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Lemma 69.3.5 (Wedge product respects de Rham cohomology)\n\nThe wedge product induces a map\n\n\[ \land : {H}_{\mathrm{{dR}}}^{k}\left( M\right) \times {H}_{\mathrm{{dR}}}^{\ell }\left( M\right) \rightarrow {H}_{\mathrm{{dR}}}^{k + \ell }\left( M\right) . \] | Proof. First, we recall that the operator \( d \) satisfies\n\n\[ d\left( {\alpha \land \beta }\right) = \left( {d\alpha }\right) \land \beta + \alpha \land \left( {d\beta }\right) . \]\n\nNow suppose \( \alpha \) and \( \beta \) are closed forms. Then from the above, \( \alpha \land \beta \) is clearly closed. Also if... | Yes |
Lemma 69.4.3 ( \( \delta \) with cup products)\n\nWe have \( \delta \left( {\phi \smile \psi }\right) = {\delta \phi } \smile \psi + {\left( -1\right) }^{k}\phi \smile {\delta \psi } \) . | Proof. Direct \( \sum \) computations. | No |
Proposition 69.4.4 (Cohomology is anticommutative)\n\n\( {H}^{ \bullet }\left( {X;R}\right) \) is an anticommutative ring, meaning \( \phi \smile \psi = {\left( -1\right) }^{k\ell }\psi \smile \phi \) . | For a proof, see [Ha02, Theorem 3.11, pages 210-212]. | No |
The cohomology ring \( {H}^{ \bullet }\left( {{S}^{1} \times {S}^{1};\mathbb{Z}}\right) \) of the torus is generated by elements \( \left| \alpha \right| = \left| \beta \right| = 1 \) which satisfy the relations \( \alpha \smile \alpha = \beta \smile \beta = 0 \), and \( \alpha \smile \beta = - \beta \smile \alpha \) .... | Thus as a \( \mathbb{Z} \) -module it is\n\n\[ \n{H}^{ \bullet }\left( {{S}^{1} \times {S}^{1};\mathbb{Z}}\right) \cong \mathbb{Z} \oplus \left\lbrack {\alpha \mathbb{Z} \oplus \beta \mathbb{Z}}\right\rbrack \oplus \left( {\alpha \smile \beta }\right) \mathbb{Z}.\n\]\n\nThis gives the expected dimensions \( 1 + 2 + 1 =... | Yes |
Consider \( {S}^{n} \) for \( n \geq 1 \) . The nontrivial cohomology groups are given by \( {H}^{0}\left( {{S}^{n};\mathbb{Z}}\right) \cong \) \( {H}^{n}\left( {{S}^{n};\mathbb{Z}}\right) \cong \mathbb{Z} \) . So as an abelian group\n\n\[ \n{H}^{ \bullet }\left( {{S}^{n};\mathbb{Z}}\right) \cong \mathbb{Z} \oplus \alp... | Now, observe that \( \left| {\alpha \smile \alpha }\right| = {2n} \), but since \( {H}^{2n}\left( {{S}^{n};\mathbb{Z}}\right) = 0 \) we must have \( \alpha \smile \alpha = 0 \) . So even more succinctly,\n\n\[ \n{H}^{ \bullet }\left( {{S}^{n};\mathbb{Z}}\right) \cong \mathbb{Z}\left\lbrack \alpha \right\rbrack /\left( ... | Yes |
Theorem 69.6.3 (Cohomology pseudo-rings of wedge sums)\n\nWe have\n\n\[ \n{\widetilde{H}}^{ \bullet }\left( {X \land Y;R}\right) \cong {\widetilde{H}}^{ \bullet }\left( {X;R}\right) \times {\widetilde{H}}^{ \bullet }\left( {Y;R}\right) \n\] \n\nas graded pseudo-rings. | This allows us to resolve the first question posed at the beginning. Let \( X = {\mathbb{{CP}}}^{2} \) and \( Y = {S}^{2} \vee {S}^{4} \) . We have that\n\n\[ \n{H}^{ \bullet }\left( {{\mathbb{{CP}}}^{2};\mathbb{Z}}\right) \cong \mathbb{Z}\left\lbrack \alpha \right\rbrack /\left( {\alpha }^{3}\right) \n\] \n\nHence thi... | Yes |
Theorem 70.4.2 (Intersections and unions of varieties)\n\n(a) The intersection of affine varieties (even infinitely many) is an affine variety.\n\n(b) The union of finitely many affine varieties is an affine variety. | In fact we have\n\n\[ \mathop{\bigcap }\limits_{\alpha }\mathcal{V}\left( {I}_{\alpha }\right) = \mathcal{V}\left( {\mathop{\sum }\limits_{\alpha }{I}_{\alpha }}\right) \;\text{ and }\;\mathop{\bigcup }\limits_{{k = 1}}^{n}\mathcal{V}\left( {I}_{k}\right) = \mathcal{V}\left( {\mathop{\bigcap }\limits_{{k = 1}}^{n}{I}_{... | Yes |
Example 70.5.3 (Irreducible varieties of \( {\mathbb{A}}^{1} \) ) | The irreducible varieties of \( {\mathbb{A}}^{1} \) are:\n\n- the empty set \( \mathcal{V}\left( 1\right) \) ,\n\n- a single point \( \mathcal{V}\left( {x - a}\right) \), and\n\n- the entire line \( {\mathbb{A}}^{1} = \mathcal{V}\left( 0\right) \) . | Yes |
Theorem 70.5.5 (Prime \( \Leftrightarrow \) irreducible)\n\nLet \( I \) be a radical ideal, and \( V = \mathcal{V}\left( I\right) \) a nonempty variety. Then \( I \) is prime if and only if \( V \) is irreducible. | Proof. First, assume \( V \) is irreducible; we’ll show \( I \) is prime. Let \( f, g \in \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) so that \( {fg} \in I \) . Then \( V \) is a subset of the union \( \mathcal{V}\left( f\right) \cup \mathcal{V}\left( g\right) \) ; actually, \( V = \left( {V \cap \... | Yes |
Proposition 70.7.1 \( (\mathcal{V}\left( -\right) \) is inclusion reversing) | If \( I \subseteq J \) then \( \mathcal{V}\left( I\right) \supseteq \mathcal{V}\left( J\right) \) . Thus \( \mathcal{V}\left( -\right) \) is inclusion-reversing. | Yes |
Let us determine the open sets of \( {\mathbb{A}}^{1} \), which as usual we picture as a straight line (ignoring the fact that \( \mathbb{C} \) is two-dimensional). | Since \( \mathbb{C}\left\lbrack x\right\rbrack \) is a principal ideal domain, rather than looking at \( \mathcal{V}\left( I\right) \) for every \( I \subseteq \mathbb{C}\left\lbrack x\right\rbrack \), we just have to look at \( \mathcal{V}\left( f\right) \) for a single \( f \) . There are a few flavors of polynomials... | Yes |
Let \( V = \mathcal{V}\left( {y - {x}^{2}}\right) \subseteq {\mathbb{A}}^{2} \) be a parabola, and let \( U = V \smallsetminus \{ \left( {1,1}\right) \} \) . We claim \( U \) is open in \( V \) . | Indeed, \( \widetilde{U} = {\mathbb{A}}^{2} \smallsetminus \{ \left( {1,1}\right) \} \) is open in \( {\mathbb{A}}^{2} \) (since it is the complement of the closed set \( \mathcal{V}\left( {x - 1, y - 1}\right) ) \), so \( U = \widetilde{U} \cap V \) is open in \( V \) . Note that on the other hand the set \( U \) is n... | Yes |
Theorem 71.4.2 (Coordinate rings correspond to ideal)\n\nLet \( I \) be a radical ideal, and \( V = \mathcal{V}\left( I\right) \subseteq {\mathbb{A}}^{n} \) . Then\n\n\[ \mathbb{C}\left\lbrack V\right\rbrack \cong \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I \] | Proof. There's a natural surjection as above\n\n\[ \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \rightarrow \mathbb{C}\left\lbrack V\right\rbrack \]\n\nand the kernel is \( I \) . | Yes |
Example 72.1.8 (Example of a graded quotient ring)\n\nLet \( R = \mathbb{C}\left\lbrack {x, y}\right\rbrack \) and set \( I = \left( {{x}^{3},{y}^{2}}\right) \) . Let \( S = R/I \) . Then | \n\[
{S}^{0} = \mathbb{C}
\]
\n\n\[
{S}^{1} = \mathbb{C}x \oplus \mathbb{C}y
\]
\n\n\[
{S}^{2} = \mathbb{C}{x}^{2} \oplus \mathbb{C}{xy}
\]
\n\n\[
{S}^{3} = \mathbb{C}{x}^{2}y
\]
\n\n\[
{S}^{d} = 0\;\forall d \geq 4.
