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Theorem 2.6. Let \( k \) be infinite. An algebraic variety \( V \) in \( {k}^{n} \) is homogeneous iff it is defined by a set of homogeneous polynomials. (We agree that the variety defined by the empty set of polynomials is \( {k}^{n} \) .) | Proof. Since the theorem is trivial if \( V = {k}^{n} \), assume \( V \subsetneqq {k}^{n} \) . \n\n\( \Leftarrow \) : Let the variety be \( V = \mathrm{V}\left( {{q}_{1},\ldots ,{q}_{r}}\right) \), where each \( {q}_{i} \in k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is homogeneous of degree \( {d}_{i} \) .... | Yes |
Lemma 2.14. Let \( {q}_{1},\ldots ,{q}_{r} \in k\left\lbrack {{X}_{1},\ldots ,{X}_{n + 1}}\right\rbrack \) be homogeneous; let \( \mathbf{V}\left( {{q}_{1},\ldots ,{q}_{r}}\right) \subset {\mathbb{P}}^{n}\left( k\right) \) be the projective variety defined by \( {q}_{1},\ldots ,{q}_{r} \) . Then \[ {\mathrm{D}}_{i}\lef... | Proof. The variety \( \left. {\mathrm{V}\left( {{\mathrm{D}}_{i}\left( {q}_{1}\right) ,\ldots ,{\mathrm{D}}_{i}\left( {q}_{r}\right) }\right) }\right) \) can be looked at as the intersection of the variety \( \mathrm{V}\left( {{q}_{1},\ldots ,{q}_{r}}\right) \) with the plane given by \( {X}_{i} = 1 \) in \( {k}^{n + 1... | Yes |
Lemma 2.15. Let \( p \in k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Then\n\n\[{\mathrm{D}}_{i}\left( {{\mathrm{H}}_{i}\left( p\right) }\right) = p.\] | Proof. Obvious from the definitions of \( {\mathrm{D}}_{i} \) and \( {\mathrm{H}}_{i} \) . | No |
Lemma 2.16. Let \( q \) be a homogeneous polynomial in \( k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . Then for any \( i = 1,\ldots, n \), it can happen that \[ {\mathrm{H}}_{i}\left( {{\mathrm{D}}_{i}\left( q\right) }\right) \neq q\text{.} \] | Proof. Let \( q\left( {{X}_{1},{X}_{2}}\right) = {X}_{1}{X}_{2} \) . Then \( {\mathrm{D}}_{1}\left( q\right) = {X}_{2} \), and \( {\mathrm{H}}_{1}\left( {{\mathrm{D}}_{1}\left( q\right) }\right) = {X}_{2} \neq \) \( {X}_{1}{X}_{2} \) . Similarly, \( {\mathrm{H}}_{2}\left( {{\mathrm{D}}_{2}\left( q\right) }\right) = {X}... | Yes |
Lemma 2.17. Let \( {\mathbb{P}}_{\infty }{}^{n - 1}\left( k\right) \) be a hyperplane at infinity of \( {\mathbb{P}}^{n}\left( k\right) \), and let \( V \subset {k}^{n} = {\mathbb{P}}^{n}\left( k\right) \smallsetminus {\mathbb{P}}_{\infty }{}^{n - 1}\left( k\right) \) . Let \( \mathrm{H}\left( V\right) \) be the projec... | Proof. We leave verification of (4) as an easy exercise. (5) follows from Lemma 2.16 by letting \( V \) be \( \mathrm{V}\left( {{X}_{1}{X}_{2}}\right) \) . More generally, if \( V \) is any variety in \( {\mathbb{P}}^{n}\left( k\right) \) not containing \( {\mathbb{P}}_{\infty }{}^{n - 1}\left( k\right) \), then (5) ho... | No |
Consider the real circle \( \mathrm{V}\left( {{X}^{2} + {Y}^{2} - 1}\right) \subset {\mathbb{R}}^{2} \). The homogenized polynomial \( {X}^{2} + {Y}^{2} - {Z}^{2} \in \mathbb{R}\left\lbrack {X, Y, Z}\right\rbrack \) determines the cone in Figure 3 as well as the circles in Figure 4. (Since antipodal points are identifi... | We may dehomogenize at an arbitrary \( {\mathbb{P}}_{\infty }{}^{1}\left( \mathbb{R}\right) \) by choosing an appropriate 2-space in \( {\mathbb{R}}_{XYZ} \). Since the intersection of the cone with a parallel translate of this 2-space yields a copy of the affine part of the curve with respect to \( {\mathbb{P}}_{\inft... | Yes |
We next consider the alpha curve \( \mathrm{V}\left( {{Y}^{2} - {X}^{2}\left( {X + 1}\right) }\right) \) \( \subset {\mathbb{R}}^{2} \) . Homogenizing the polynomial gives | \[ {Y}^{2}Z - {X}^{2}\left( {X + Z}\right) = 0 \] the intersection of \( \mathrm{V}\left( {{Y}^{2}Z - {X}^{2}\left( {X + Z}\right) }\right) \) with a sphere centered at \( \left( {0,0,0}\right) \) is shown in Figure 10. Figure 9a-c show the affine parts after dehomo-genizing at \( Z, Y \), and \( X \), respectively; Fi... | Yes |
Let \( V \subset {\mathbb{R}}^{2} \) consist of \( n \) distinct parallel lines. If the lines are \( {L}_{1},\ldots ,{L}_{n} \), given by, say, \( Y = 1, Y = 2,\ldots, Y = n \), then the union of these lines is given by \( p\left( {X, Y}\right) = \left( {Y - 1}\right) \left( {Y - 2}\right) \cdot \ldots \cdot \left( {Y ... | To explore these cases, we homogenize and then dehomogenize so the intersection point (at infinity) becomes the new origin.\n\nHomogenizing \( p\left( {X, Y}\right) \) gives us\n\n\[ \n{\mathrm{H}}_{Z}\left( p\right) = \left( {Y - Z}\right) \cdot \left( {Y - {2Z}}\right) \cdot \ldots \cdot \left( {Y - {nZ}}\right) = 0 ... | Yes |
Theorem 3.6. Let \( p\left( {X, Y}\right) \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) satisfy\n\n(3.6.1) \( p\left( {0,0}\right) = 0 \), and\n\n(3.6.2) \( {p}_{Y}\left( {0,0}\right) \neq 0 \) .\n\nThen within some neighborhood of \( \left( {0,0}\right) \), those points \( \left( {x, y}\right) \) satisfying \( p\l... | In proving this theorem we assume the following standard integral theorems of complex variables. For our purposes it suffices to state them \ | No |
Lemma 3.11. Let \( p\left( {X, Y}\right) \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \smallsetminus \mathbb{C} \) . Then \( C = \mathrm{V}\left( {p\left( {X, Y}\right) }\right) \) is a real-analytic manifold at any point \( \left( {{x}_{0},{y}_{0}}\right) \) where either \( {p}_{X}\left( {{x}_{0},{y}_{0}}\right) \ne... | Proof. Suppose without loss of generality that \( {p}_{Y}\left( {{x}_{0},{y}_{0}}\right) \neq 0 \) . Writing \( p\left( {X, Y}\right) \) as \( {p}_{1}\left( {X, Y}\right) + i{p}_{2}\left( {X, Y}\right) \) gives\n\n\[ {p}_{1}\left( {{X}_{1} + i{X}_{2},{Y}_{1} + i{Y}_{2}}\right) = {p}_{2}\left( {{X}_{1} + i{X}_{2},{Y}_{1... | Yes |
Let \( D \) be any unique factorization domain. Let two polynomials in \( D\left\lbrack X\right\rbrack \) be\n\n\[ f\left( X\right) = {a}_{0}{X}^{m} + \ldots + {a}_{m}, \]\n\n\[ g\left( X\right) = {b}_{0}{X}^{n} + \ldots + {b}_{n} \]\n\nWe assume that at least one of \( {a}_{0},{b}_{0} \) is nonzero. Then \( f\left( X\... | Proof. Since \( D \) is a unique factorization domain, so is \( D\left\lbrack X\right\rbrack \), by Gauss’ lemma. Let the unique factorizations of \( f \) and \( g \) be\n\n\[ f = d \cdot {f}_{1}^{{m}_{1}} \cdot \ldots \cdot {f}_{r}^{{m}_{r}},\;g = e \cdot {g}_{1}{}^{{n}_{1}} \cdot \ldots \cdot {g}_{s}{}^{{n}_{s}}, \]\... | Yes |
Lemma 4.7. Let \( D \) be any unique factorization domain of characteristic zero. Then \( f \in D\left\lbrack X\right\rbrack \) has a repeated (nonconstant) factor iff \( f \) and \( {f}^{\prime } \) have a common factor. Thus\n\n\[ \text{fhas a repeated factor iff}\mathcal{D}\left( f\right) = 0\text{.} \] | Proof. First, suppose that \( f \) has no repeated factors. Then \( f = {p}_{1}{p}_{2},\ldots ,{p}_{r} \) , where the \( {p}_{i} \) are distinct irreducible polynomials. Differentiating, we obtain\n\n\[ {f}^{\prime } = {p}_{1}^{\prime }{p}_{2},\ldots ,{p}_{r} + {p}_{1}{p}_{2}^{\prime },\ldots ,{p}_{r} + \ldots + {p}_{1... | Yes |
