[T_0, \dots, T_n]/I)\) as a scheme
2. Every closed subscheme \(X \subset \P^n_R\) is equal to \(V_+(I)\) as a scheme for some \(I\)

Def: A morphism \(f: X \rightarrow Y\) of schemes is locally of finite type iff for all \(U \subset X, V \subset Y\) affine open \(f(U) \subset V\) the corresponding ring map \((f^\#)^U_V: \mathcal O_y(V) \rightarrow \mathcal{O}_X(U)\) as a ring map of finite type.

Lemma: If for all \(x \in X\), there is some \(x \in U \subset X\), \(V \subset Y\) affine open, \(f(U) \subset V\) such that if \((f^\#)^U_V: \O_Y(V) \rightarrow \O_X(U)\) is of finite type, then \(f\) is of finite type.

Proof: Let \(U \subset X, V \subset Y\) be affine open, \(f(U) \subset V\). We have the following steps:

1. For all \(x \in U\) pick \(x \in U_x \subset X, V_x \subset Y\) affine open, \(f(U_x) \subset V_x\) and \((f^\#)^{U_x}_{V_x}: \O_Y(V_x) \rightarrow \O_X(U_x)\) finite type.

2. We may assume \(U_X \subset U\).

   Proof: Write \(U_x = \Spec(B_x)\). Then we may pick \(b_x \in B_x = \O_X(U_x)\) such that \(x in D(b_x) \subset U_x \cap U \subset U_x\). But we know that the mapping \(\O_Y(V_x) \rightarrow B_x\) is finite type, and the mapping \(B_x \rightarrow \O_X(D(b_x)) = (B_x)_{b_x}\) is a finite ring map due to localization.

3. Write \(U = \Spec(B)\). We may assume \(U_x = D(b_x)\) for some \(b_x \in B\).

   Proof: Pick \(b_x \in B\) such that \(D(b_x) \subset U_x \subset U\). Correspondingly, we have some mapping \(b_x \in B \mapsto \overline{b}_x\), so \(D(\overline{b}_x = D(b_x)\), so we get that \(B_{b_x} \cong (B_x)_{\overline{b}_x}\).

4. Write \(V = \Spec(A)\). Then, we may assume \(V_x = D(a_x)\) for some \(a_x \in A\).

   Proof: Repeat steps 2, 3 but now on \(Y\) and use the fact that \(\varphi: A \rightarrow B\) satisfies that \(\Spec(\varphi^-1(D(a))) = D(\varphi(a))\)

5. Put everything together: for all \(x \in U = \Spec(B)\), we have the following diagram:

   

   We have another algebra fact: \(A \rightarrow B\) ring map for all \(\mathfrak q \subset B\) prime there exists \(b \in B, b \not\in \mathfrak q\) such that \(A \rightarrow B_b\) finite type then \(A \rightarrow B\) is of finite type.

   The ideal for the algebra fact is that \(S = \{b: B_b \text{ is finite type over } A\}\) generates the unit ideal in \(B\), so pick \(b_1, \dots, b_m \in S\) and \(c_1, \dots, c_m \in B\) such that \(1 = b_1c_1 + \dots + b_mc_m\). For each \(j\) we have that there is some \(b_{j_1}, \dots, b_{j_{n_j}}\) such that \(B_{b_j}\) is generated by \(b_j, \frac{1}{b_j}, b_{j_1}, \dots, b_{j_{n_j}}\). Set now \(B'\) to be the \(A\)-subalgebra of \(B\) generated by \(b_1c_1, \dots, b_nc_n, b_{j_i}\). Then, we have

   

Note: the above holds for locally of finite presentation, locally quasi-finite, flat, smooth, etale, unramified, lci, etc.

Fix \(V \subset \P^n\) as above, \(H \subset \P^n\), given by \(\ell = 0\) for \(\ell \in k[T_0, \dots, T_n]_1\).

Assume now that \(V \not\subset H\), and set theoretically we have a decomposition \[ \P^n \supset V \cap H = W_1 \cup \dots \cup W_T \] where \(W_i\) are irreducible components; by the Krull Hauptidealsatz, eahc \(W_i \subset V\) has codimension 1. Now set \(m_i\) to be the multiplicity of \(W_i\) as a component of the scheme theoretic intersection \(V \cap H\): \[ \text{length}(\mathcal O_{V \cap H, \xi_i}) \] where \(\xi_i \in W_i\) is a generic point.

