diff --git a/database/data/categories/Fld.yaml b/database/data/categories/Fld.yaml index 2c0bb2835..4dd09eb54 100644 --- a/database/data/categories/Fld.yaml +++ b/database/data/categories/Fld.yaml @@ -11,13 +11,14 @@ tags: related: - CRing + - Fld_0 comments: - Limits and colimits are discussed in MSE/359352. satisfied_properties: - property: locally small - proof: There is a forgetful functor $\Fld \to \Set$ and $\Set$ is locally small. + proof: There is a forgetful functor $\Fld \to \Set$, and $\Set$ is locally small. - property: inhabited proof: This is trivial. @@ -26,10 +27,10 @@ satisfied_properties: proof: It is well-known that every field homomorphism is injective and hence a monomorphism. - property: well-copowered - proof: Epimorphisms are the purely inseparable field extensions. If $K \to L$ is purely inseparable, then for all $x \in L$ there is some $n \in \IN$ with $x^n \in L$. An element of $K$ has at most $n$ $n$th-roots. So we can bound the size of $L$. + proof: Epimorphisms are the purely inseparable field extensions (see below). If $K \to L$ is purely inseparable, then for every $x \in L$ there is some $n \in \IN$ such that $x^n \in K$. An element of $K$ has at most $n$ $n$th roots. Hence, we can bound the cardinality of $L$. - property: multi-algebraic - proof: See Eg. 4.3(1) in [AR01]. + proof: Example 4.3(1) in [AR01] presents an FPC-sketch that models fields. unsatisfied_properties: - property: skeletal @@ -39,34 +40,41 @@ unsatisfied_properties: proof: There are infinitely many homomorphisms $\IQ(X) \to \IQ(X)$, $X \mapsto X^k$. - property: connected - proof: A field of characteristic $0$ cannot be connected with a field of characteristic $p > 0$. in fact, the connected components of $\Fld$ are the subcategories $\Fld_p$ of fields of characteristic $p$, where $p$ is a prime or $0$. + proof: A field of characteristic $0$ cannot be connected to a field of characteristic $p > 0$. In fact, the connected components of $\Fld$ are the subcategories $\Fld_p$ of fields of characteristic $p$, where $p$ is a prime or $0$. - property: balanced proof: Every non-trivial purely inseparable field extension, such as $\IF_p(X^p) \to \IF_p(X)$, provides a counterexample by the descriptions of special morphisms below. - property: core-thin - proof: If this category was core-thin, Galois theory would not exist. Specifically, the conjugation $\IC \to \IC$, $z \mapsto \overline{z}$ is a non-trivial automorphism. + proof: If this category were core-thin, Galois theory would not exist. Specifically, the conjugation $\IC \to \IC$, $z \mapsto \overline{z}$ is a non-trivial automorphism. - property: multi-terminal object - proof: Every field has a non-trivial extension, for instance, the rational function field over itself in one variable. Hence, a multi-terminal object never exists. + proof: Every field has a non-trivial extension, for instance, the rational function field in one variable over itself. Hence, a multi-terminal object never exists. - property: generator - proof: Assume that $G$ is a generator, say of characteristic $p$. Then for all $q \neq p$ all homomorphisms between two fields of characteristic $q$ would be equal, which is absurd. + proof: Assume that $G$ is a generator, say of characteristic $p$. Then for all $q \neq p$, all homomorphisms between two fields of characteristic $q$ would be equal, which is absurd. - property: cogenerating set - proof: 'We apply this lemma to the collection of fields: Any homomorphism of fields is injective. For every infinite cardinal $\kappa$ the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swap two variables).' + proof: 'We apply this lemma to the collection of fields: Any homomorphism of fields is injective. For every infinite cardinal $\kappa$, the field of rational functions in $\kappa$ variables has cardinality $\geq \kappa$ and a non-trivial automorphism (swapping two variables).' + label: Fld_no_cogenerating_set - property: binary powers - proof: 'Assume that the product $P \coloneqq \IQ(\sqrt{2}) \times \IQ(\sqrt{2})$ exists. This field is isomorphic to a subfield of $\IQ(\sqrt{2})$, hence $P \cong \IQ$ or $P \cong \IQ(\sqrt{2})$. In the first case, the two projections $P \rightrightarrows \IQ(\sqrt{2})$ must be equal, which means that every two homomorphisms $K \rightrightarrows \IQ(\sqrt{2})$ are equal, which is absurd (take $K = \IQ(\sqrt{2})$ and its two automorphisms). In the second case, the projections induce for every field $K$ a bijection $\Hom(K,\IQ(\sqrt{2})) \cong \Hom(K,\IQ(\sqrt{2}))^2$, which however fails for $K = \IQ(\sqrt{2})$: the left hand side has $2$ elements, the right hand side has $4$ elements. A more general result about products in $\Fld$ can be found at MSE/359352.' + proof: 'Assume that the product $P \coloneqq \IQ(\sqrt{2}) \times \IQ(\sqrt{2})$ exists. This field is isomorphic to a subfield of $\IQ(\sqrt{2})$, hence $P \cong \IQ$ or $P \cong \IQ(\sqrt{2})$. In the first case, the two projections $P \rightrightarrows \IQ(\sqrt{2})$ must be equal, which means that every two homomorphisms $K \rightrightarrows \IQ(\sqrt{2})$ are equal, which is absurd (take $K = \IQ(\sqrt{2})$ and its two automorphisms). In the second case, the projections induce for every field $K$ a bijection $\Hom(K,\IQ(\sqrt{2})) \cong \Hom(K,\IQ(\sqrt{2}))^2$, which, however, fails for $K = \IQ(\sqrt{2})$: the left-hand side has $2$ elements, while the right-hand side has $4$ elements. A more general result about products in $\Fld$ can be found at MSE/359352.' + label: Fld_no_binary_powers - property: locally cartesian closed - proof: 'Assume that $K$ is a field such that $\Fld / K$ is cartesian closed. This slice category is equivalent to the poset of subfields of $K$. This poset is a lattice, and our assumption implies that it is distributive (see here). But this is quite rare: Consider $K = \IQ(\sqrt{2}, \sqrt{3})$. By Galois theory, the lattice of subfields is isomorphic to the diamond lattice $M_3$ which is not distributive. Specifically, $(\IQ(\sqrt{2}) \wedge \IQ(\sqrt{6})) \vee (\IQ(\sqrt{3}) \wedge \IQ(\sqrt{6})) = \IQ \vee \IQ = \IQ$, while $(\IQ(\sqrt{2}) \vee \IQ(\sqrt{3})) \wedge \IQ(\sqrt{6}) = \IQ(\sqrt{2},\sqrt{3}) \wedge \IQ(\sqrt{6}) = \IQ(\sqrt{6})$.' + proof: >- + Assume that $K$ is a field such that $\Fld / K$ is cartesian closed. This slice category is equivalent to the poset of subfields of $K$. This poset is a lattice, and our assumption implies that it is distributive (see here). But this is quite rare: Consider $K = \IQ(\sqrt{2}, \sqrt{3})$. By Galois theory, the lattice of subfields is isomorphic to the diamond lattice $M_3$, which is not distributive. Specifically, + $$(\IQ(\sqrt{2}) \wedge \IQ(\sqrt{6})) \vee (\IQ(\sqrt{3}) \wedge \IQ(\sqrt{6})) = \IQ \vee \IQ = \IQ,$$ + while + $$(\IQ(\sqrt{2}) \vee \IQ(\sqrt{3})) \wedge \IQ(\sqrt{6}) = \IQ(\sqrt{2},\sqrt{3}) \wedge \IQ(\sqrt{6}) = \IQ(\sqrt{6}).$$ + label: Fld_not_lcc - property: cofiltered-limit-stable epimorphisms proof: >- - Inside of $\IF_p(X)$ consider the descending sequence of subfields + Inside $\IF_p(X)$, consider the descending sequence of subfields $$\IF_p(X) \supseteq \IF_p(X^p) \supseteq \IF_p(X^{p^2}) \supseteq \cdots,$$ - whose intersection is $\IF_p$. Each $\IF_p(X^{p^n}) \hookrightarrow \IF_p(X)$ is purely inseparable, hence an epimorphism, but in the limit we get $\IF_p \hookrightarrow \IF_p(X)$, which is not even algebraic. + whose intersection is $\IF_p$. Each $\IF_p(X^{p^n}) \hookrightarrow \IF_p(X)$ is purely inseparable, hence an epimorphism, but in the limit we obtain $\IF_p \hookrightarrow \IF_p(X)$, which is not even algebraic. special_objects: {} diff --git a/database/data/categories/Fld_0.yaml b/database/data/categories/Fld_0.yaml new file mode 100644 index 000000000..5a7f4e0b4 --- /dev/null +++ b/database/data/categories/Fld_0.yaml @@ -0,0 +1,82 @@ +id: Fld_0 +name: category of fields of characteristic zero +notation: $\Fld_0$ +objects: fields of characteristic $0$ +morphisms: field homomorphisms (i.e., ring homomorphisms) +description: This is the full subcategory of $\Fld$ consisting of fields of characteristic $0$. It is isomorphic to the coslice category $\IQ / \Fld$. +nlab_link: https://ncatlab.org/nlab/show/Field + +tags: + - algebra + +related: + - Fld + - CRing + +comments: + - Limits and colimits are discussed in MSE/359352. + +satisfied_properties: + - property: locally small + proof: It is a full subcategory of $\Fld$, which is locally small. + + - property: initial object + proof: The field $\IQ$ is an initial object; see for example here. + + - property: left cancellative + proof: This property is inherited from $\Fld$. + + - property: quotient-trivial + proof: 'An epimorphism of fields of