Lemma 4.8. Suppose \( p\left( {X, Y}\right) \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) satisfies Assumption 4.2, p having (total) degree \( n \) . Then the points \( {x}_{0} \in {\mathbb{C}}_{X} \) at which \( p\left( {{x}_{0}, Y}\right) \) has fewer than \( n \) zeros are precisely the zeros of the polynomial \... | Proof. Let \( {x}_{0} \in {\mathbb{C}}_{X} \) . Then \( \deg p\left( {{x}_{0}, Y}\right) = n \), and from the form of the resultant in (14) it is evident that\n\n\[ \n{\mathcal{D}}_{Y}{\left( p\left( X, Y\right) \right) }_{X = {x}_{0}} = \mathcal{D}\left( {p\left( {{x}_{0}, Y}\right) }\right) .\n\]\n\nThis, together wi... | Yes |
Corollary 4.18. Let \( {x}_{0} \) be an arbitrary point of \( {\mathbb{C}}_{X} \) . Each of the \( n \) series in (20) converges in a neighborhood of \( {x}_{0} \) . | Proof. Each of the \( m \) series in (16) converges in a neighborhood of \( {x}_{0} \) . Of course these \( m \) series are only the ones corresponding to an \( m \) -fold ramp at \( \left( {{x}_{0},{y}_{0}}\right) \) . Considering now the totality of all the series analogous to (16) corresponding to all the roots of \... | Yes |
Lemma 5.8. Let \( p\left( {X, Y}\right) \) be a polynomial with no repeated factors,\n\n\[ p\left( {X, Y}\right) = {a}_{0}\left( X\right) {Y}^{n} + {a}_{1}\left( X\right) {Y}^{n - 1} + \ldots + {a}_{n}\left( X\right) ,\]\n\nwhere \( {a}_{i}\left( X\right) \in \mathbb{C}\left\lbrack X\right\rbrack ,{a}_{0} \neq 0 \), an... | Proof. As we saw in the last section, the discriminant \( {\mathcal{D}}_{Y}\left( p\right) \in \mathbb{C}\left\lbrack X\right\rbrack \) is not the zero-polynomial since \( p \) has no repeated factors; hence the discriminant variety \( \mathrm{V}\left( {{\mathcal{D}}_{Y}\left( p\right) }\right) \subset {\mathbb{C}}_{X}... | Yes |
Lemma 5.11. Let \( T \) be a compact Hausdorff space, and let \( P \) be any point of \( T \) . Then\n\n\[{\left( T\smallsetminus \{ P\} \right) }^{ * } = T\] | The proof is a strightforward exercise and is left to the reader. | No |
What is the nature of \( C \) above each of these two points? | Let us first expand \( {X}^{2} + {Y}^{2} - 1 \) about the point \( X = 1, Y = 0 \), or, what is the same, set \( {X}^{\prime } = \) \( X - 1 \) and \( {Y}^{\prime } = Y \) and expand about \( {X}^{\prime } = 0,{Y}^{\prime } = 0 \) . This gives \( {\left( {X}^{\prime } + 1\right) }^{2} + \) \( {\left( {Y}^{\prime }\righ... | Yes |
The representation of a given curve \( C \) as a near covering can change markedly as we vary \( {\mathbb{P}}^{1}\left( \mathbb{C}\right) \) and \( {P}_{\infty } \) . | For instance in Example 5.13, one might choose for \( {P}_{\infty } \) a point in the circle. This can be done, for example, by picking coordinates in \( {\mathbb{C}}_{XYZ} \) so that dehomogenizing at \( Z \) gives, in affine space \( {\mathbb{C}}_{XY} \), the complex parabola \( \mathbf{V}\left( {Y - {X}^{2}}\right) ... | Yes |
What about above the infinite point \( {\mathbb{P}}^{1}\left( \mathbb{C}\right) \smallsetminus {\mathbb{C}}_{X} \) ? | Dehomogenizing \( {Y}^{2} - {XZ} \) at \( X = 1 \) places this point at the origin, the new affine representative being given by \( {Y}^{2} - Z = 0 \) ; in \( {\mathbb{C}}_{YZ} \) it is \( {\mathbb{C}}_{Z} \) whose completion is \( {\mathbb{P}}^{1}\left( \mathbb{C}\right) \) . Then \( {Y}^{2} = Z \) describes another 2... | No |
One can derive the topology of a curve in \( {\mathbb{P}}^{2}\left( \mathbb{C}\right) \) by looking at any affine part of it (though if the line at infinity is in the curve, one must put it back again after analyzing the topological closure of the affine part). The alpha curve \( C \) defined in \( {\mathbb{C}}_{XYZ} \... | Dehomogenizing \( {Y}^{2}Z - {X}^{2}\left( {X + Z}\right) \) at \( Y \) gives \( Z - {X}^{3} - {X}^{2}Z \) . Equating this to zero yields\n\n\[ Z = \frac{{X}^{3}}{1 - {X}^{2}} \]\n\n(22)\n\nThe real part of the graph of (22) appears in Figure 9b. Let \( {\mathbb{P}}^{1}\left( \mathbb{C}\right) \) be the projective comp... | Yes |
Lemma 6.4. A polynomial \( p \in \mathbb{C}\left\lbrack {{X}_{0},\ldots ,{X}_{r}}\right\rbrack \) is homogeneous of degree \( m \) (or else is the zero polynomial) iff for a new indeterminate \( T \) , \n\n\[ \np\left( {T{X}_{0},\ldots, T{X}_{r}}\right) = {T}^{m}p\left( {{X}_{0},\ldots ,{X}_{r}}\right) \]\n\nholds in \... | ## Proof\n\n\( \Rightarrow \) : Obvious.\n\n\( \Leftarrow \) : Assume \( p \) is not the zero polynomial; suppose \( p \) has degree \( k \) and that \( p \) satisfies (23). Write\n\n\[ \np = {p}_{0} + {p}_{1} + \ldots + {p}_{k} \]\n\nwhere \( {p}_{k} \neq 0 \) and \( {p}_{i} = {p}_{i}\left( {{X}_{0},\ldots ,{X}_{r}}\r... | Yes |
Lemma 6.5. Let \( p, q \in \mathbb{C}\left\lbrack {{X}_{0},\ldots ,{X}_{r}}\right\rbrack \left( {r \geq 2}\right) \) be nonconstant homogeneous polynomials. Then \( p \) and \( q \) have a common zero other than \( \left( {0,\ldots ,0}\right) \) . | Proof. With notation as before, let\n\n\[ p = \mathop{\sum }\limits_{{i = 0}}^{m}{p}_{i}{X}_{r}^{m - i},\;q = \mathop{\sum }\limits_{{i = 0}}^{n}{q}_{i}{X}_{r}^{n - i}. \]\n\nBy performing a linear change of coordinates if necessary, we may assume that \( {p}_{0} \neq 0 \) and \( {q}_{0} \neq 0 \) . (The argument is es... | Yes |
Theorem 7.4. Let \( p\left( {X, Y}\right) \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) have no repeated factors. Then \( \mathrm{V}\left( p\right) \subset {\mathbb{C}}_{XY} \) is smooth at \( P \in \mathrm{V}\left( p\right) \) iff at least one of the following holds:\n\n\[ \frac{\partial p}{\partial X}\left( P\rig... | Proof of Theorem 7.4\n\n\( \Leftarrow \) : This is just Corollary 3.9.\n\n\( \Rightarrow \) : We prove this half by contradiction. The strategy is this: Assume that \( \mathrm{V}\left( p\right) \) is smooth at \( P \) and that \( \left( {\partial p/\partial x}\right) \left( P\right) = \left( {\partial p/\partial y}\rig... | No |
Theorem 8.4. Any complex algebraic curve \( C \subset {\mathbb{P}}^{2}\left( \mathbb{C}\right) \) is connected. | Besides assuming \( q\left( {X, Y, Z}\right) \) is irreducible, we may further reduce the problem to considering only affine varieties: If we dehomogenize with respect to a projective line containing a point of \( {\mathbb{P}}^{2}\left( \mathbb{C}\right) \) not in \( C \) (say, without loss of generality, at \( Z \) ),... | No |
Theorem 8.5. Let\n\n\[ p\left( {X, Y}\right) = {Y}^{n} + {a}_{1}\left( X\right) {Y}^{n - 1} + \ldots + {a}_{n}\left( X\right) \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \;\left( {n \geq 1}\right) \]\n\nbe irreducible. Then \( \mathbf{V}\left( p\right) \subset {\mathbb{C}}_{XY} \) is connected. | Our general strategy in proving Theorem 8.5 is this: We prove that for a particular finite set of points \( \left\{ {P}_{i}\right\} ,\mathrm{V}\left( p\right) \smallsetminus \left\{ {P}_{i}\right\} \) is connected (which implies, by Lemma 8.3, that the closure \( \mathrm{V}\left( p\right) \subset {\mathbb{C}}_{XY} \) i... | Yes |