Recall that if \(Z_1 \rightarrow S\), \(Z_2 \rightarrow S\) are two closed subschemes of a scheme, then there is a scheme theoretic intersection \(Z_1 \cap Z_2 \rightarrow S\), which is made by, affine locally, setting \[ S = \Spec(A), \ Z_1 = \Spec(A/I_1), \ Z_2 = \Spec(A/I_2) \implies Z_1 \cap Z_2 = \Spec(A/(I_1 + I_2)) \]

Example: Let \(V = V_+(T_0T_1 - T_2^2) \subset \P^2\), and \(H = V_+(T_1) \subset \P^2\). Then, set theoretically, \[ V \cap H = \{(1: 0 : 0)\} \subset D_+(T_0) = \A^2\} \] and similarly, \(V \cap \A^2 \) is \(x - y^2 = 0\), \(V \cap H\) is \(x = 0\). Then, we have that \[ V \cap H \cap A^2 = (x - y^2, x) \subset k[x,y] \] and so \[ m = \text{length}\left(\frac{k[x,y]}{(x, x-y^2)}\right)_{(x,y)} = 2 \] And in general, since \(\xi_i \in W_i \subset V \cap H\) is a generic point of an irreducible component of the Neotherian scheme \(V \cap H\) we have \[ \mathcal O_{V \cap H, \xi_i} \text{ is Artinian } \implies m_i \geq 1, m_i < \infty \] Then Bezout gives us that (dim(V) > 0), \[ \deg(V) = \sum_{i=1}^r m_i \deg(W_i) \] if \(H\) is linear.

The sketch of the proof goes as follows: \(V \cap H\) is defined by \((I_V, \ell) \subset S = k[T_0, \dots, T_n]\) and we get \[ 0 \rightarrow S_V(-1) \rightarrow S_V \rightarrow S/(I_V, \ell) \rightarrow 0 \] and so we check that \(\deg(V)\) is the normalzied leading coefficient of the Hilbert polynomial of \(S/(I_V, \ell)\), and then you check that there is a correspondence \[ W_1, \dots, W_r \text{ irreducible components of } V \cap H \leftrightarrow I_{W_1}, \dots, I_{W_r} \text{ minimal primes of } S \] and further \[ I_{W_1}, \dots, I_{W_r} \text{ minimal primes of } S \leftrightarrow I_{W_1}, \dots, I_{W_r} \text{ generic points of } \text{Supp}(S/(I_V, \ell)) \] The last thing to check is that \[ m_i = \text{length}(S/(I_V, \ell))_{I_{W_i}} \] where the last ting is Artinian local.

The last thing we need is a commutative algebra fact, namely that given a finite graded \(S\)-module \(M\) we ca find a filtration \[ 0 \subset M_1 \subset \dots \subset M_t = M \] by graded submodules, such that for all \(i\), \(M_i / M_{i-1} \cong (S / \mathfrak p_i)(f_i)\) and moreover of \(\mathfrak p \in \text{Supp}(M)\) is a generic point of \(\text{Supp}(M)\). Then, \[ \# \{i \mid \mathfrak p_i = \mathfrak p\} = \text{length}(M_{\mathfrak p}) \]

Putting everything together, if we apply this to \(M = S/(I_V, \ell)\), we see that the Hilbert polynomial of \(M\) is just \[ \sum \text{ Hilbert polynomials of } (S / \mathfrak p_i)(f_i) \] And once we compare leading coefficients, we see that this shows \[ \frac{\deg(V)}{(\dim(V) - 1)!}t^{\dim(V)-1} = \sum m_i \frac{\deg(W_i)}{(\dim W_i)!}t^{\dim W_i} \]

Fact: Given \(V \subset \P^n\) a subvariety, \(\dim V \geq 1\) and for and \(d\) we can always pick \(H\) such that \(m_i = 1\) for all \(i\) (if \(k = \overline{k}\)). This is a Bertini theorem! And if we do this for \(d = 1\) we get a hyperplane \(H\) such that \[ \deg(V) = \sum \deg(W_i) \] and if \(\dim(V) = 1\), we see that \(\deg(V) = \#(V \cap H)\) for such an \(H\).

For \(\dim(V) = 2\), we can do it again so that \[ \deg(W_i) = \# (W_i \cap H') \] so \[ \deg(V) = \sum \deg(W_i) = \#(V \cap H \cap H') \]

Thm: What happens when you intersect \(V\) with 2 hypersurfaces \(H_1, H_2\) of degrees \(d_1, d_2\) where we assume the dimensions are OK? In general, intersecting things can give you wonky things; this is called a proper intersection. That is,

1. \(\dim(V) \geq 2\)
2. \(\dim(V \cap H_1 \cap H_2) = \bigcup_{i=1}^t Z_j\) (set theoretically) is the decomposition into irreducible components.