characteristic $0$ is clearly also an epimorphism of fields, and is therefore purely inseparable by MSE/687869. In characteristic $0$, a purely inseparable field homomorphism is an isomorphism.' + + - property: multi-algebraic + proof: 'Example 4.3(1) in [AR01] presents an FPC-sketch that models fields. It is an extension of the FP-sketch that models commutative rings. To model fields of characteristic $0$, we make the same construction with the FP-sketch that models commutative $\IQ$-algebras.' + +unsatisfied_properties: + - property: skeletal + proof: This is trivial. + + - property: locally finite + proof: There are infinitely many homomorphisms $\IQ(X) \to \IQ(X)$, $X \mapsto X^k$. + + - property: core-thin + proof: If this category were core-thin, Galois theory would not exist. Specifically, the conjugation $\IC \to \IC$, $z \mapsto \overline{z}$ is a non-trivial automorphism. + + - property: semi-strongly connected + proof: There is no homomorphism from $\IQ(\sqrt{2})$ to $\IQ(\sqrt{3})$, since a direct calculation shows that $\IQ(\sqrt{3})$ has no element $a$ with $a^2=2$. Similarly, there is no homomorphism from $\IQ(\sqrt{3})$ to $\IQ(\sqrt{2})$. See also MSE/1069387. + + - property: cogenerating set + proof: We can copy the proof from $\Fld$. + references: + - Fld_no_cogenerating_set + + - property: binary powers + proof: We can copy the proof from $\Fld$ that the product $\IQ(\sqrt{2}) \times \IQ(\sqrt{2})$ does not exist. + references: + - Fld_no_binary_powers + + - property: locally cartesian closed + proof: We can copy the proof from $\Fld$ that $\Fld_0 / \IQ(\sqrt{2}, \sqrt{3})$ is not cartesian closed. + references: + - Fld_not_lcc + + - property: mono-regular + proof: >- + The following example comes from MSE/5129895, where a more general classification of regular monomorphisms is also given. Assume that $\IQ \hookrightarrow \IR$ is a regular monomorphism, i.e. the equalizer of two field homomorphisms $f,g : \IR \rightrightarrows K$. Consider the real numbers + $$x_1 = \sqrt{2}, \quad x_2 = \sqrt{3}, \quad x_3 = \sqrt{6}.$$ + We have $x_1 x_2 = x_3$ and $f(x_i) = \pm g(x_i)$. Therefore, for some $i$ we have $f(x_i)=g(x_i)$. But $x_i \notin \IQ$. + + - property: generator + proof: Assume that $G$ is a generator of $\Fld_0$. It must distinguish the two homomorphisms $\IQ(\sqrt{2}) \rightrightarrows \IQ(\sqrt{2})$. Since $\IQ(\sqrt{2})$ has only two subfields, we see that $G \cong \IQ(\sqrt{2})$. But $G$ must also distinguish the two homomorphisms $\IQ(\sqrt{3}) \rightrightarrows \IQ(\sqrt{3})$, so we likewise obtain $G \cong \IQ(\sqrt{3})$. This contradicts $\IQ(\sqrt{2}) \not\cong \IQ(\sqrt{3})$. + +special_objects: + initial object: + description: $\IQ$ + +special_morphisms: + isomorphisms: + description: bijective field homomorphisms + proof: It is a full subcategory of $\CRing$, for which we know that isomorphisms are bijective homomorphisms. + regular monomorphisms: + description: A Galois extension is a regular monomorphism iff it is procyclic, and the general case can be reduced to this situation; see MSE/5129895 for details. + proof: See MSE/5129895. diff --git a/database/data/category-implications/subobject-trivial.yaml b/database/data/category-implications/subobject-trivial.yaml index e701a27ed..bf79b6338 100644 --- a/database/data/category-implications/subobject-trivial.yaml +++ b/database/data/category-implications/subobject-trivial.yaml @@ -1,5 +1,12 @@ # results on subobject-trivial categories +- id: well_powered_trivial_criterion + assumptions: + - subobject-trivial + conclusions: + - well-powered + proof: This is trivial. + - id: thin_regular-subobject-trivial assumptions: - equalizers diff --git a/database/data/special-morphism-rules.yaml b/database/data/special-morphism-rules.yaml index 349546d7f..6b0634a5b 100644 --- a/database/data/special-morphism-rules.yaml +++ b/database/data/special-morphism-rules.yaml @@ -63,6 +63,16 @@ description: same as epimorphisms proof: The category is epi-regular. +- property: subobject-trivial + type: monomorphisms + description: same as isomorphisms + proof: The category is subobject-trivial. + +- property: quotient-trivial + type: epimorphisms + description: same as isomorphisms + proof: The category is quotient-trivial. + - property: regular-subobject-trivial type: regular monomorphisms description: same as isomorphisms