Lemma 8.8. Any chainwise connected topological space \( S \) is connected. | Proof. If \( S \) is not connected, then for two nonempty subsets \( B \) and \( C \) we have \( S = B \cup C \), where \( \bar{B} \cap C = B \cap \bar{C} = \varnothing \) . Let \( b \in B, c \in C \), and let \( \left( {{\mathcal{O}}_{1},\ldots ,{\mathcal{O}}_{m}}\right) \) be a chain from \( b \) to \( c \) . Then \(... | Yes |
Lemma 8.13. Relative to \( \left( {\mathrm{V}\left( p\right) \smallsetminus {\pi }_{Y}{}^{-1}\left( \mathcal{D}\right) ,\mathbb{C} \smallsetminus \mathcal{D},{\pi }_{Y}}\right) \), any simply connected open subset of \( \mathbb{C} \smallsetminus \mathcal{D} \) is allowable. | This is an immediate consequence of the familiar \ | No |
Theorem 8.14 (Monodromy theorem). Let \( \Omega \) be a simply connected open set in \( \mathbb{C} \), and suppose an analytic function element \( \mathcal{Q} \) is a lifting of a connected open set \( \mathcal{O} \subset \Omega \) . If \( \mathcal{Q} \) can be analytically continued along any polygonal path in \( \Ome... | For a proof of Theorem 8.14, see, e.g., [Ahlfors, Chapter VI, Theorem 2]. | No |
Theorem 10.1 (Genus formula). Let \( C \subset {\mathbb{P}}^{2}\left( \mathbb{C}\right) \) be a nonsingular projective curve defined by the irreducible polynomial \( p\left( {X, Y}\right) \) . If \( \deg p = n \), then the genus \( g \) of \( C \) is\n\n\[ g = \frac{\left( {n - 1}\right) \left( {n - 2}\right) }{2}. \] | The basic outline of the proof is this:\n\nFirst, we note that any compact connected orientable 2-manifold \( M \) may be looked at, topologically, as a polyhedron having \( g \) handles.\n\nSecond, we recall the basic fact that one can compute \( g \) from \( M \) looked at as any polyhedron having \( V \) vertices, \... | No |
Lemma 10.2. Let \( M \) be any polyhedron in the above sense, having \( V \) vertices, \( E \) edges, and \( F \) faces. Suppose that \( M \) has genus \( g \) . Then \[ V - E + F = 2 - {2g}. \] (Or equivalently, \( g = 1 - \left( \frac{1}{2}\right) \left( {V - E + F}\right) \) .) | Proof. We first consider the case \( g = 0 \) . Therefore, assume \( M \) is a sphere. Let \( e \) be any edge of \( M;e \) belongs to some closed polygonal curve \( C \) consisting of edges and vertices of \( M \) . The union of \( e \) and the two faces on either side of \( e \) is a single connected open set, which ... | No |
Lemma 10.4. Let \( f : {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{n}} \rightarrow \mathbb{C} \) be complex analytic at \( a = \left( {{a}_{1},\ldots ,{a}_{n}}\right) \) , and suppose it has order \( s \) in \( {X}_{i} \) at a. Then there is an open polydisk about \( a \) in \( {\mathbb{C}}^{n},\Delta \left( a\right) = \Delta =... | Proof. The proof exactly parallels part of the proof of the implicit function theorem, Theorem 3.6: The hypothesis that \( f \) has order \( s \) in \( {X}_{i} \) at \( a \) just says that \( f\left( {{a}_{1}^{\prime },\ldots ,{a}_{i - 1}^{\prime },{X}_{i},{a}_{i + 1}^{\prime },\ldots ,{a}_{n}^{\prime }}\right) \) has ... | Yes |
Lemma 2.6. Any lattice \( \left( {L, \leq ,\vee , \land }\right) \) together with a closure map \( a \rightarrow \bar{a} \) on the underlying p.o. set determines a new lattice \( \left( {{L}^{\prime }, \leq ,{ \vee }^{\prime },{ \land }^{\prime }}\right) ;{L}^{\prime } \) is a sub-p.o. set of \( L \) (but not a sublatt... | Proof. \( { \vee }^{\prime } \) : Let \( c \) be any closed upper bound of \( a \) and \( b \) . Then \( a \vee b \leq c \) , therefore \( \overline{a \vee b} \leq \bar{c} = c \) . But \( \overline{a \vee b} \) is itself an upper bound of \( a \) and \( b \), so it is the least closed one-that is, \( a{ \vee }^{\prime ... | Yes |
Theorem 2.11 (Basic decomposition theorem for lattices). Let \( \left( {L,\vee , \land }\right) \) be a lattice.\n\n(2.11.1) If \( L \) satisfies the a.c.c., then there exists an irredundant representation of any \( a \in L \) as the meet \( a = {a}_{1} \land \ldots \land {a}_{m} \) of \( \land \) -irreducible elements... | ## Proof of the basic decomposition theorem (Theorem 2.11)\n\n(2.11.1): By symmetry it suffices to prove only one of the statements in (2.11.1). Therefore assume that \( L \) satisfies the a.c.c. If \( a \) is \( \land \) -irreducible we are done; if not we may write \( a = {a}_{1} \land {a}_{2} \), where \( a < {a}_{1... | No |
Lemma 2.12. Let \( \mathfrak{a},\mathfrak{b} \in \mathcal{I} \), and suppose that \( \mathrm{V}\left( \mathfrak{a}\right) \rightarrow \mathfrak{b} \). Then \( \mathfrak{a} \subset \mathfrak{b} \), and \( \mathfrak{b} \rightarrow \;\mathsf{V}\left( \mathfrak{a}\right) \rightarrow \mathfrak{b} \rightarrow \;\mathsf{V}\le... | Proof. That \( \mathfrak{a} \subset \mathfrak{b} \) is obvious. This in turn implies \( \mathrm{V}\left( \mathfrak{a}\right) \supset \mathrm{V}\left( \mathfrak{b}\right) \). But from \( \mathrm{V}\left( \mathfrak{a}\right) \rightarrow \mathfrak{b}, s \in \mathrm{V}\left( \mathfrak{a}\right) \) implies \( f\left( s\righ... | Yes |
Let \( S = {\mathbb{R}}^{2}, k = \mathbb{R} \), and \( R = \mathbb{R}\left\lbrack {X, Y}\right\rbrack \) . We show that there are ideals \( {\mathfrak{a}}_{1},{\mathfrak{a}}_{2} \) in \( \mathcal{J} \) such that \( {\mathfrak{a}}_{1} + {\mathfrak{a}}_{2} \notin \mathcal{J} \) . Let \( {\mathfrak{a}}_{1} = \left( Y\righ... | Now \( Y \) and \( Y - {X}^{2} \) are both irreducible in \( \mathbb{R}\left\lbrack {X, Y}\right\rbrack \) ; this implies that \( \left( Y\right) \) and \( \left( {Y - {X}^{2}}\right) \) are the largest defining ideals of their varieties, for if a polynomial \( p \) vanishing on all of, say, \( \mathbf{V}\left( Y\right... | Yes |
Lemma 2.18. The map \( \mathfrak{a} \rightarrow \overline{\mathfrak{a}} = \mathcal{J}\left( {\mathcal{V}\left( \mathfrak{a}\right) }\right) \) is a closure map on the p.o. set \( \left( {\mathcal{I}, \subset }\right) \) . | Proof. It follows immediately from Lemma 2.12 that \( \mathfrak{a} \subset \overline{\mathfrak{a}} \) and \( \overline{\mathfrak{a}} = \overline{\overline{\mathfrak{a}}} \) . If \( {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \), then \( \mathrm{V}\left( {\mathfrak{a}}_{2}\right) \subset \mathrm{V}\left( {\mathfrak{a}... | Yes |
Theorem 2.19. J and V are lattice-reversing isomorphisms between \( \left( {\mathcal{V}, \subset ,\cap , \cup }\right) \) and \( \left( {\mathcal{J}, \subset ,\cap , + }\right) \) . | Proof. We need only show that for \( {\mathfrak{c}}_{1},{\mathfrak{c}}_{2} \in \mathcal{J} \), (2.19.1) \( V\left( {{\mathfrak{c}}_{1} \cap {\mathfrak{c}}_{2}}\right) = V\left( {\mathfrak{c}}_{1}\right) \cup V\left( {\mathfrak{c}}_{2}\right) \) (2.19.2) \( V\left( {{c}_{1} + {c}_{2}}\right) = V\left( {c}_{1}\right) \ca... | Yes |