Then, we have that \[ d_1 d_2 \deg(V) = \sum_{j=1}^t \mu_j \deg(Z_j) \] for some integers \(m_j \geq 1\).

Proof: Let \(V \cap H_1 = W_1 \cup \dots \cup W_r\) be the irreducible components, and the irreducible components of \(W_i \cap H_2\) are a subset of \(\{Z_1, \dots, Z_t\}\). Further, it must be true that every \(Z_j\) occurs as an irreducible component of \(W_i \cap H_2\) for some \(i\). Now, \[ d_1 \deg(V) = \sum_{i=1}^r m_i \deg(W_i) \] and so \[ d_2 \deg(W_i) =\sum_{j, \ Z_j \subset W_i} m_{i, j} \deg(Z_j) \] and combining the two gives us what we need. But what are the \(\mu_j\)? Consider \(\xi_j \in Z_j\) a gneeric point, so that \(\mathcal O_{V, \xi_j} = R\) is a Noetherian local domain of dimension 2. Then, if \(f_1, f_2 \in \mathcal m\) are the local equations of \(H_1, H_2\). Say that \(Z_j \subset W_1, \dots, W_s\) and nothing else after renumbering; then the geenric points \(\xi_i \in W_i\) for \(1 \leq i \leq s\) correspond to prime ideals \(\mathfrak p_i \subset R\).

Algebraically, we have that if \((f_1) \subset \mathfrak p_1 \dots, \mathfrak p_s\), minimal over \((f_1)\), then \[ m_i = \text{length}((R/f_1R)_{\mathfrak p_i}) \] and \[ m_{i,j} = \text{length}((R/ \mathfrak p_i) / f_2(R / \mathfrak p_i)) = \text{length})(R / (\mathfrak p_i + f_2R)) \]

So there is a question, which is \[ \length(R/(f_1, f_2)) = \sum_{f_1 \subset \mathfrak p \subset R} \length((R / f_1)\mathfrak p) \length(R/\mathfrak p + f_2) \] But unfortunately this is true iff \(R\) is Cohen-Macaulay.

Algebraically, consider the exact sequence \[ R/f_1 R \rightarrow R/f_1R \rightarrow R/(f_1, f_2) \rightarrow 0 \] and if multiplication by \(f_2\) is injective in \(R/ f_1 R\), then it is also true for \(f_2^n\) as well. But then, it is a fact that \[ \length((R/f_1 R)/(f_2^n(R/f_1 R))) = n \length (R/(f_1, f_2)) \] so now \(R/ f_1 R\) has a filtration whose successive quotients are

1. \(R / \mathfrak p_i\) each \(m_i\) times
2. \(R / \mathfrak m\) a finite amount of times

and we see \[ \length((R/f_1R)/f_2^n) = \sum_{i=1}^s \length((R/ \mathfrak p_i)/f_2^n) + C \] and so considering the s.e.s. \[ 0 \rightarrow R/\mathfrak p_i \rightarrow R/\mathfrak p_2 \rightarrow (R/\mathfrak p_2)/f_2^n \rightarrow 0 \] we see that \[ \length((R/\mathfrak p_i)/f_2^n) = n \length(R/(\mathfrak p_2, f_2)) = nm_{i,j} \] And we can show that \(R\) is CM iff \(f_2: R/f_1R \rightarrow R/f_1R\) is injective. Overall, we have that \[ \length(\mathcal O_{V \cap H_1 \cap H_2, \xi_j}) \geq \mu_j \geq 1 \] where equality holds in the first inequality iff \(\mathcal O_{V, \xi_j}\) is CM.

For a reference, see Fulton, Intersection Teory, 8.4.5.

Def: The (ruled) join \(J(V, W) \subset \P^{2n + 1}\), where

\begin{gather*} V \leftrightarrow \mathcal P \subset k[T_0, \dots, T_n] \\ W \leftrightarrow \mathcal Q \subset k[T_0, \dots, T_n] \\ \end{gather*}

is the closed subscheme cut out by the ideal \[ \mathcal A = \langle f(X), f \in \mathcal P, g(Y), g \in \mathcal Q \rangle \subset k[x_0, \dots, x_n, y_0, \dots y_n] \]

Set theoretically, on \(k\)-rational points, \(J(V, W) = \{[x_0 : \dots : x_n : y_0 : \dots : y_n] \in \P^{2n + 1}\}\) such that \((x_0 : \dots : x_n) \in V\) or \(x_0 = \dots = x_n = 0\) and \((y_0 : \dots : y_n) \in W\) or \(y_0 = \dots = y_n = 0\).