\]
\n\nSo in fact \( S = R/I \) is graded, and is a six-dimensional \( \mathbb{C} \) -vector space. | Yes |
Example 72.2.4 (Colloquially, \( {\mathbb{{CP}}}^{1} = {\mathbb{A}}^{1} \cup \{ \infty \} \) ) | The space \( {\mathbb{{CP}}}^{1} \) consists of pairs \( \left( {s : t}\right) \), which you can think of as representing the complex number \( z/1 \) . In particular \( {U}_{1} = \{ \left( {z : 1}\right) \} \) is basically another copy of \( {\mathbb{A}}^{1} \) . There is only one new point, \( \left( {1 : 0}\right) \... | Yes |
If \( V \) is the single point \( \left( {0 : 0 : 1}\right) \), with ideal \( {\mathcal{I}}_{\text{rad }}\left( V\right) = \left( {x, y}\right) \), then what is the Hilbert function \( {h}_{I}\left( d\right) \) of the quotient ring? | \[ \mathbb{C}\left\lbrack {x, y, z}\right\rbrack /\left( {x, y}\right) \cong \mathbb{C}\left\lbrack z\right\rbrack \cong \mathbb{C} \oplus z\mathbb{C} \oplus {z}^{2}\mathbb{C} \oplus {z}^{3}\mathbb{C}\ldots \] which has dimension 1 in all degrees. Consequently, we have \[ {h}_{I}\left( d\right) \equiv 1\text{.} \] | Yes |
Proposition 73.2.7 (Hilbert functions of \( I \cap J \) and \( I + J \) ) Let \( I \) and \( J \) be homogeneous ideals in \( \mathbb{C}\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack \) . Then \[ {h}_{I \cap J} + {h}_{I + J} = {h}_{I} + {h}_{J} \] | Proof. Consider any \( d \geq 0 \) . Let \( S = \mathbb{C}\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack \) for brevity. Then \[ 0 \rightarrow {\left\lbrack S/\left( I \cap J\right) \right\rbrack }^{d} \hookrightarrow {\left\lbrack S/I\right\rbrack }^{d} \oplus {\left\lbrack S/J\right\rbrack }^{d} \rightarrow {\le... | Yes |
Example 73.2.8 (Hilbert function of two points in \( {\mathbb{{CP}}}^{1} \) ) | In \( {\mathbb{{CP}}}^{1} \) with coordinate ring \( \mathbb{C}\left\lbrack {s, t}\right\rbrack \), consider \( I = \left( s\right) \) the ideal corresponding to the point \( \left( {0 : 1}\right) \) and \( J = \left( t\right) \) the ideal corresponding to the point \( \left( {1 : 0}\right) \) . Then \( I \cap J = \lef... | Yes |
Theorem 73.2.9 (Hilbert functions of zero-dimensional varieties)\n\nLet \( V \) be a projective variety consisting of \( m \) points (where \( m \geq 0 \) is an integer). Then\n\n\[ \n{h}_{V}\left( d\right) = m\text{for}d \gg 0\text{.} \n\] | Proof. We already did \( m = 0 \), so assume \( m \geq 1 \) . Let \( I = {\mathcal{I}}_{\text{rad }}\left( V\right) \) and for \( k = 1,\ldots, m \) let \( {I}_{k} = {\mathcal{I}}_{\text{rad }}\left( {k\text{th point of}V}\right) \) .\n\nExercise 73.2.10. Show that \( {h}_{{I}_{k}}\left( d\right) = 1 \) for every \( d ... | No |
Corollary 73.2.11 \( \left( {h}_{I}\right. \) eventually constant when \( \left. {\dim {\mathcal{V}}_{\mathrm{{pr}}}\left( I\right) = 0}\right) \) | Let \( I \) be an ideal, not necessarily radical, such that \( {\mathcal{V}}_{\mathrm{{pr}}}\left( I\right) \) consists of finitely many points. Then the Hilbert \( {h}_{I} \) is eventually constant.\n\nProof. Induction on the number of points, \( m \geq 1 \) . The base case \( m = 1 \) was essentially done in Example ... | No |
The Hilbert function of \( {\mathbb{{CP}}}^{n} \) is | \[ {h}_{{\mathbb{{CP}}}^{n}}\left( d\right) = \left( \begin{matrix} d + n \\ n \end{matrix}\right) = \frac{1}{n!}\left( {d + n}\right) \left( {d + n - 1}\right) \ldots \left( {d + 1}\right) \] by a \ | Yes |
Consider the parabola \( {zy} - {x}^{2} \) in \( {\mathbb{{CP}}}^{2} \) with coordinates \( \mathbb{C}\left\lbrack {x, y, z}\right\rbrack \) . Then\n\n\[ \mathbb{C}\left\lbrack {x, y, z}\right\rbrack /\left( {{zy} - {x}^{2}}\right) \cong \mathbb{C}\left\lbrack {y, z}\right\rbrack \oplus x\mathbb{C}\left\lbrack {y, z}\r... | A combinatorial computation gives that\n\n\[ {h}_{\left( zy - {x}^{2}\right) }\left( 0\right) = 1 \]\nBasis 1\n\n\[ {h}_{\left( zy - {x}^{2}\right) }\left( 1\right) = 3 \]\nBasis \( x, y, z \)\n\n\[ {h}_{\left( zy - {x}^{2}\right) }\left( 2\right) = 5\;\text{Basis}{xy},{xz},{y}^{2},{yz},{z}^{2}\text{.} \]\n\nWe thus in... | No |
Theorem 73.3.3 (Hilbert polynomial)\n\nLet \( I \subseteq \mathbb{C}\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack \) be a homogeneous ideal, not necessarily radical. Then\n\n(a) There exists a polynomial \( {\chi }_{I} \) such that \( {h}_{I}\left( d\right) = {\chi }_{I}\left( d\right) \) for all \( d \gg 0 \) .\... | Proof. The base case was addressed in the previous section.\n\nFor the inductive step, consider \( {\mathcal{V}}_{\mathrm{{pr}}}\left( I\right) \) with dimension \( m \) . Consider a hyperplane \( H \) such that no irreducible component of \( {\mathcal{V}}_{\mathrm{{pr}}}\left( I\right) \) is contained inside \( H \) (... | Yes |
Theorem 73.4.3 (Bézout's theorem)\n\nLet \( I \) be a homogeneous ideal of \( \mathbb{C}\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack \), such that \( \dim {\mathcal{V}}_{\mathrm{{pr}}}\left( I\right) \geq 1 \) . Let \( f \in \mathbb{C}\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack \) be a homogeneous polyno... | Proof. Let \( S = \mathbb{C}\left\lbrack {{x}_{0},\ldots ,{x}_{n}}\right\rbrack \) again. This time the exact sequence is\n\n\[ 0 \rightarrow {\left\lbrack s/I\right\rbrack }^{d - k} \leftrightarrow {\left\lbrack S/I\right\rbrack }^{d} \rightarrow {\left\lbrack S/\left( I + \left( f\right) \right) \right\rbrack }^{d} \... | Yes |
Corollary 73.5.1 (Hypersurfaces: the degree deserves its name)\n\nLet \( V \) be a hypersurface, i.e. \( {\mathcal{I}}_{\text{rad }}\left( V\right) = \left( f\right) \) for \( f \) a homogeneous polynomial of degree \( k \) . Then \( \deg V = k \) . | Proof. Recall \( \deg \left( 0\right) = \deg {\mathbb{{CP}}}^{n} = 1 \) . Take \( I = \left( 0\right) \) in Bézout’s theorem. | No |