Lemma 3.3. R satisfies the a.c.c. iff every ideal of \( R \) has a finite basis. | Proof. \( \Rightarrow \) : Suppose some ideal \( \mathfrak{a} \) did not have a finite basis. Then one could find a sequence of elements \( {a}_{1},{a}_{2},\ldots \left( {{a}_{k} \in \mathfrak{a}}\right) \) such that\n\n\[ \left( {a}_{1}\right) \subsetneqq \left( {{a}_{1},{a}_{2}}\right) \subsetneqq \ldots ,\]\n\nand \... | Yes |
Corollary 3.6. If \( k \) is a field, then \( k\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is Noetherian. | Proof. Certainly \( k \) satisfies the a.c.c. since it has only two ideals. Then by repeated application of Theorem 3.5, \( k\left\lbrack {X}_{1}\right\rbrack, k\left\lbrack {X}_{1}\right\rbrack \left\lbrack {X}_{2}\right\rbrack = k\left\lbrack {{X}_{1},{X}_{2}}\right\rbrack ,\ldots \) , \( k\left\lbrack {{X}_{1},\ldot... | Yes |
Lemma 4.8. Every irreducible closed ideal in \( \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) is prime. | Proof. Let \( \mathfrak{c} \) be irreducible and closed in \( \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \), and suppose it is not prime-say \( {a}_{1},{a}_{2} \in \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \smallsetminus \mathfrak{c} \) satisfy \( {a}_{1} \cdot {a}_{2} \in \mathfr... | Yes |
Corollary 5.4. There is a 1:1-onto correspondence between points of \( {\mathbb{C}}^{n} \) and maximal ideals of \( \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) . The maximal ideal corresponding to \( \left( {{c}_{1},\ldots ,{c}_{n}}\right) \in {\mathbb{C}}^{n} \) is \( \left( {\left( {{X}_{1} - {c}... | Proof of Corollary 5.4. First note that in \( {\mathbb{C}}^{n} \) any single point is irreducible, but any finite union of two or more points is reducible. Then Theorem 2.19 implies that \( J \) restricted to points defines a \( 1 : 1 \) onto correspondence between points of \( {\mathbb{C}}^{n} \) and closed maximal id... | Yes |
Lemma 5.7. Let \( R \) be any ring and let \( \mathfrak{a} \) be any ideal of \( R \) . Then\n\n\[ \sqrt{\mathfrak{a}} = \mathop{\bigcap }\limits_{{\mathfrak{p} \supset \mathfrak{a}}}\mathfrak{p} \] | Proof\n\n\( \sqrt{\mathfrak{a}} \subset \mathop{\bigcap }\limits_{{\mathfrak{p} \supset \mathfrak{a}}}\mathfrak{p} \) : If \( r \in \sqrt{\mathfrak{a}} \), then \( {r}^{n} \in \mathfrak{a} \) for some \( n \), so \( {r}^{n} \) is in each \( \mathfrak{p} \supset \mathfrak{a} \) . Since \( \mathfrak{p} \) is prime, \( r ... | Yes |
Theorem 5.11. In any ring, every prime ideal is irreducible. | Proof. Suppose \( \mathfrak{a} = {\mathfrak{a}}_{1} \cap {\mathfrak{a}}_{2},{\mathfrak{a}}_{1} \gneqq \mathfrak{a} \), and \( {\mathfrak{a}}_{2} \supsetneqq \mathfrak{a} \) . Surely we cannot have \( {\mathfrak{a}}_{1} \subset {\mathfrak{a}}_{2} \) or \( {\mathfrak{a}}_{2} \subset {\mathfrak{a}}_{1} \) . Hence let \( a... | Yes |
Lemma 6.2. Let \( D \subset {D}^{ * } \) be Noetherian integral domains. The elements of \( {D}^{ * } \) which are integral over D form an integral domain. | Proof of Lemma 6.2. It is clearly enough to show that if elements \( a, b \in {D}^{ * } \) are integral over \( D \), then so are \( a - b \) and \( {ab} \) . For this, note that \( w \) is integral over \( D \) if \[ {w}^{n} = {a}_{1}{w}^{n - 1} + \ldots + {a}_{n} \] that is, \( w \) is integral over \( D \) if some (... | Yes |
Lemma 7.4. If \( {h}_{\mathfrak{p}} \) is the natural homomorphism\n\n\[ \n{h}_{\mathfrak{p}} : \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \rightarrow \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p},\n\]\n\nthen \( {h}_{\mathfrak{p}}{}^{-1} \) induces a natural lattice-emb... | Proof. Let \( {\mathfrak{a}}_{1} \neq {\mathfrak{a}}_{2} \) be distinct ideals of \( \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p} \) ; say \( p \in {\mathfrak{a}}_{1} \) but \( p \notin {\mathfrak{a}}_{2} \) . Then for any \( q \in \left\{ {{h}_{\mathfrak{p}}{}^{-1}\left( p\right) }\right... | Yes |
Lemma 7.6. \( {h}_{\mathfrak{p}}{}^{-1} \) defines a natural lattice-embedding of \n\n\[ \n\left( {\mathcal{J}\left( {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p}}\right) ,\cap , + }\right) \n\]\n\ninto \( \left( {\mathcal{J}\left( {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\r... | Proof. Clearly \( {h}_{\mathfrak{p}}{}^{-1} \) defines a set-injection of \( \mathcal{J}\left( {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack /\mathfrak{p}}\right) \) into \( \mathcal{I}\left( {\mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack }\right) \) . That this injection is actually in... | Yes |
Let\n\n\[ p : \left( {{X}_{1},{X}_{2}}\right) \rightarrow \left( {{X}_{1},{X}_{1}{}^{2} + {X}_{2}}\right) = \left( {{Y}_{1},{Y}_{2}}\right) \]\n\nmap \( {\mathbb{R}}_{{X}_{1}{X}_{2}} \) to \( {\mathbb{R}}_{{Y}_{1}{Y}_{2}} \) . Then | \[ {p}^{-1} : \left( {{Y}_{1},{Y}_{2}}\right) \rightarrow \left( {{Y}_{1}, - {Y}_{1}^{2} + {Y}_{2}}\right) = \left( {{X}_{1}, - {X}_{1}^{2} + \left( {{X}_{1}^{2} + {X}_{2}}\right) }\right) = \left( {{X}_{1},{X}_{2}}\right) \]\n\nis the inverse of \( p \), so \( p \) is a polynomial isomorphism. Under \( p \), the horiz... | Yes |
Theorem 8.7. Two irreducible affine varieties \( V \subset {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{n}} \) and \( W \subset \) \( {\mathbb{C}}_{{Y}_{1},\ldots ,{Y}_{m}} \) are isomorphic iff their coordinate rings are \( \mathbb{C} \) -isomorphic. | Proof of Theorem 8.7\n\n\( \Leftarrow : \) Let \( \mathbb{C}\left\lbrack x\right\rbrack = \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \simeq \mathbb{C}\left\lbrack {{y}_{1},\ldots ,{y}_{m}}\right\rbrack = \mathbb{C}\left\lbrack y\right\rbrack \) be a given isomorphism. Then of course each \( {y}_{i} \... | Yes |
The natural embedding \( h : \mathbb{C}\left\lbrack X\right\rbrack \subset \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) . First note that for any embedding \( R \subset {R}^{ * } \), contraction becomes just intersection with \( R \) -that is, \( {\mathfrak{a}}^{ * } \subset {R}^{ * } \) implies \( {h}^{-1}\left( {\ma... | Now in \( \mathbb{C}\left\lbrack {X, Y}\right\rbrack \), any maximal ideal is of the form \( \left( {X - c, Y - d}\right) \) where \( c, d \in \mathbb{C} \), so the contraction \( {\left( X - c, Y - d}\right) }^{c} \) is \( \left( {X - c, Y - d}\right) \cap \mathbb{C}\left\lbrack X\right\rbrack = \) \( \left( {X - c}\r... | Yes |
Lemma 9.4. For any two embedded coordinate rings \( R \subset {R}^{ * } \), if \( {\mathfrak{m}}^{ * } \) is maximal in \( {R}^{ * } \), then \( {\mathfrak{m}}^{ * } \cap R = \mathfrak{m} \) is maximal in \( R \) . | Proof. Write \( R = \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) and \( {R}^{ * } = \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) where \( m \geq n \) . Then \( {m}^{ * } \) is of the form\n\n\[ \left( {{x}_{1} - {a}_{1},\ldots ,{x}_{m} - {a}_{m}}\right) \]\n\nso \( {R}^{ * }/{\ma... | Yes |