Morally, this is some linear interpolation between points of \(V\) and points of \(W\)!

Furthermore, there is another “diagonal” copy of \(\P^n\) in \(\P^{2n + 1}\)

\begin{align*} \delta : \P^n &\rightarrow \P^{2n + 1}\\ (t_0 : \dots : t_n) &\mapsto (t_0 : \dots : t_n : t_0 : \dots : t_n) \end{align*}

which is cut out by the equations \(X_i - Y_i = 0\) (that is, \(\delta(\P^n) = V(X_i - Y_i)\)).

Lemma: As schemes, we have that \[ V \cap W = \delta(\P^n) \cap J(V, W) = J(V, W) = J(V, W) \cap H_0 \cap \dots \cap H_n \] where \(H_i = V(X_i - Y_i)\). So we can understand the general intersection theory of two subvarieties by reducing to the hyperplane case!

Lemma: Further, the degree of \(J(V, W)\) is \(\deg(V) \cdot \deg(W)\). But it is clear that \[ k[x_0, \dots, x_n, y_0, \dots, y_n] / \mathcal A \cong k[x_0, \dots, x_n] / \mathcal P \otimes k[y_0, \dots, y_n] / \mathcal Q \] By the fact that the RHS is a tensor product of domains, we can conclude that \(J(V, W)\) is a subvariety and that \(\mathcal A\) is its homogeneous prime ideal. Then if \(P\) is the Hilbert polynomial of \(J(V, W)\), and \(P_1, P_2\) the Hilbert polynomials of \(V, W\). Then, \[ P(d) = \sum_{i=0}^d P_1(i)P_2(d-i) \] and we can just find the leading coefficient.

Thm: Let \(Z_1, \dots, Z_i\) be the irreducible components of \(V \cap W\), which we identify using \(\delta: \P^n \rightarrow \P^{2n + 1}\) with the irreducible components of \(J(V, W) \cap H_0 \cap \dots \cap H_n\). Then,

1. \(\deg(V) \cdot \deg(W) \geq \sum_{i=1}^r \deg(Z_i)\)
2. If each \(Z_i\) has the “correct” dimension, i.e. \(\dim(Z_i) = \dim(V) + \dim(W) - n > 0\), then there exist integers \(m_i \geq 1\) such that \(\sum_{i=1}^r m_i \deg(Z_i) = \deg(V) \cdot \deg(W)\).

Proof: Since \(Z_1, \dots, Z_r\) are the irreducible components of \(J(V, W) \cap H_0 \cap \dots \cap H_n\) with \(\deg(H_i) = 1\), (1) follows from the following lemma: for a subvariety \(V \subset \P^n\), and \(H_1, \dots, H_r\) hyperplanes in \(\P^n\), we always have that \(\deg(V) \geq \sum \deg(Z)\) where \(Z\) ranges over the irreducible components of \(V \cap H_1 \cap \dots \cap H_r\).

We show this by induction on \(r\); if \(r = 1\), either \(V \subset H_1\), and this is trivially true, or \(V \not\subset H_1\), and we know that \(\deg(V) = \sum m_i Z_i \geq \sum Z_i\).

If \(r > 1\), by induction \(\deg(V) \geq \sum \deg(Z')\) where \(Z' \subset V \cap H_1 \cap \dots \cap H_n\) is an irreducible component, and furthermore each irreducible component \(Z_i\) of \(V \cap H_1 \cap \dots \cap H_n\) is an irreducible component of \(Z' \cap H_r\) for some \(Z'\). But by \(r = 1\), we know that \(\deg(Z') \geq \sum \deg(Z'')\), where the sum spans irreducible components of \(Z'\).

The proof of (2) is similar, but you can make everything strict; further each \(m_i\) is the “Samuel multiplicity” of the ideal generated by \(\mathcal O_{J(V, W), \delta(\xi)}\), where \(\xi \in Z_i\) is a generic point.

Application: if \(C_1, C_2 \subset \P^n\) are curves of degrees \(d_1, d_2, C_1 \neq C_2\), then \(\#(C_1 \cap C_2) \leq d_1d_2\). And this is strict, since you can just pick two lines that meet.

Another application: \(V \subset \P^n\) subvariety, \(d = \deg(V)\), not contained in a hyperplane, satisfies that \[ \dim(V) + d \geq n + 1 \] and the quick argument is to reduce to the case of a curve. Then we just need to show that \(d \geq n\).