Theorem 73.5.3 (Pascal's theorem)\n\nLet \( A, B, C, D, E, F \) be six distinct points which lie on a conic \( \mathcal{C} \) in \( {\mathbb{{CP}}}^{2} \). Then the points \( {AB} \cap {DE},{BC} \cap {EF},{CD} \cap {FA} \) are collinear. | Proof. Let \( X \) be the variety equal to the union of the three lines \( {AB},{CD},{EF} \), hence \( X = {\mathcal{V}}_{\mathrm{{pr}}}\left( f\right) \) for some cubic polynomial \( f \) (which is the product of three linear ones). Similarly, let \( Y = {\mathcal{V}}_{\mathrm{{pr}}}\left( g\right) \) be the variety e... | Yes |
Example 74.1.4 (The pullback of \( \frac{1}{y - {25}} \) under \( t \mapsto {t}^{2} \) ) | The map\n\n\[ f : X = {\mathbb{A}}^{1} \rightarrow Y = {\mathbb{A}}^{1}\;\text{ by }\;t \mapsto {t}^{2} \]\n\nis a morphism of varieties. For example, consider the regular function \( \varphi = \frac{1}{y - {25}} \) on the open set \( Y \smallsetminus \{ {25}\} \subseteq Y \) . The \( f \) -inverse image is \( X \small... | Yes |
Theorem 74.2.2 (Regular maps of affine varieties are globally polynomials)\n\nLet \( X \subseteq {\mathbb{A}}^{m} \) and \( Y \subseteq {\mathbb{A}}^{n} \) be affine varieties. Every morphism \( f : X \rightarrow Y \) of varieties is given by\n\n\[ x = \left( {{x}_{1},\ldots ,{x}_{m}}\right) \overset{f}{ \mapsto }\left... | Proof. It's not too hard to see that all such functions work, so let's go the other way. Let \( f : X \rightarrow Y \) be a morphism.\n\nFirst, remark that \( {f}^{\text{pre }}\left( Y\right) = X \) . Now consider the regular function \( {\pi }_{1} \in {\mathcal{O}}_{Y}\left( Y\right) \), given by the projection \( \le... | Yes |
Example 74.2.3 (Projective map which is not globally polynomial)\n\nLet \( V = {\mathcal{V}}_{\mathrm{{pr}}}\left( {{xy} - {z}^{2}}\right) \subseteq {\mathbb{{CP}}}^{2} \). Then the map\n\n\[ V \rightarrow {\mathbb{{CP}}}^{1}\;\text{ by }\;\left( {x : y : z}\right) \mapsto \left\{ \begin{array}{ll} \left( {x : z}\right... | This is well defined just because \( \left( {x : z}\right) = \left( {z : y}\right) \) if \( x, y \neq 0 \) ; this should feel reminiscent of the definition of regular function. | No |
Example 74.2.5 (Example of an isomorphism)\n\nIn fact, the map \( V = {\mathcal{V}}_{\mathrm{{pr}}}\left( {{xy} - {z}^{2}}\right) \rightarrow {\mathbb{{CP}}}^{1} \) is an isomorphism. | The inverse map \( {\mathbb{{CP}}}^{1} \rightarrow V \) is given by\n\n\[ \left( {s : t}\right) \mapsto \left( {{s}^{2} : {st} : {t}^{2}}\right) \]\n\nThus actually \( V \cong {\mathbb{{CP}}}^{1} \). | Yes |
Example 74.3.3 (The parabola is quasi-projective) | Consider the parabola \( V = \mathcal{V}\left( {y - {x}^{2}}\right) \subset {\mathbb{A}}^{2} \) . We take the projective variety \( W = \) \( {\mathcal{V}}_{\mathrm{{pr}}}\left( {{zy} - {x}^{2}}\right) \) and look at the standard affine chart \( D\left( z\right) \) . Then there is an isomorphism\n\n\[ \nV \rightarrow D... | Yes |
Let \( X = {\mathbb{A}}^{1} \smallsetminus \{ 0\} \) be an quasi-projective variety. We claim that in fact we have an isomorphism\n\n\[ X \cong V = \mathcal{V}\left( {{xy} - 1}\right) \subseteq {\mathbb{A}}^{2} \] | The maps are\n\n\[ X \leftrightarrow V \]\n\n\[ t \mapsto \left( {t,1/t}\right) \]\n\n\[ x \leftarrow \left( {x, y}\right) \text{.} \] | No |
Theorem 74.4.2 (Distinguished open subsets of affines are affine)\n\nConsider \( X = D\left( f\right) \subseteq V = \mathcal{V}\left( {{f}_{1},\ldots ,{f}_{m}}\right) \subseteq {\mathbb{A}}^{n} \), where \( V \) is an affine variety, and the distinguished open set \( X \) is thought of as a quasi-projective variety. De... | For lack of a better name, I will dub this the hyperbola effect, and it will play a significant role later on. | No |
Theorem 75.3.7 (Stalks of \( {\mathcal{O}}_{V} \) )\n\nLet \( V \subseteq {\mathbb{A}}^{n} \) be a variety, and assume \( p \in V \) is a point. Then\n\n\[ \n{\mathcal{O}}_{V, p} \cong \left\{ {\frac{f}{g} \mid f, g \in \mathbb{C}\left\lbrack V\right\rbrack, g\left( p\right) \neq 0}\right\} .\n\] | Proof. A regular function \( \varphi \) on \( U \subseteq V \) is supposed to be a function on \( U \) that \ | No |
(a) Let \( V = {\mathbb{A}}^{1} \) ; then the stalk of \( {\mathcal{O}}_{V} \) at each point \( p \in V \) is\n\n\[{\mathcal{O}}_{V, p} = \left\{ {\left. {\frac{f\left( x\right) }{g\left( x\right) } \mid g\left( p\right) \neq 0}\right| \;}\right\} . | Examples of elements are \( {x}^{2} + 5,\frac{1}{x + 1} \) if \( p \neq 1,\frac{x + 7}{{x}^{2} - 9} \) if \( p \neq \pm 3 \), and so on. | No |
Lemma 75.6.5 (Stalks preserved by sheafification)\n\nLet \( \\mathcal{F} \) be a pre-sheaf and \( {\\mathcal{F}}^{\\text{sh }} \) its sheafification. Then for any point \( q \), there is an isomorphism\n\n\[ \n{\\left( {\\mathcal{F}}^{\\mathrm{{sh}}}\\right) }_{q} \\cong {\\mathcal{F}}_{q} \n\] | Proof. A germ in \( {\\left( {\\mathcal{F}}^{\\text{sh }}\\right) }_{q} \) looks like \( \\left( {{\\left( {g}_{p}\\right) }_{p \\in U}, U}\\right) \), where \( {g}_{p} = \\left( {{s}_{p},{U}_{p}}\\right) \) are themselves germs of \( {\\mathcal{F}}_{p} \), and \( q \\in U \) . Then the isomorphism is given by\n\n\[ \n... | Yes |
Example 76.2.3 (Localizations of \( \mathbb{C}\left\lbrack x\right\rbrack \) ) Let \( A = \mathbb{C}\left\lbrack x\right\rbrack \). (a) Suppose we let \( S = \left\{ {1, x,{x}^{2},{x}^{3},\ldots }\right\} \) be the powers of \( x \). Then | \[ {S}^{-1}A = \left\{ {\left. {\frac{f\left( x\right) }{{x}^{n}} \mid f \in \mathbb{C}\left\lbrack x\right\rbrack, n \in {\mathbb{Z}}_{ \geq 0}}\right| \;}\right\} . \] In other words, we get the Laurent polynomials in \( x \). You might recognize this as \[ {\mathcal{O}}_{V}\left( U\right) \text{where}V = {\mathbb{A}... | Yes |