Can this map be looked at as a projection, as in Examples 9.2 and 9.3? | To answer this, let us write \( R \subset {R}^{ * } \) as\n\n\[ \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \subset \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \;\left( {m \geq n}\right) . \]\n\nBy so writing \( R \) and \( {R}^{ * } \) we have, of course, selected affine models \( {... | No |
Lemma 9.7. Let \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \subset \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{m}}\right\rbrack \) where \( m \geq n \). For any maximal ideal \( {\mathfrak{m}}^{ * } = \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{m} - {c}_{m}}\right) \subset \mathbb{C}\left\lbrack {{x}_{... | Proof. Since \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /\left( {x - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \simeq \mathbb{C}\left\lbrack {{c}_{1},\ldots ,{c}_{n}}\right\rbrack = \mathbb{C} \), \( \left( {{x}_{1} - {c}_{1},\ldots ,{x}_{n} - {c}_{n}}\right) \) is clearly maximal in \( \mathbb{C}... | Yes |
Theorem 10.1. Let \( R \subset {R}^{ * } \) be coordinate rings, and let \( \mathfrak{a} \) be any ideal of \( R \) . The set of all points in \( {\mathrm{Y}}_{{R}^{ * }} \) lying above \( \mathrm{Y}\left( \mathrm{a}\right) \subset {\mathrm{Y}}_{R} \) is the variety \( \mathrm{Y}\left( {\mathrm{a}}^{e}\right) \) . (Hen... | Proof. First recall from Definition 8.9 that for any ideal \( b \) in any coordinate ring \( R, y\left( b\right) \) is the subset of \( {y}_{R} \) consisting of all maximal ideals in \( R \) containing b. In particular,\n\n(10.2) \( y\left( {\mathfrak{a}}^{e}\right) \) is the set \( \left\{ {\mathfrak{m}}^{ * }\right\}... | Yes |
Theorem 10.8. Let \( R \subset {R}^{ * } \) be coordinate rings, and let \( \underline{\pi } \) be the natural projection \( {\mathfrak{m}}^{ * } \rightarrow {\mathfrak{m}}^{ * } \cap R \) from \( {\mathrm{Y}}_{{R}^{ * }} \) to \( {\mathrm{Y}}_{R} \). (10.8.1) If \( \mathfrak{a} \) and \( {\mathfrak{a}}^{ * } \) are id... | In proving Theorem 10.8 we shall assume the following fact: Lemma 10.9. Let \( V \) | No |
Consider the hyperbola \( V = \mathrm{V}\left( {{XY} - 1}\right) \subset {\mathbb{C}}_{XY} \). There is no point of \( V{\pi }_{Y} \) -lying over the origin of \( {\mathbb{C}}_{X} \); this fits in with the fact that \( {XY} - 1 = 0 \) does not define an integral equation for \( Y \) over \( \mathbb{C}\left\lbrack X\rig... | Tilting the \( Y \)-axis in effect adds a leading \( {Y}^{\prime 2} \)-term to \( {X}^{\prime }{Y}^{\prime } - 1 \), thus making \( {Y}^{\prime } \) integral over \( \mathbb{C}\left\lbrack {X}^{\prime }\right\rbrack \). | Yes |
Lemma 11.5 (Normalization lemma). If \( R = \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) has transcendence degree \( d \) over \( \mathbb{C} \), then there are elements \( {y}_{1},\ldots ,{y}_{d} \) in \( R \) such that \( R \) is integral over \( \mathbb{C}\left\lbrack {{y}_{1},\ldots ,{y}_{d}}\rig... | Proof. If \( d = n \), then we may take \( {y}_{i} = {x}_{i} \), and the lemma is trivially true. Thus suppose without loss of generality that \( {x}_{n} \) is algebraic over \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n - 1}}\right\rbrack \) . Let \( q\left( {{X}_{1},\ldots ,{X}_{n}}\right) \) be a polynomial of l... | Yes |
Lemma 2.6. With notation as immediately above and as in Notation 2.2, we have\n\n\[ \n\\operatorname{rank}\\left( {J{\\left( \\mathrm{H}\\left( V\\right) \\right) }_{{L}_{P} \\cap H}}\\right) = \\operatorname{rank}\\left( {J{\\left( \\mathrm{H}\\left( V\\right) \\cap H\\right) }_{{L}_{P} \\cap H}}\\right) ,\n\]\n\nwher... | Proof. Since these ranks are unaffected by any nonsingular change of coordinates in \( {\\mathbb{C}}^{n + 1} \), we may assume that \( H \) is defined by \( {X}_{n + 1} = 1 \), and that \( {L}_{P} \\cap H = \\left( {0,\\ldots ,0,1}\\right) \). Since relative to these coordinates, any Jacobian array for \( \\mathrm{H}\\... | Yes |
Theorem 2.13. Let \( V \subset {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{n}} \) be a nonempty irreducible variety. Suppose \( V \) ’s coordinate ring \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) has transcendence base \( \left\{ {{x}_{1},\ldots ,{x}_{d}}\right\} \) . Let (0) be a typical point of \( V \... | Proof. Note that if \( V \) is of codimension 1, the theorem follows immediately from Lemma 10.4 of Chapter II. For arbitrary codimension, the proof can easily be reduced to the codimension-one case, as follows: First, the standard proof of the theorem of the primitive element (as given, for instance, in [van der Waerd... | Yes |
Theorem 2.23. A variety in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or \( {\mathbb{C}}^{n} \) is a hypersurface \( \Leftrightarrow \) it is of pure dimension \( n - 1 \) . | Proof. Since any variety in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) is represented by a homogeneous variety in \( {\mathbb{C}}^{n + 1} \), it suffices to prove the result in the affine case.\n\n\( \Rightarrow : \) Suppose \( V = \mathbf{V}\left( p\right) \subset {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{n}} \), where \... | Yes |
(2.24.1) The set-theoretic product \( V \times W \subset {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{m},{Y}_{1},\ldots ,{Y}_{n}} \) is a variety. (We call it a product variety.) | Proof. The proof of (2.24.1) may be reduced to the case when \( V \) and \( W \) are both irreducible, since obviously \( \left( {\mathop{\bigcup }\limits_{i}{V}_{i}}\right) \times \left( {\mathop{\bigcup }\limits_{j}{W}_{j}}\right) = \mathop{\bigcup }\limits_{{i, j}}{V}_{i} \times {W}_{j} \) . This case then follows a... | No |
Theorem 3.8. If \( {V}_{1} \) and \( {V}_{2} \) are irreducible varieties in \( {\mathbb{C}}^{n} \), then each component of \( {V}_{1} \cap {V}_{2} \) has codimension at most \( \operatorname{cod}{V}_{1} + \operatorname{cod}{V}_{2} \) . | Our proof will essentially consist in looking at one component of \( {V}_{1} \cap {V}_{2} \) at a time; we do this by \ | No |
Theorem 6.12. Let \( {V}^{s} \subset {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{n}} \) be an irreducible variety of dimension \( s \) variety-theoretically projecting onto \( {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{s}} \), and let \( W \) be an irreducible subvariety of \( {V}^{s} \) . Then for each integer \( k \geq 0 \), the set ... | Proof. Suppose, first, that \( {V}^{s} \) is a hypersurface. Without loss of generality, let \( p \in \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) be irreducible; then from Theorem 6.6.2, the order with respect to \( {\mathbb{C}}_{{X}_{n}} \) of \( {V}^{s} = \mathbf{V}\left( p\right) \) at \( \left(... | Yes |
Theorem 6.13. Let \( X \) be an irreducible subvariety of a pure-dimensional variety \( V \) in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or \( {\mathbb{C}}^{n} \). For almost every point \( P \) on \( X, m\left( {V;P}\right) \) has a common, fixed value. | Proof. The proof is essentially the same as the proof of Theorem 6.8.1; assume without loss of generality that \( X \) is affine, and in place of the translate \( S \times \{ P\} \times \{ 1\} \) ), use \( S \times X \times \{ 1\} \) . This gives an \ | No |