An action of \(G\) on \(X\) over \(S)\) is a morphism \(\alpha: G \times_S X \rightarrow X\) such that

Example: \(X = \A^n_S\), \(w_1, \dots, w_n \in \Z^n\) weights. Then \[ \mathbb G_{m, S} \times_S \A^n_S \rightarrow \A^n_S, (t, (x_1, \dots, x_n)) \rightarrow (t^{w_1}x_1, \dots, t^{w_n}x_n) \]

We define equivariant modules \(\mathcal F\) quasi-coherent on \(X\) together with \(\alpha: a^*\mathcal F \rightarrow pr_2^*\mathcal F\) an isomorphism of \(\mathcal O_{G \times_S X}\)-modules such that the cocycle condition holds.

We have some facts in the affine case: let \(S = \Spec(R), X = \Spec(A), G = \mathbb G_{m, S}\) and in this case the category of \(\mathbb G_m\)-equivalence quasi-coherent \(\mathcal O_X\)-modules is equivalent to the category of \(Z\)-graded \(A = \oplus A_n\)-modules, and \(M = \oplus M_n\). Thus \(G\) actions on \(X\) just correspond to \(\Z\)-gradings \(A = \oplus_{n \in \Z}A_n\) on \(A)\) as an \(R\)-algebra.

Now, if \(\mathcal L \in \Pic(X)\) when is \(\mathcal L\) \(G\)-equivariant; that is,

1. When is \(a^* \mathcal A\) isomorphic to \(pr_2^* \mathcal L\) on \(G \times X_S X\)?
2. And if yes, when can we pick \(\alpha\) such that the cocycle condition holds?

(1) is always OK if \(X\) is Noetherian integral normal; but (2) is more unclear. But this is OK if (for example) \(S = \Spec(k), H^0(X, \mathcal O_X) = k\), and \(X\) has a \(k\)-rational \(G\)-fixed point, and \(G\) is affine. But (2) usually isn’t a problem anyway, so it should be OK.

Given an action \(a: G \times_S X \rightarrow X\) an invariant morphism (of schemes over \(S\)) is a morphism \(f: X \rightarrow Y\) such that the following diagram commutes:

So we’re trying to understand what it would mean that \(Y\) is a quotient of \(X\) by \(G\); there are a few possible options:

1. \(f: X \rightarrow Y\) is a categorical quotient.
2. \(Y\) is an orbit space.
3. \(\mathcal O_Y = (f_* \mathcal O_X)^G\), the \(G\)-invariants.
4. \(f\) is surjective.
5. \(f\) is submersive (\(f\) surjective and \(T \subset Y\) closed iff \(f^{-1}(T) \subset X\) closed).
6. If \(Z \subset X\) is closed and \(G\)-invariant, then \(f(Z)\) is closed.
7. The last condition, but also \(f(Z_1 \cap Z_2) = f(Z_1) \cap f(Z_2)\).

We say a condition \(P\) like above holds universally if for any

we have \(P\) for \(f'\); \(P\) holds uniformly if the same, but \(g\) is flat.

People also use the term good quotients to mean affine, (4), (3) universally, and (2) universally; a geometric quotient is (2), (5) universally, (3).

The main idea here is that quotients in algebraic geometry should be taking the spectrum of invariants.

In particular, we will consider \(S = \Spec(k), G = \mathbb G_{m,S}, X = \Spec(A)\), \(A = \oplus_{n \in \Z} A_n\), and \(Y = \Spec(A_0) = X / \mathbb G_m\).

Example: \(X = \mathbb G_{m, k}\); \(A = \Gamma(X, \mathcal O_X) = k[t, t^{-1}] \supset k = A_0\). Then \(Y = \Spec(k) \xleftarrow{f} X\).

Example: \(X = \Spec(k[x,y]) \cong \A^2\), with the grading \(wt(x) = 1, wt(y) = -1\), and \(t \cdot (x,y) = (tx, t^{-1}y)\). So \(Y = \Spec(k[xy]) \cong \A^1\).

So what about the properties?

1. Yes.
2. No (see last example).
3. Yes.
4. Yes. Certainly in the examples, but we need to check that \(\Spec(A) \rightarrow \Spec(A_0)\) is surjective, but take any \(\mathfrak p \subset A_0\) prime, and look at \(A_0 \cap \mathfrak p A = \mathfrak p\), so that \(A_0 / \mathfrak p \rightarrow A/\mathfrak p A\), which is true iff dominant on \(\Spec\), so \(\mathfrak p\) is in the image.
5. Yes. Implied by (5) + (4).
6. Yes. Say \(Z \subset X = \Spec(A)\) is \(G\)-invariant closed; then \(Z = \Spec(B)\) with \(A \rightarrow B\) surjective, and do some annoying diagram chasing or something.
7. Yes.

Also, everything above holds universally.