Example 76.3.3 (Some arithmetic examples of localizations)\n\n(a) We localize \( \\mathbb{Z} \) away from 6 :\n\n\[ \n\\mathbb{Z}\\left\\lbrack {1/6}\\right\\rbrack = \\left\\{ {\\frac{m}{{6}^{n}} \\mid m \\in \\mathbb{Z}, n \\in {\\mathbb{Z}}_{ \\geq 0}}\\right\\} .\n\] | So \( A\\left\\lbrack {1/6}\\right\\rbrack \) consist of those rational numbers whose denominators have only powers of 2 and 3 . For example, it contains \( \\frac{5}{12} = \\frac{15}{36} \) . | No |
Example 76.3.5 (An example with zero-divisors)\n\nLet \( A = \mathbb{C}\left\lbrack {x, y}\right\rbrack /\left( {xy}\right) \) (which intuitively is the coordinate ring of two axes). Suppose we localize at \( x \) : equivalently, allowing denominators of \( x \) . Since \( {xy} = 0 \) in \( A \), we now have \( 0 = {x}... | \[ A\left\lbrack {1/x}\right\rbrack \cong \mathbb{C}\left\lbrack {x,1/x}\right\rbrack \] | Yes |
Proposition 76.5.1 (The prime ideals of \( {S}^{-1}A \) )\n\nLet \( A \) be a ring and \( S \subseteq A \) a multiplicative set. Then there is a natural inclusion-preserving bijection between:\n\n- The set of prime ideals of \( {S}^{-1}A \), and\n\n- The set of prime ideals of \( A \) not intersecting \( S \) . | Proof. Consider the homomorphism \( \iota : A \rightarrow {S}^{-1}A \) . For any prime ideal \( \mathfrak{q} \subseteq {S}^{-1}A \), its pre-image \( {\iota }^{\text{pre }}\left( \mathfrak{q}\right) \) is a prime ideal of \( A \) (by Problem \( 5{\mathrm{C}}^{ \star } \) ). Conversely, for any prime ideal \( \mathfrak{... | No |
Corollary 76.5.2 (Spectrums of localizations)\n\nLet \( A \) be a ring.\n\n(a) If \( \mathfrak{p} \) is a prime ideal of \( A \), then the prime ideals of \( A\left\lbrack {1/f}\right\rbrack \) are naturally in bijection with prime ideals of \( A \) do not contain the element \( f \) .\n\n(b) If \( \mathfrak{p} \) is a... | Proof. Part (b) is immediate; a prime ideal doesn’t meet \( A \smallsetminus \mathfrak{p} \) exactly if it is contained in \( \mathfrak{p} \) . For part (a), we want prime ideals of \( A \) not containing any power of \( f \) . But if the ideal is prime and contains \( {f}^{n} \), then it should contain either \( f \) ... | No |
Example 76.5.3 (Prime ideals of \( \mathbb{Z}\left\lbrack {1/6}\right\rbrack \) )\n\nSuppose we localize \( \mathbb{Z} \) away from the element 6, i.e. consider \( \mathbb{Z}\left\lbrack {1/6}\right\rbrack \) . As we saw,\n\n\[ \mathbb{Z}\left\lbrack {1/6}\right\rbrack = \left\{ {\left. \frac{n}{{2}^{x}{3}^{y}}\right| ... | But (2) and (3) no longer correspond to prime ideals; in fact in \( {A}_{6} \) we have \( \left( 2\right) = \left( 3\right) = \left( 1\right) \), the whole ring. | Yes |
Example 76.5.4 (Prime ideals of \( {A}_{\left( 5\right) } \) ) | Suppose we localize \( \mathbb{Z} \) at the prime (5). As we saw, \[ {\mathbb{Z}}_{\left( 5\right) } = \left\{ {\left. {\frac{m}{n} \mid m, n \in \mathbb{Z},5}\right| \;n}\right\} \] consist of those rational numbers whose denominators are not divisible by 5 . This is an integral domain, so (0) is still a prime ideal. ... | Yes |
Proposition 76.6.1 (The prime ideals of \( A/I \) )\n\nIf \( A \) is a ring and \( I \) is any ideal (not necessarily prime) then the prime (resp. maximal) ideals of \( A/I \) are in bijection with prime (resp. maximal) ideals of \( A \) which are supersets of \( I \) . This bijection is inclusion-preserving. | Proof. Consider the quotient homomorphism \( \psi : A \rightarrow A/I \) . For any prime ideal \( \mathfrak{q} \subseteq A/I \) , its pre-image \( {\psi }^{\text{pre }}\left( \mathfrak{q}\right) \) is a prime ideal (by Problem \( 5{\mathrm{C}}^{ \star } \) ). Conversely, for any prime ideal \( \mathfrak{p} \) with \( I... | No |
Theorem 76.7.1 (Localization commutes with quotients)\n\nLet \( S \) be a multiplicative set of a ring \( A \), and \( I \) an ideal of \( A \) . Let \( \bar{S} \) be the image of \( S \) under the projection map \( A \rightarrow A/I \) . Then\n\n\[ \n{\bar{S}}^{-1}\left( {A/I}\right) \cong {S}^{-1}A/{S}^{-1}I \n\]\n\n... | Proof. Omitted; Atiyah-Macdonald is the right reference for these type of things in the event that you do care. | No |
Let \( I \subseteq \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) be an ideal. By Proposition 76.6.1, the set \[ \operatorname{Spec}\mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack /I \] consists of those prime ideals of \( \mathbb{C}\left\lbrack {{x}_{1},\ldots ,{x}_{n}}\right\rbrack \) ... | So in addition to the \ | No |
Proposition 77.3.12 (Spectrums of integral domains are irreducible)\n\nIf \( A \) is an integral domain, then \( \operatorname{Spec}A \) is irreducible. | Proof. Just note (0) is a prime ideal, and in every open set. | No |
Theorem 77.3.14 (Points are in bijection with irreducible closed sets)\n\nConsider \( X = \operatorname{Spec}A \) . For every irreducible closed set \( Z \), there is exactly one point \( \mathfrak{p} \) such that \( Z = \overline{\{ \mathfrak{p}\} } \) . (In particular points of \( X \) are in bijection with closed su... | Idea of proof. The point \( \mathfrak{p} \) corresponds to the closed set \( \mathcal{V}\left( \mathfrak{p}\right) \), which one can show is irreducible. | No |
Proposition 77.4.1 \( \left( {\mathcal{V}\left( \sqrt{I}\right) = \mathcal{V}\left( I\right) }\right) \) | Proof. We have \( \sqrt{I} \supseteq I \) . Hence automatically \( \mathcal{V}\left( \sqrt{I}\right) \subseteq \mathcal{V}\left( I\right) \). Conversely, if \( \mathfrak{p} \in \mathcal{V}\left( I\right) \), then \( I \subseteq \mathfrak{p} \), so \( \sqrt{I} \subseteq \sqrt{\mathfrak{p}} = \mathfrak{p} \) (by Proposit... | Yes |