Theorem 6.18. Let \( {V}_{1} \) and \( {V}_{2} \) in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or \( {\mathbb{C}}^{n} \) be of pure dimensions \( r \) and \( s \) , respectively, and let \( {L}^{\left( {n - r}\right) + \left( {n - s}\right) } = L \) be linear of dimension \( {2n} - r - s \) . If \( {V}_{1},{V}_{2}... | Proof. The proof is entirely analogous to that of Theorem 6.15, except we use Theorem 6.10 instead of Theorem 11.1.1 of Chapter III. | No |
Theorem 6.20. Let \( {V}_{1} \) and \( {V}_{2} \) in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or \( {\mathbb{C}}^{n} \) be of pure dimensions \( r \) and \( s \) , respectively. If they intersect properly at a point \( P \), then for almost every \( P \) -containing transform \( {L}^{\prime } \) of a linear varie... | Proof. The proof is similar to that of Theorem 6.8.1. The \ | No |
Theorem 6.22. Let \( {V}_{1} \) and \( {V}_{2} \) in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or \( {\mathbb{C}}^{n} \) be of pure dimension, and suppose they intersect properly. If \( C \) is an irreducible component of \( {V}_{1} \cap {V}_{2} \), then at almost every point \( P \in C, i\left( {{V}_{1},{V}_{2};P... | Proof. The proof is like that of Theorem 6.20; assume \( C \) is affine, and in place of \( S \times \{ P\} \times \{ 1\} \) for \( W \), use \( S \times C \times \{ 1\} \) . This gives the \ | No |
Theorem 7.1 (Bézout’s theorem). Suppose two pure-dimensional varieties \( V \) and \( W \) in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) intersect properly. Then\n\n\[ \deg \left( {V \cdot W}\right) = \deg V \cdot \deg W. \] | We shall prove this theorem by showing that it can be reduced to the case when one variety, say \( V \), is of an especially simple form, namely when it is a union of \( \deg V \) distinct projective subspaces of dimension \( \dim V \) . From Theorem 6.2 and Definition 6.3 we know that \( W \), almost every projective ... | Yes |
Lemma 7.3. With respect to appropriate coordinates, any pure-dimensional variety \( V \subset {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) can be reduced to a union of \( \deg V \) distinct subspaces of dimension \( r = \dim V \) by means of a projective transformation defined by an \( \left( {n + 1}\right) \times \left... | Proof. Let \( \mathrm{H}\left( V\right) \subset {\mathbb{C}}_{{X}_{1},\ldots ,{X}_{n + 1}} \) be the homogeneous variety representing \( V \) . Without loss of generality we may suppose coordinates have been chosen so that \( {\mathbb{C}}_{{X}_{r + 1},\ldots ,{X}_{n + 1}} \) intersects \( \mathrm{H}\left( V\right) \) i... | Yes |
Let \( V = {\mathbb{C}}_{XY} \). The function field of \( V \) is \( \mathbb{C}\left( {X, Y}\right) \). What is the value of \( Y/X \) at the origin? | We cannot directly assign a definite value, even infinity, to \( 0/0 \). The origin is then a point of indeterminancy for \( Y/X \). We can, however, approach \( \left( {0,0}\right) \) along different directions in \( {\mathbb{C}}_{XY} \); for various directions we get different values. For instance, approaching \( \le... | Yes |
Theorem 2.12. Let \( K \) and \( k \) be fields; if each element in \( K \) is assigned a value in \( k \cup \{ \infty \} \), and if this assignment satisfies properties (2.5.1) and (2.5.2), then the set of elements assigned finite values forms a subring \( R \) of \( K \), and for each \( a \in K, a \notin R \) implie... | Proof. The first half is obvious. For the converse, assume without loss of generality that \( R \neq K \), and let \( m \) be the set of elements \( a \) of \( R \) such that \( 1/a \notin R \) . We show that \( \mathfrak{m} \) is a maximal ideal in \( R \) . Then properties (2.5.1) and (2.5.2) follow at once for the f... | Yes |
Lemma 2.18. Any proper valuation ring \( R\left( {\mathbb{C} \subset R}\right) \) of a field \( K = \mathbb{C}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) having transcendence degree one over \( \mathbb{C} \) is discrete rank one. | The idea of the proof is quite simple, and is given in Exercise 2.2. | No |
Lemma 2.20. Let \( R \) be a discrete rank one valuation ring of a field. For any ideal \( \mathfrak{a} \subset R \) , \[ \mathfrak{a} = \{ a \in R \mid \operatorname{ord}\left( a\right) \geq \operatorname{ord}\left( \mathfrak{a}\right) \} \] | Proof. First, for each nonnegative integer \( n \), a contains at least one element of order \( \operatorname{ord}\left( \mathfrak{a}\right) + n \) . For let \( a \) be any element of \( \mathfrak{a} \) such that \( \operatorname{ord}\left( a\right) = \operatorname{ord}\left( \mathfrak{a}\right) \) , and let \( b \) be... | Yes |
Corollary 2.21. Every discrete rank one valuation ring is a principal ideal ring. | Proof. Let \( \mathfrak{a} \) be any nonzero ideal of \( R \), and let \( a \) be an element of least order in a. From the proof of Lemma 2.20 we see that \( \left( a\right) \) consists of all elements of \( R \) of order \( \geq \operatorname{ord}\left( \mathfrak{a}\right) \) ; but so does \( \mathfrak{a} \), so \( \m... | Yes |
Corollary 2.22. Every discrete rank one valuation ring is Noetherian. | Proof. Every principal ideal ring \( R \) is Noetherian. (Let \( {\mathfrak{a}}_{1} \subseteq {\mathfrak{a}}_{2} \subseteq \cdots \) be an ascending sequence of ideals of \( R \) . Then \( \mathop{\bigcup }\limits_{i}{a}_{i} = \left( a\right) \) for some \( a \in R \) . But \( a \in {\mathfrak{a}}_{i} \), for some \( i... | No |
Corollary 2.23. Let \( \mathfrak{a} \) be any proper ideal in a discrete rank one valuation ring \( R \) .\n\nThen \( \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{\mathfrak{a}}^{m} = \left( 0\right) \) . | Proof. If \( \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{\mathfrak{a}}^{m} \neq \left( 0\right) \), then \( \mathop{\bigcap }\limits_{{m = 1}}^{\infty }{\mathfrak{a}}^{m} = \left( b\right) \) for some \( b \neq 0 \) . If \( \operatorname{ord}\left( \mathfrak{a}\right) = r \) and \( \operatorname{ord}\left( b\right) = ... | Yes |
Each element \( x \) of \( \mathbb{C}\left( {{x}_{1},\ldots ,{x}_{n}}\right) \) has a Laurent series development \( x = \mathop{\sum }\limits_{{i = {n}_{0}}}^{\infty }{c}_{i}{T}^{i}\left( {{n}_{0} \in \mathbb{Z}}\right) \) convergent, except possibly at zero, in some neighborhood of \( \left( 0\right) \in {\mathbb{C}}_... | Proof. Write \( x \) as \( y/z \), where \( y \) and \( z \) are in \( R \) . For some \( n \geq 0, z = \) \( {c}_{n}{T}^{n} + {c}_{n + 1}{T}^{n + 1} + \ldots = {T}^{n}\left( {{c}_{n} + {c}_{n + 1}T + \ldots }\right) \) where \( {c}_{n} \neq 0 \) ; since the reciprocal of \( {c}_{n} + {c}_{n + 1}T + \ldots \) is analyt... | Yes |
