The category of fields of characteristic zero

database/data/categories/Fld.yaml

@@ -11,13 +11,14 @@ tags:
 
 related:
   - CRing
+  - Fld_0
 
 comments:
   - Limits and colimits are discussed in <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.
 
 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 <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01]</a>.
+    proof: Example 4.3(1) in <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01]</a> 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 <a href="/content/missing_cogenerating_sets">this lemma</a> 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 <a href="/content/missing_cogenerating_sets">this lemma</a> 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 <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.'
+    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 <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.'
+    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 <a href="/category-implication/distributive_criterion">here</a>). 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 <a href="/category-implication/distributive_criterion">here</a>). 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: {}
 

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 <a href="https://en.wikipedia.org/wiki/Characteristic_(algebra)" target="_blank">characteristic</a> $0$. 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 <a href="https://math.stackexchange.com/questions/359352" target="_blank">MSE/359352</a>.
+
+satisfied_properties:
+  - property: locally small
+    proof: It is a full subcategory of <a href="/category/Fld">$\Fld$</a>, which is locally small.
+
+  - property: initial object
+    proof: The field $\IQ$ is an initial object; see for example <a href="https://proofwiki.org/wiki/Field_of_Characteristic_Zero_has_Unique_Prime_Subfield" target="_blank">here</a>.
+
+  - property: left cancellative
+    proof: This property is inherited from <a href="/category/Fld">$\Fld$</a>.
+
+  - property: quotient-trivial
+    proof: 'An epimorphism of fields of characteristic $0$ is clearly also an epimorphism of fields, and is therefore purely inseparable by <a href="https://math.stackexchange.com/questions/687869" target="_blank">MSE/687869</a>. In characteristic $0$, a purely inseparable field homomorphism is an isomorphism.'
+
+  - property: multi-algebraic
+    proof: 'Example 4.3(1) in <a href="http://www.tac.mta.ca/tac/volumes/8/n3/8-03abs.html" target="_blank">[AR01]</a> 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 <a href="https://math.stackexchange.com/questions/1069387">MSE/1069387</a>.
+
+  - property: cogenerating set
+    proof: We can copy the proof from <a href="/category/Fld">$\Fld$</a>.
+    references:
+      - Fld_no_cogenerating_set
+
+  - property: binary powers
+    proof: We can copy the proof from <a href="/category/Fld">$\Fld$</a> 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 <a href="/category/Fld">$\Fld$</a> 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 <a href="https://math.stackexchange.com/questions/5129895" target="_blank">MSE/5129895</a>, 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 <a href="https://math.stackexchange.com/questions/5129895" target="_blank">MSE/5129895</a> for details.
+    proof: See <a href="https://math.stackexchange.com/questions/5129895" target="_blank">MSE/5129895</a>.

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

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