Theorem 77.4.2 (Radical is intersection of primes)\n\nLet \( I \) be an ideal of a ring \( A \) . Then\n\n\[ \sqrt{I} = \mathop{\bigcap }\limits_{{\mathfrak{p} \supseteq I}}\mathfrak{p} \] | Proof. This is a famous statement from commutative algebra, and we prove it here only for completeness. It is \ | No |
Example 77.4.4 \( \left( {\sqrt{\left( {2016}\right) } = \left( {42}\right) \text{in}\mathbb{Z}}\right) \) | In the ring \( \mathbb{Z} \), we see that \( \sqrt{\left( {2016}\right) } = \left( {42}\right) \), since the distinct primes containing (2016) are (2), (3), (7). | No |
Theorem 77.4.5 (Radical ideals correspond to closed sets)\n\nLet \( I \) and \( J \) be ideals of \( A \), and considering the space \( \operatorname{Spec}A \) . Then\n\n\[ \mathcal{V}\left( I\right) = \mathcal{V}\left( J\right) \Leftrightarrow \sqrt{I} = \sqrt{J} \]\n\nIn particular, radical ideals exactly correspond ... | Proof. If \( \mathcal{V}\left( I\right) = \mathcal{V}\left( J\right) \), then \( \sqrt{I} = \mathop{\bigcap }\limits_{{\mathfrak{p} \in \mathcal{V}\left( I\right) }}\mathfrak{p} = \mathop{\bigcap }\limits_{{\mathfrak{p} \in \mathcal{V}\left( J\right) }}\mathfrak{p} = \sqrt{J} \) as needed.\n\nConversely, suppose \( \sq... | Yes |
Theorem 78.2.2 (Distinguished open sets form a base)\n\nThe distinguished open sets \( D\left( f\right) \) form a basis for the Zariski topology: any open set \( U \) is a union of distinguished open sets. | Proof. Let \( U \) be an open set; suppose it is the complement of closed set \( V\left( I\right) \) . Then verify that\n\n\[ U = \mathop{\bigcup }\limits_{{f \in I}}D\left( f\right) \] | Yes |
Theorem 78.2.3 (Sections of \( D\left( f\right) \) are localizations away from \( f \) ) Let \( A \) be a ring and \( f \in A \) . Then\n\n\[ \n{\mathcal{O}}_{\operatorname{Spec}A}\left( {D\left( f\right) }\right) \cong A\left\lbrack {1/f}\right\rbrack \n\] | Proof. Omitted, but similar to Theorem 71.6.1. | No |
Example 78.2.4 (The punctured line is isomorphic to a hyperbola) | \[ {\mathcal{O}}_{\operatorname{Spec}\mathbb{C}\left\lbrack x\right\rbrack }\left( {D\left( x\right) }\right) = \mathbb{C}\left\lbrack {x,{x}^{-1}}\right\rbrack \cong \mathbb{C}\left\lbrack {x, y}\right\rbrack /\left( {{xy} - 1}\right) . \] | Yes |
Corollary 78.2.5 ( \( A \) is the ring of global sections)\n\nThe ring of global sections of \( \operatorname{Spec}A \) is \( A \) . | Proof. By previous theorem, \( {\mathcal{O}}_{\operatorname{Spec}A}\left( {\operatorname{Spec}A}\right) = {\mathcal{O}}_{\operatorname{Spec}A}\left( {D\left( 1\right) }\right) = A\left\lbrack {1/1}\right\rbrack = A \) . | Yes |
Theorem 78.3.1 (Stalks of \( \operatorname{Spec}A \) are \( {A}_{\mathfrak{p}} \) )\n\nLet \( A \) be a ring and let \( \mathfrak{p} \in \operatorname{Spec}A \) . Then\n\n\[ \n{\mathcal{O}}_{\text{Spec }A,\mathfrak{p}} \cong {A}_{\mathfrak{p}} \n\]\n\nIn particular \( X \) is a locally ringed space. | Proof. Since sheafification preserved stalks, it’s enough to check it for \( \mathcal{F} \) the pre-sheaf of globally rational functions in our definition. The proof is basically the same as Theorem 75.3.7: there is an obvious map \( {\mathcal{F}}_{\mathfrak{p}} \rightarrow {A}_{\mathfrak{p}} \) on germs by\n\n\[ \n\le... | Yes |
What is the unique maximal ideal \( \mathfrak{p}{A}_{\mathfrak{p}} \) of the local ring \( {A}_{\mathfrak{p}} \) in \( \operatorname{Spec}\mathbb{C}\left\lbrack {x, y}\right\rbrack \) at the origin? | Proof of Theorem 78.4.1. One may check set \( I = \mathfrak{p}{A}_{\mathfrak{p}} \) is an ideal of \( {A}_{\mathfrak{p}} \) . Moreover, \( 1 \notin I \) , so \( I \) is proper.\n\nTo prove it is maximal and unique, it suffices to prove that any \( f \in {A}_{\mathfrak{p}} \) with \( f \notin I \) is a unit of \( {A}_{\... | Yes |
Theorem 78.4.7 (The germ-to-value square)\n\nLet \( A \) be a ring and \( \mathfrak{p} \) a prime ideal. The following diagram commutes: \n\nIn particular, \( \kappa \left( \mathfrak{p}\right) \) can also be describe... | So for example, if \( A = \mathbb{C}\left\lbrack {x, y}\right\rbrack \) and \( \mathfrak{p} = \left( {x, y}\right) \), then \( A/\mathfrak{p} = \mathbb{C} \) and \( \operatorname{Frac}\left( {A}_{\mathfrak{p}}\right) = \operatorname{Frac}\left( \mathbb{C}\right) = \mathbb{C} \) , as we expected. In practice, \( \operat... | Yes |
Example 78.6.1 (On the double point, all multiples of \( x \) are zero at all points) | The space \( \operatorname{Spec}\mathbb{C}\left\lbrack x\right\rbrack /\left( {x}^{2}\right) \) has only one point, \( \left( x\right) \) . The functions 0 and \( x \) (and for that matter \( {2x},{3x},\ldots \) ) all vanish on it. This shows that functions are not determined uniquely by values in general. | Yes |
Proposition 80.2.1 (Induced stalk morphisms)\n\nIf \( \pi : X \rightarrow Y \) is a map of ringed spaces sending \( \pi \left( p\right) = q \), then we get a map\n\n\[ \n{\pi }_{p}^{\sharp } : {\mathcal{O}}_{Y, q} \rightarrow {\mathcal{O}}_{X, p}\n\]\n\nwhenever \( \pi \left( p\right) = q \) . | Proof. If \( \left( {s, U}\right) \) is a germ at \( q \), then \( \left( {{\pi }^{\sharp }\left( s\right) ,{\pi }^{\text{pre }}\left( U\right) }\right) \) is a germ at \( p \), and this is a well-defined morphism because of compatibility with restrictions. | Yes |