Theorem 2.31. Let \( P = \left( {0,0}\right) \) be an arbitrary nonsingular point of an irreducible plane curve \( C = \mathrm{V}\left( p\right) \subset {\mathbb{C}}_{XY} \), and suppose \( \mathrm{V}\left( X\right) \left( { = {\mathbb{C}}_{Y} \subset {\mathbb{C}}_{XY}}\right) \) is not tangent to \( C \) at \( P \) . ... | Proof. Assume without loss of generality that coordinates have been chosen so the tangent line to \( C \) at \( \left( {0,0}\right) \) is \( {\mathbb{C}}_{X}\left( { = \mathrm{V}\left( Y\right) }\right) \) . Then \( {p}_{X}\left( {0,0}\right) = 0 \), so by nonsingularity, \( {p}_{Y}\left( {0,0}\right) \neq 0 \) . By Th... | Yes |
Theorem 2.32. Let \( P \) be an arbitrary point of an irreducible plane curve \( C \subset {\mathbb{C}}_{XY} \) having coordinate ring \( \mathbb{C}\left\lbrack {x, y}\right\rbrack \) and function field \( {K}_{C} = \mathbb{C}\left( {x, y}\right) \). Then there is an evaluation of the elements of \( {K}_{C} \) satisfyi... | Proof. If at \( P \) we could express both \( x \) and \( y \) as convergent power series in an element \( T \in {K}_{C} \) of order one, then every element \( f \in {K}_{C} \) would be a Laurent series in \( T, f\left( {x, y}\right) = {c}_{m}{T}^{m} + {c}_{m + 1}{T}^{m + 1} + \ldots \). If \( \left( {x, y}\right) \) e... | Yes |
Theorem 2.33. Let \( C \subset {\mathbb{C}}_{XY} \) be an irreducible curve with coordinate ring \( \mathbb{C}\left\lbrack {x, y}\right\rbrack \) and function field \( \mathbb{C}\left( {x, y}\right) \). Let \( P \) be a given point of \( C \), let \( R \) be a valuation ring with center \( P \) on \( C \), and let \( {... | Proof. Assume without loss of generality that \( P = \left( {0,0}\right) \in {\mathbb{C}}_{XY} \) and that \( x \notin \mathbb{C} \). Then the order in \( R \) of \( x \) is \( m \geq 1 \). From Theorem 2.26 we have, for some \( U \) of order 1 in \( R \),\n\n\[ x = {U}^{m}\left( {{c}_{0} + {c}_{1}U + \ldots }\right) \... | Yes |
Let \( \operatorname{ord}\left( {p\left( {X, Y}\right) }\right) \) denote the total order at \( \left( {0,0}\right) \) of a polynomial \( p \in \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) ; as in (1), if \( f\left( {X, Y}\right) = p\left( {X, Y}\right) /q\left( {X, Y}\right) \) is any element of \( \mathbb{C}\left( {... | Now not only does every nonzero element \( c \in \mathbb{C} \) have order 0, but so also does \( X/Y \), for example. The elements \( X/Y \) and \( c \) cannot possibly represent the same coset in \( R/m \), for their difference \( \left( {X/Y}\right) - c \) would then be in \( \mathfrak{m} \) ; this is not so since \(... | Yes |
In \( \mathbb{C}\left( {X, Y}\right) \), let \( R = \{ p\left( {X, Y}\right) /q\left( {X, Y}\right) \mid q\left( {X,0}\right) \neq 0\} (p, q \) relatively prime in \( \mathbb{C}\left\lbrack {X, Y}\right\rbrack \) ). The ring \( R \) consists of the set of all elements in \( \mathbb{C}\left( {X, Y}\right) \) having nonn... | In the above example, we approached first in the \( Y \) -direction, then in \( X \) -direction to get a value in \( \mathbb{C} \cup \{ \infty \} \) . It is reasonable to next ask whether we get the same value by approaching first in the \( X \) -direction, then in the \( Y \) -direction. In general, we do not. | Yes |
Theorem 3.1. Let \( {V}_{1} \) and \( {V}_{2} \) be varieties in \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or in \( {\mathbb{C}}^{n} \), each of whose irreducible components contains a given point \( P \) ; if there is an open neighborhood \( U \) of \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) or \( {\mathbb{C}... | Proof. If \( {V}_{1} \) and \( {V}_{2} \) are both irreducible, then the theorem follows at once from Theorem 2.11 of Chapter IV. In the general case, we note that each irreducible component of \( {V}_{1} \) contains a point \( P \in U \) which is in no other irreducible component of \( {V}_{1} \), and in precisely one... | Yes |
Lemma 3.9. The ring \( {R}_{\mathfrak{p}} \) is Noetherian. | Proof. \( {\left( \;\right) }^{c} \) is 1 : 1 onto the set of contracted ideals of \( R \) ; since \( {\left( \;\right) }^{c} \) preserves inclusion, any infinite strictly ascending sequence of ideals in \( {R}_{\mathfrak{p}} \) would map, under \( {\left( \;\right) }^{c} \), to an infinite strictly ascending sequence ... | No |
Theorem 3.10. Let \( R \) and \( \mathfrak{p} \) be as above. Then \( \mathfrak{q} = {\mathfrak{q}}^{ec} \) for any irreducible ideal \( \mathfrak{q} \subset \mathfrak{p} \) . | Proof. Let \( x \) be any element of \( R \), and let \( \left( x\right) \) be the principal ideal of \( R \) generated by \( x \) ; define the quotient ideal \( \mathfrak{q} : \left( x\right) \) to be \( \mathfrak{q} : \left( x\right) = \{ r \in R \mid {xr} \in \mathfrak{q}\} \) . (This is a special case of the quotie... | No |
In \( {\mathbb{C}}_{XY} \) if \( P = \left( {0,0}\right) \), if \( V = \mathrm{V}\left( {Y - X}\right) \), and if \( {V}^{\prime } = \mathrm{V}\left( {Y - {X}^{2}}\right) \), then \( {V}_{P} = \left( {\mathrm{V}\left( {Y - X}\right) ,\left( {0,0}\right) }\right) \) and \( {V}_{P}^{\prime } = \left( {\mathrm{V}\left( {Y... | We can then translate this geometric fact into ideal language (Cf. Theorem 3.14): Let \( {\mathfrak{a}}^{ * } \) and \( {\mathfrak{b}}^{ * } \) be the principal ideals \( \left( {Y - X}\right) \) and \( \left( {Y - {X}^{2}}\right) \) in the localization \( \mathbb{C}{\left\lbrack X, Y\right\rbrack }_{\left( X, Y\right)... | Yes |
For each ideal \( \mathfrak{a} \in \mathcal{J}\left( R\right) ,{\mathfrak{a}}^{ec} \) is the intersection of those prime ideals which contain \( \mathfrak{a} \) and which are contained in \( \mathfrak{p} \) . (The intersection of an empty set of prime ideals is defined to be \( R \) .) Thus if \( \mathfrak{a} \subset R... | Proof. Write \( \mathfrak{a} = {\mathfrak{p}}_{1} \cap \ldots \cap {\mathfrak{p}}_{r} \) where each \( {\mathfrak{p}}_{i} \) is prime in \( R \) . It follows at once from Theorem 3.10 that \( {\mathfrak{p}}_{i}{}^{ec} = {\mathfrak{p}}_{i} \) iff \( {\mathfrak{p}}_{i} \subset \mathfrak{p} \) . If \( {\mathfrak{p}}_{i} ⊄... | Yes |
Theorem 3.16. Let \( {\mathfrak{a}}^{ * },{\mathfrak{b}}^{ * } \) be any two ideals in \( \mathcal{J}\left( {R}_{\mathfrak{p}}\right) \), and \( {X}_{W},{Y}_{W} \) any two elements of \( \mathcal{G}\left( {R}_{\mathfrak{p}}\right) \) . Then\n\n\[ \text{(3.16.1)}{G}^{ * }\left( {{\mathfrak{a}}^{ * } \cap {\mathfrak{b}}^... | Proof (3.16.1): This is easy, since \( {\left( \;\right) }^{c} \) preserves \( \cap , \vee \) is lattice-reversing, and ( \( \;{)}_{w} \) preserves \( \cup \) . | No |
Theorem 4.2. Let \( W \) be an irreducible subvariety of an \( r \) -dimensional irreducible variety \( V \) in \( {\mathbb{C}}^{n} \) or \( {\mathbb{P}}^{n}\left( \mathbb{C}\right) \) . Then \( W \) is nonsingular in \( V \) iff almost every point \( P \in W \) is nonsingular in \( V \) . | Proof. It suffices to assume that \( V \) is affine. Let \( \left( y\right) \) be a generic point of \( W \) . \( \Rightarrow \) : Some \( \left( {n - r}\right) \times \left( {n - r}\right) \) submatrix of \( J{\left( V\right) }_{\left( y\right) } \) has nonzero determinant. This determinant is an element of \( \mathbb... | Yes |