Proposition 80.2.2 (Uniqueness of morphisms via stalks)\n\nConsider a map of ringed spaces \( \\left( {\\pi ,{\\pi }^{\\sharp }}\\right) : \\left( {X,{\\mathcal{O}}_{X}}\\right) \\rightarrow \\left( {Y,{\\mathcal{O}}_{Y}}\\right) \) and the corresponding map \( {\\pi }_{p}^{\\sharp } \) of stalks. Then \( {\\pi }^{\\sh... | Proof. Given a section \( s \\in {\\mathcal{O}}_{Y}\\left( U\\right) \), let\n\n\[ t = {\\pi }_{U}^{\\sharp }\\left( s\\right) \\in {\\mathcal{O}}_{X}\\left( {{\\pi }^{\\text{pre }}\\left( U\\right) }\\right) \]\n\n\ndenote the image under \( {\\pi }^{\\sharp } \) .\n\nWe know \( {t}_{p} \) for each \( p \\in {\\pi }^{... | No |
Example 80.4.4 (Constant map to \( \left( {y - 3}\right) \) )\n\nWe analyze scheme morphisms\n\n\[ \nX = \operatorname{Spec}\mathbb{R}\left\lbrack x\right\rbrack \xrightarrow[]{\pi }\operatorname{Spec}\mathbb{R}\left\lbrack y\right\rbrack = Y \n\] \n\nwhich send all points of \( X \) to \( \mathfrak{m} = \left( {y - 3}... | This example is simple enough that we can even do it by sections, as much as I think stalks are simpler. Let \( U \) be any open subset of \( Y \), then we need to specify a map\n\n\[ \n{\pi }_{U}^{\sharp } : {\mathcal{O}}_{Y}\left( U\right) \rightarrow {\mathcal{O}}_{X}\left( {{\pi }^{\text{pre }}\left( U\right) }\rig... | Yes |
Example 80.4.5 (Constant map to \( \left( {{y}^{2} + 1}\right) \) does not exist)\n\nLet’s see if there are constant maps \( X = \operatorname{Spec}\mathbb{R}\left\lbrack x\right\rbrack \rightarrow \operatorname{Spec}\mathbb{R}\left\lbrack y\right\rbrack = Y \) which send everything to \( \left( {{y}^{2} + 1}\right) \)... | Copying the previous example, we see that we want\n\n\[{\mathcal{O}}_{Y}\left( U\right) \rightarrow {\mathcal{O}}_{X}\left( X\right) = \mathbb{R}\left\lbrack x\right\rbrack\]\n\nWe find that \( y \) and \( 1/y \) has nowhere to go: the same argument as last time shows that \( y - c \) should be a unit of \( \mathbb{R}\... | Yes |
Example 80.4.6 (The generic point repels smaller points)\n\nChanging the tune, consider maps \( \operatorname{Spec}\mathbb{C}\left\lbrack x\right\rbrack \rightarrow \operatorname{Spec}\mathbb{C}\left\lbrack y\right\rbrack \) . We claim that if \( \mathfrak{m} \) is a maximal ideal (closed point) of \( \mathbb{C}\left\l... | For otherwise, we would get a local ring homomorphism\n\n\[ \mathbb{C}\left( y\right) \cong {\mathcal{O}}_{\operatorname{Spec}k\left\lbrack y\right\rbrack ,\left( 0\right) } \rightarrow {\mathcal{O}}_{\operatorname{Spec}\mathbb{C}\left\lbrack x\right\rbrack ,\mathfrak{m}} \cong \mathbb{C}{\left\lbrack x\right\rbrack }_... | Yes |
Example 80.4.7 (The map \( t \mapsto {t}^{2} \) ) | We consider a map\n\n\[ \pi : X = \operatorname{Spec}\mathbb{C}\left\lbrack x\right\rbrack \rightarrow \operatorname{Spec}\mathbb{C}\left\lbrack y\right\rbrack = Y \]\n\ndefined on points as follows:\n\n\[ \pi \left( \left( 0\right) \right) = \left( 0\right) \]\n\n\[ \pi \left( \left( {x - a}\right) \right) = \left( {y... | Yes |
Proposition 80.5.1 (Affine reconstruction)\n\nLet \( \pi : \operatorname{Spec}A \rightarrow \operatorname{Spec}B \) be a map of schemes. Let \( \psi : B \rightarrow A \) be the ring homomorphism obtained by taking global sections, i.e.\n\n\[ \psi = {\pi }_{B}^{\sharp } : {\mathcal{O}}_{\operatorname{Spec}B}\left( {\ope... | Proof. This requires two parts.\n\n- We need to check that the maps agree on points; surprisingly this is the harder half. To see how this works, let \( \mathfrak{q} = \pi \left( \mathfrak{p}\right) \) . The key fact is that a function \( f \in B \) vanishes on \( \mathfrak{q} \) if and only if \( {\pi }_{B}^{\sharp }\... | Yes |
Theorem 80.5.2 (Spec \( A \rightarrow \operatorname{Spec}B \) is just \( B \rightarrow A \) ) | These two construction gives a bijection between ring homomorphisms \( B \rightarrow A \) and \( \operatorname{Spec}A \rightarrow \operatorname{Spec}B \) .\n\nProof. We have seen how to take each \( \pi : \operatorname{Spec}A \rightarrow \operatorname{Spec}B \) and get a ring homomorphism \( \psi \) . Proposition 80.5.... | Yes |
Theorem 80.7.1 (Distinguished open sets are isomorphic to affine schemes) Let \( A \) be a ring and \( f \) an element. Then\n\n\[ \operatorname{Spec}A\left\lbrack {1/f}\right\rbrack \cong D\left( f\right) \subseteq \operatorname{Spec}A. \] | Proof. Annoying check, not included yet. (We have already seen the bijection of prime ideals, at the level of points.) | No |
Corollary 80.7.2 (Open subsets are schemes)\n\n(a) Any nonempty open subset of an affine scheme is itself a scheme.\n\n(b) Any nonempty open subset of any scheme (affine or not) is itself a scheme. | Proof. Part (a) has essentially been done already:\n\nQuestion 80.7.3. Combine Theorem 78.2.2 with the previous proposition to deduce (a).\n\nPart (b) then follows by noting that if \( U \) is an open set, and \( p \) is a point in \( U \), then we can take an affine open neighborhood \( \operatorname{Spec}A \) at \( p... | No |
Proposition 80.7.5 (Famous example: punctured plane isn't affine)\n\nThe punctured plane \( U = \left( {U,{\mathcal{O}}_{U}}\right) \), obtained by deleting \( \left( {x, y}\right) \) from \( \operatorname{Spec}k\left\lbrack {x, y}\right\rbrack \), is not isomorphic to any affine scheme \( \operatorname{Spec}B \) . | Proof. We already know \( {\mathcal{O}}_{U}\left( U\right) = k\left\lbrack {x, y}\right\rbrack \) and we have a good handle on it. For example, \( y \in {\mathcal{O}}_{U}\left( U\right) \) is a global section which vanishes on what looks like the \( y \) -axis. Similarly, \( x \in {\mathcal{O}}_{X}\left( X\right) \) is... | Yes |
Theorem 81.5.1 (Zorn's lemma)\n\nLet \( \mathcal{P} \) be a nonempty partially ordered set. If every chain has an upper bound, then \( \mathcal{P} \) has a local maximum. | Proof. Look at the poset whose elements are sets of independent real numbers. Every chain \( {S}_{0} \varsubsetneq {S}_{1} \varsubsetneq \ldots \) has an upper bound \( \bigcup {S}_{\alpha } \) (which you have to check is actually an element of the poset). Thus by Zorn, there is a local maximum \( S \) . Then \( S \) m... | No |
Lemma 82.2.1 (Cantor's diagonal argument)\n\nFor any set \( X \), it’s impossible to construct an injective map \( \iota : \mathcal{P}\left( X\right) \hookrightarrow X \) . | Proof. Assume for contradiction \( \iota \) exists.\n\nExercise 82.2.2. Show that if, \( \iota \) exists, then there exists a surjective map \( j : X \rightarrow \mathcal{P}\left( X\right) \) . (This is easier than it appears, just \ | No |
Theorem 82.7.2 (There is no set of all sets) \( V \) is a proper class. | Proof. Assume not, and \( V \) is a set. Then \( V \in V \), which violates Foundation. (In fact, \( V \) cannot be a set even without Foundation, as we saw earlier). | Yes |
Example 83.3.2 (7 is transitive) | The set 7 is transitive: for example, \( 2 \in 5 \in 7 \Rightarrow 2 \in 7 \) . | No |
Example 83.3.4 \( \\left( { \\in \\text{is a well-ordering on}\\omega \\cdot 3}\\right) \) | In \( \\omega \\cdot 3 \), we have an ordering\n\n\[ \n0 \\in 1 \\in 2 \\in \\cdots \\in \\omega \\in \\omega + 1 \\in \\cdots \\in \\omega \\cdot 2 \\in \\omega \\cdot 2 + 1 \\in \\ldots .\n\]\n\nwhich has no infinite descending chains. Indeed, a typical descending chain might look like\n\n\[ \n\\omega \\cdot 2 + 6 \\... | Yes |
Theorem 83.3.9 (Ordinals are strictly ordered)\n\nGiven any two ordinal numbers \( \alpha \) and \( \beta \), either \( \alpha < \beta ,\alpha = \beta \) or \( \alpha > \beta \) . | Proof. Surprisingly annoying, thus omitted. | No |
Theorem 83.4.1 (There is no set of all ordinals) On is a proper class. | Proof. Assume for contradiction not. Then On is well-ordered by \( \in \) and transitive, so On is an ordinal, i.e. On \( \in \) On, which violates Foundation. | Yes |
Theorem 83.4.3 (Sets of ordinals are bounded)\n\nLet \( A \subseteq \) On. Then there is some ordinal \( \alpha \) such that \( A \subseteq \alpha \) (i.e. \( A \) must be bounded). | Proof. Otherwise, look at \( \bigcup A \) . It is a set. But if \( A \) is unbounded it must equal On, which is a contradiction. | Yes |
Theorem 83.5.2 (Transfinite recursion)\n\nTo define a sequence \( {x}_{\alpha } \) for every ordinal \( \alpha \), it suffices to\n\n- define \( {x}_{0} \), then\n\n- for any \( \beta \), define \( {x}_{\beta } \) for any \( \alpha < \beta \). | So a transfinite induction or recursion is very often broken up into three cases. In the induction phrasing, it looks like\n\n- (Zero Case) First, resolve \( P\left( 0\right) \).\n\n- (Successor Case) Show that from \( P\left( \alpha \right) \) we can get \( P\left( {\alpha + 1}\right) \).\n\n- (Limit Case) For \( \lam... | Yes |
Example 83.6.1 \( \left( {2 + 3 = 5}\right) \) | Under the explicit construction for \( \alpha = 2 \) and \( \beta = 3 \), we get the set\n\n\[ \nX = \{ \left( {0,0}\right) < \left( {0,1}\right) < \left( {1,0}\right) < \left( {1,1}\right) < \left( {1,2}\right) \} \n\]\n\nwhich is isomorphic to 5. | No |
Example 83.6.5 (Ordinal multiplication is not commutative)\n\nWe have \( \omega \cdot 2 = \omega + \omega \), but \( 2 \cdot \omega = \omega \) . | Exercise 83.6.6. Prove this. | No |
Theorem 83.7.7 (The von Neumann hierachy is complete)\n\nThe class \( V \) is equal to \( \mathop{\bigcup }\limits_{{\alpha \in \mathrm{{On}}}}{V}_{\alpha } \) . In other words, every set appears in some \( {V}_{\alpha } \) . | Proof. Assume for contradiction this is false. The key is that because \( \in \) satisfies Foundation, we can take a \( \in \) -minimal counterexample \( x \) . Thus \( \operatorname{rank}\left( y\right) \) is defined for every \( y \in x \), and we can consider (by Replacement) the set\n\n\[ \n\{ \operatorname{rank}\l... | Yes |
We have \( \left| X\right| < \left| {\mathcal{P}\left( X\right) }\right| \) for every set \( X \) . | Proof. There is an injective map \( X \hookrightarrow \mathcal{P}\left( X\right) \) but there is no injective map \( \mathcal{P}\left( X\right) \hookrightarrow X \) by Lemma 82.2.1. | Yes |
Lemma 84.3.5 (Aleph numbers constitute all infinite cardinals)\n\nIf \( \kappa \) is a cardinal then either \( \kappa \) is finite (i.e. \( \kappa \in \omega \) ) or \( \kappa = {\aleph }_{\alpha } \) for some \( \alpha \in \) On. | Proof. Assume \( \kappa \) is infinite, and take \( \alpha \) minimal with \( {\aleph }_{\alpha } \geq \kappa \) . Suppose for contradiction that we have \( {\aleph }_{\alpha } > \kappa \) . We may assume \( \alpha > 0 \), since the case \( \alpha = 0 \) is trivial.\n\nIf \( \alpha = \bar{\alpha } + 1 \) is a successor... | Yes |
Theorem 84.4.5 (Infinite cardinals squared)\n\nLet \( \kappa \) be an infinite cardinal. Then \( \kappa \cdot \kappa = \kappa \) . | Proof. Obviously \( \kappa \cdot \kappa \geq \kappa \), so we want to show \( \kappa \cdot \kappa \leq \kappa \) .\n\nThe idea is to try to repeat the same proof that we had for \( {\aleph }_{0} \cdot {\aleph }_{0} = {\aleph }_{0} \), so we re-iterate it here. We took the \ | No |
Theorem 84.4.6 (Infinite cardinal arithmetic is trivial)\n\nGiven cardinals \( \kappa \) and \( \mu \), one of which is infinite, we have\n\n\[ \kappa \cdot \mu = \kappa + \mu = \max \{ \kappa ,\mu \} . \] | Proof. The point is that both of these are less than the square of the maximum. Writing out the details:\n\n\[ \max \{ \kappa ,\mu \} \leq \kappa + \mu \]\n\n\[ \leq \kappa \cdot \mu \]\n\n\[ \leq \max \{ \kappa ,\mu \} \cdot \max \{ \kappa ,\mu \} \]\n\n\[ = \max \{ \kappa ,\mu \} \text{.} \] | Yes |