Lemma 4.6. Let \( W \subset V \) be irreducible varieties of \( {\mathbb{C}}^{n} \) of dimensions \( s \) and \( r \) respectively, and let \( m \) be the maximal ideal of the local ring of \( W \) in \( V \). Then there exist \( r - s \) elements \( {a}_{1},\ldots ,{a}_{r - s} \) of \( m \) such that \(\\sqrt{\\left( ... | Proof. It suffices to find \( r - s \) polynomials \( {p}_{1},\ldots ,{p}_{r - s} \in \mathbb{C}\left\lbrack {{X}_{1},\ldots ,{X}_{n}}\right\rbrack \) such that \( V \cap \mathrm{V}\left( {p}_{1}\right) \cap \ldots \cap \mathrm{V}\left( {p}_{r - s}\right) \) is an \( s \) -dimensional variety containing \( W \). For th... | Yes |
Theorem 4.8. Let \( W \subset V \) be irreducible varieties in \( {\mathbb{C}}^{n} \) of dimensions \( s \) and \( r \) , respectively, and let \( \mathfrak{m} \) be the maximal ideal of the local ring \( R = \mathfrak{o}\left( {W;V}\right) \) of \( V \) at \( W \) . Then \( W \) is nonsingular in \( V \) iff \( m \) i... | Proof of THEOREM 4.8. We first establish the theorem for \( V = {\mathbb{C}}^{n} \) ; we then use this result to prove the full theorem.\n\nFirst, note from Definition 4.1 that any irreducible variety \( W \subset {\mathbb{C}}^{n} \) is always nonsingular in \( {\mathbb{C}}^{n} \) ; in fact, each point of \( W \) is no... | Yes |
Let us consider the complex circle \( C = \mathrm{V}\left( {{X}^{2} + {Y}^{2} - 2}\right) \subset \) \( {\mathbb{C}}_{XY} \), its coordinate ring being \( \mathbb{C}\left\lbrack {x, y}\right\rbrack = \mathbb{C}\left\lbrack {X, Y}\right\rbrack /\left( {{X}^{2} + {Y}^{2} - 2}\right) \) . Any nonzero ideal of \( \mathbb{C... | The totality of these four points is \( \mathrm{V}\left( {a}_{1}\right) \cup \mathrm{V}\left( {a}_{2}\right) = \mathrm{V}\left( {{a}_{1} \cap {a}_{2}}\right) = \) \( \mathrm{V}\left( {{\mathfrak{a}}_{1} \cdot {\mathfrak{a}}_{2}}\right) \), and may also be looked at as the union \( \mathrm{V}\left( {\mathfrak{b}}_{1}\ri... | Yes |
Lemma 5.7. Suppose \( R \) is a Noetherian domain with quotient field \( K \), and let \( \mathfrak{m} \) be any maximal ideal of \( R \) ; if the maximal ideal \( \mathfrak{M} \) of \( {R}_{\mathfrak{m}} \) is principal, then \( {R}_{\mathrm{m}} \) is a valuation ring. | Proof. Suppose \( \mathfrak{M} = \left( m\right) \) . We first show this:\n\n(5.8) Each element \( a \) of \( {R}_{\mathrm{m}} \) can be written as \( a = u{m}^{n} \) for some unit \( u \in {R}_{\mathrm{m}} \smallsetminus \mathfrak{M} \) and some nonnegative integer \( n \) .\n\nIf \( m \) does not divide \( a \) (that... | Yes |
Lemma 5.11. Let \( {R}_{m} \) be the valuation ring of Lemma 5.7. Then \( {R}_{m} \) is integrally closed in its quotient field \( K \) . | Proof. If \( {R}_{\mathrm{m}} \) ’s maximal ideal is \( \left( m\right) \), let \( a = u{m}^{-n} \), where \( n > 0 \), be a typical element of \( K \smallsetminus {R}_{\mathfrak{m}}\left( u\right. \), a unit in \( \left. {R}_{\mathfrak{m}}\right) \) . If \( a \) were integral over \( {R}_{\mathfrak{m}} \), there would... | Yes |
Theorem 5.16. Let \( \left( 0\right) \) be an arbitrary point of a nonsingular curve \( C \subset {\mathbb{C}}_{{X}_{1},{X}_{2}}\; \) defined \( \; \) by \( \; \) an \( \; \) irreducible \( \; \) polynomial \( \;q\left( {{X}_{1},{X}_{2}}\right) = \) \( q \in \mathbb{C}\left\lbrack {{X}_{1},{X}_{2}}\right\rbrack \), and... | Proof. The following conditions are satisfied for all but finitely many choices of linear coordinates about \( \left( 0\right) \in {\mathbb{C}}_{{X}_{1}{X}_{2}} \) : (a) The leading terms of \( p \) and \( q \) are \( {X}_{2}{}^{\deg p} \) and \( {X}_{2}{}^{\deg q} \), respectively; (b) \( \left( 0\right) = C \cap \mat... | Yes |
Lemma 6.16. \( {\operatorname{ord}}_{P}\omega \) is independent of the choice of uniformizing variable. | Proof. Let \( z \) and \( w \) be any two uniformizing variables in \( \mathfrak{o}\left( {P;C}\right) \) ; then \( {fdz} = f\left( {{dz}/{dw}}\right) {dw} \) . We wish to prove \( {\operatorname{ord}}_{P}\left( {{dz}/{dw}}\right) = 0 \) . Now \( z = {uw} \) where \( u \) is a unit in \( \mathfrak{o}\left( {P;C}\right)... | Yes |
Lemma 7.1. For each divisor \( D \) on \( C,\dim L\left( D\right) \) is finite. | Proof. Since \( \deg \operatorname{div}\left( f\right) = 0, - D \leq \operatorname{div}\left( f\right) \) implies that \( \deg D \geq 0 \) . Hence if \( \deg D \leq 0 \), then \( \dim L\left( D\right) = 0 \) . Now suppose \( \deg D \geq 0 \), and write \( D = {P}_{1} + \ldots + {P}_{n} - {Q}_{1} - \ldots - {Q}_{{n}^{\p... | Yes |
Lemma 7.2. \( {D}_{1} \simeq {D}_{2} \) implies \( \dim L\left( {D}_{1}\right) = \dim L\left( {D}_{2}\right) \) . | Proof. If \( {D}_{1} \simeq {D}_{2} \), then \( {D}_{1} - {D}_{2} = \operatorname{div}\left( f\right) \) for some \( f \in {K}_{C} \), so\n\n\[\n\deg {D}_{1} - \deg {D}_{2} = \deg \operatorname{div}\left( f\right) = 0.\n\] | No |
Lemma 7.3. \( {D}_{1} \simeq {D}_{2} \) implies \( \dim L\left( {D}_{1}\right) = \dim L\left( {D}_{2}\right) \) . | Proof. For some \( f \in {K}_{C} \smallsetminus \{ 0\} ,{D}_{1} = {D}_{2} + \operatorname{div}\left( f\right) \) . Since \( f \) is not the zero function, \( g \rightarrow {fg} \) is \( 1 : 1 \) and linear from \( L\left( {D}_{1}\right) \) onto \( L\left( {D}_{2}\right) \) . Hence \( \dim L\left( {D}_{1}\right) = \) \(... | Yes |
Theorem 7.12 (Riemann’s theorem). Let \( C \subset {\mathbb{P}}^{2}\left( \mathbb{C}\right) \) be a nonsingular curve of genus \( g \) . Then for all divisors \( D \) on \( C \) , \[ \dim L\left( D\right) \geq \deg D - \left( {g - 1}\right) \] \( g - 1 \) is the smallest constant \( c \) for which \( \dim L\left( D\rig... | Proof. As noted earlier, we want to find divisors \( D \) of arbitrary large degree for which we can compute \( L\left( D\right) \) . We choose these as follows: Take any line in \( {\mathbb{P}}^{2}\left( \mathbb{C}\right) \) intersecting \( C \) in \( n = \deg C \) distinct points \( {P}_{1},\ldots ,{P}_{n} \) ; let \... | Yes |
For every prime \( p \), the group \( {\mathbb{J}}_{p} \) of \( p \) -adic integers (see §A.4.2), considered as the ring of all endomorphisms of the Prüfer group \( \mathbb{Z}\left( {p}^{\infty }\right) \), embeds into the product \( \mathbb{Z}{\left( {p}^{\infty }\right) }^{\mathbb{Z}\left( {p}^{\infty }\right) } \) .... | The same topology on \( {\mathbb{J}}_{p} \) is induced by the product topology of \( \mathop{\prod }\limits_{{n \in \mathbb{N}}}\mathbb{Z}/{p}^{n}\mathbb{Z} \), when we consider \( {\mathbb{J}}_{p} \) as the inverse limit \( \mathop{\lim }\limits_{{ \leftarrow n \in \mathbb{N}}}\mathbb{Z}/{p}^{n}\mathbb{Z} \) (for more... | Yes |
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