Example 84.6.5 ( \( {\aleph }_{0} \) is regular) | \( \operatorname{cof}\left( {\aleph }_{0}\right) = {\aleph }_{0} \), because no finite subset of \( {\aleph }_{0} = \omega \) can reach arbitrarily high. | Yes |
Example 84.6.6 \( \left( {\aleph }_{1}\right. \) is regular) | \( \operatorname{cof}\left( {\aleph }_{1}\right) = {\aleph }_{1} \) . Indeed, assume for contradiction that some countable set of ordinals \( A = \left\{ {{\alpha }_{0},{\alpha }_{1},\ldots }\right\} \subseteq {\aleph }_{1} \) reaches arbitrarily high inside \( {\aleph }_{1} \) . Then \( \Lambda = \cup A \) is a counta... | Yes |
Theorem 84.6.8 (Successor cardinals are regular)\n\nIf \( \kappa = {\bar{\kappa }}^{ + } \) is a successor cardinal, then it is regular. | Proof. We copy the proof that \( {\aleph }_{1} \) was regular.\n\nAssume for contradiction that for some \( \mu \leq \bar{\kappa } \), there are \( \mu \) sets reaching arbitrarily high in \( \kappa \) as a cardinal. Observe that each of these sets must have cardinality at most \( \bar{\kappa } \) . We take the union o... | Yes |
Let’s take \( \mathcal{M} = \left( {M, E}\right) = \left( {\omega , \in }\right) \). This is not a very good model of ZFC, but let’s see if we can make sense of some of the first few axioms.\n\n(a) \( \mathcal{M} \) satisfies Extensionality, which is the sentence\n\n\[ \forall x\forall y\forall a : \left( {a \in x \Lef... | This just follows from the fact that \( E \) is actually \( \in \) . | No |
Lemma 85.4.3 (Tarski-Vaught)\n\nLet \( {\mathcal{M}}_{1} \subseteq {\mathcal{M}}_{2} \) . Then \( {\mathcal{M}}_{1} \prec {\mathcal{M}}_{2} \) if and only if: For every sentence \( \phi \left( {x,{x}_{1},\ldots ,{x}_{n}}\right) \) and parameters \( {b}_{1},\ldots ,{b}_{n} \in {M}_{1} \) : if there is a witness \( \wide... | Proof. Easy after the above discussion. To formalize it, use induction on formula complexity. | No |
Lemma 85.6.1 (Mostowski collapse lemma)\n\nLet \( X = \left( {X, \in }\right) \) be a model, where \( X \) is a set (possibly not transitive). Then there exists an isomorphism \( \pi : X \rightarrow M \) for a transitive model \( M = \left( {M, \in }\right) \) . | Proof. The idea behind the proof is very simple. Since \( \in \) is well-founded and extensional (satisfies Foundation and Extensionality, respectively), we can look at the \( \in \) -minimal element \( {x}_{\varnothing } \) of \( X \) with respect to \( \in \) . Clearly, we want to send that to \( 0 = \varnothing \) .... | No |
Theorem 85.7.1 (Countable transitive model)\n\nAssume \( {\mathrm{{ZFC}}}^{ + } \) . Then there exists a transitive model \( X \) of \( \mathrm{{ZFC}} \) such that \( X \) is a countable set. | Proof. Fasten your seat belts.\n\nFirst, since we assumed \( {\mathrm{{ZFC}}}^{ + } \), we can take \( {V}_{\kappa } = \left( {{V}_{\kappa }, \in }\right) \) as our model of ZFC. Start with the set \( {X}_{0} = \varnothing \) . Then for every integer \( n \), we do the following to get \( {X}_{n + 1} \).\n\n- Start wit... | Yes |
Theorem 85.7.3 (Downward Löwenheim-Skolem theorem)\n\nLet \( \\mathcal{M} = \\left( {M, E}\\right) \) be a model, and \( A \\subseteq M \) . Then there exists a set \( B \) (called the Skolem hull of \( A \) ) with \( A \\subseteq B \\subseteq M \), such that \( \\left( {B, E}\\right) \\prec \\mathcal{M} \), and\n\n\[ ... | Question 85.7.4. Prove this. (Exactly the same proof as before.) | No |
Lemma 86.2.1 (Rasiowa-Sikorski lemma)\n\nSuppose \( M \) is a countable transitive model of ZFC and \( \mathbb{P} \) is a partial order. Then there exists an \( M \) -generic filter \( G \) . | Proof. Essentially, hit them one by one. Problem 86B. | No |
Example 86.2.3 (Infinite binary tree is (very) splitting) | The infinite binary tree is about as splitting as you can get. Given \( p \in {2}^{ < \omega } \), just consider the two elements right under it. | No |
Lemma 86.2.4 (Splitting posets omit generic sets)\n\nSuppose \( \mathbb{P} \) is splitting. Then if \( F \subseteq \mathbb{P} \) is a filter such that \( F \in M \), then \( \mathbb{P} \smallsetminus F \) is dense. In particular, if \( G \subseteq \mathbb{P} \) is generic, then \( G \notin M \) . | Proof. Consider \( p \notin \mathbb{P} \smallsetminus F \Leftrightarrow p \in F \) . Then there exists \( q, r \leq p \) which are not compatible. Since \( F \) is a filter it cannot contain both; we must have one of them outside \( F \), say \( q \) . Hence every element of \( p \in \mathbb{P} \smallsetminus \left( {\... | Yes |
Let us compute\n\n\[ \n{\\text{Name}}_{0} = \\varnothing \n\] \n\n\[ \n{\\text{Name}}_{1} = \\mathcal{P}\\left( {\\varnothing \\times \\mathbb{P}}\\right) \n\] \n\n\[ \n= \\{ \\varnothing \\} \n\] \n\n\[ \n{\\text{Name}}_{2} = \\mathcal{P}\\left( {\\{ \\varnothing \\} \\times \\mathbb{P}}\\right) \n\] \n\n\[ \n= \\math... | Compare the corresponding von Neuman universe.\n\n\[ \n{V}_{0} = \\varnothing ,{V}_{1} = \\{ \\varnothing \\} ,{V}_{2} = \\{ \\varnothing ,\\{ \\varnothing \\} \\} .\n\] | No |
Example 86.3.4 (Example of an interpretation)\n\nAs we said earlier, \( {\operatorname{Name}}_{1} = \{ \varnothing \} \) . Now suppose\n\n\[ \tau = \left\{ {\left\langle {\varnothing ,{p}_{1}}\right\rangle ,\left\langle {\varnothing ,{p}_{2}}\right\rangle ,\ldots ,\left\langle {\varnothing ,{p}_{n}}\right\rangle }\righ... | \[ {\tau }^{G} = \{ \varnothing \mid \langle \varnothing, p\rangle \in \tau \text{ and }p \in G\} = \left\{ \begin{array}{ll} \{ \varnothing \} & \text{ if some }{p}_{i} \in G \\ \varnothing & \text{ otherwise. } \end{array}\right. \]\n\nIn particular, remembering that \( G \) is nonempty we see that\n\n\[ \left\{ {{\t... | Yes |
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