avewright/proofwiki-math huggingface dataset:
Theorem: Tensor Product of Free Modules is Free

Let $A$ be a commutative ring with unity.
Let $F$ and $F'$ be free $A$-modules.
Then the tensor product $F \otimes_A F'$ is a free $A$-module.

Proof:
By Free Module is Isomorphic to Free Module on Set there are sets $I$ and $I'$ and isomorphisms $\Psi : A^{\paren I} \to F$ and $\Psi' : A^{\paren {I'} } \to F'$.
By Tensor Product Distributes over Direct Sum, there is an isomorphism:
:$\ds A^{\paren I} \otimes_A A^{\paren {I'} } \cong \bigoplus_{i \mathop \in I} \bigoplus_{i' \mathop \in I'} A$
By Direct Sum of Direct Sums is Direct Sum, there is an isomorphism:
:$\ds \bigoplus_{i \mathop \in I} \bigoplus_{i' \mathop \in I'} A \cong \bigoplus_{\tuple {i, i'} \mathop \in I \times I'} A$
Hence $A^{\paren I} \otimes_A A^{\paren {I'} } \cong A^{\paren {I \times I'} }$ by definition of $A^{\paren {I \times I'} }$.
By Free Module on Set is Free $A^{\paren {I \times I'} }$ is free.
By Module Isomorphic to Free Module is Free $A^{\paren I} \otimes_A A^{\paren {I'} }$ is free.
{{qed}}
Category:Commutative Algebra
Category:Free Modules
Category:Homological Algebra
Category:Module Theory
pld9bidlgtw8w3zu3edfavkdo5fc8h7
Theorem: Equivalence of Definitions of Kernel of Morphism

Let $\mathbf C$ be a category with zero object $0$.
Let $f : A \to B$ be a morphism in $\mathbf C$.
Then the following definitions of kernel of $f$ are equivalent.

Proof:
=== Definition 1 implies Definition 2 ===
By definition of Zero Object, $0$ is an initial object, so Definition 1 is possible.
Let $f : A \to B$ be a morphism in $\mathbf C$.
Let $k : K \to A$ be a pullback of $f$ along the zero morphism $0 : 0 \to B$.
We check the universal property of the equalizer of $f$ and the zero morphism $0 : A \to B$.
Suppose $T$ is any object and $h : T \to A$ is any morphism with $f \circ h$ = $0 \circ h$.
By Composition with Zero Morphism is Zero Morphism $f \circ h = 0 : T \to B$.
By Definition it follows, that $h : T \to A$ and $0 : T \to 0$ is a cone on the pullback diagram defined by $f$ and $0 : 0 \to B$.
There is a unique morphism $t : T \to K$ with $0 \circ t = 0 : T \to 0$ and $k \circ t = h$.
It follows, that $k$ is an equalizer $\map {\mathrm {Eq} } {f, 0}$.
{{qed|lemma}}
Theorem: Zero Morphism does not Depend on Zero Object

Let $\mathbf C$ be a category.
Let $A$ and $B$ be objects of $\mathbf C$.
Let $0_1$ and $0_2$ be zero objects of $\mathbf C$.
Then the morphism defined as the composition 
:$\beta \circ \alpha : A \to 0_1 \to B$ 
of the unique morphism $\alpha : A \to 0_1$ and the unique morphism $\beta : 0_1 \to B$ is equal to the morphism defined as the composition 
:$\delta \circ \gamma : A \to 0_2 \to B$ 
of the unique morphism $\gamma : A \to 0_2$ and the unique morphism $\delta : 0_2 \to B$.

Proof:
There are unique morphisms $\epsilon : 0_1 \to 0_2$ and $\zeta : 0_2 \to 0_1$.
Since $0_1$ is terminal, we have
: $\zeta \circ \epsilon = \operatorname{id}_{0_1}$
: $\beta \circ \zeta = \delta$
Since $0_2$ is terminal, we have
: $\epsilon \circ \alpha = \gamma$
Hence
{{begin-eqn}}
{{eqn | l = \beta \circ \alpha
      | r = \beta \circ \operatorname{id}_{0_1} \circ \alpha
      | c =
}}
{{eqn | l = 
      | r = \beta \circ \zeta \circ \epsilon \circ \alpha
      | c = 
}}
{{eqn | l = 
      | r = \delta \circ \gamma
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Freyd-Mitchell Embedding Theorem

Let $\AA$ be a small abelian category.
Then there exists a ring with unity $R$ and a fully faithful and exact functor $F : \AA \to R \text{-} \mathbf{Mod}$ to the category of left $R$-modules.

Proof:
{{ProofWanted}}
{{Namedfor|Peter John Freyd|name2 = Barry M. Mitchell|cat = Mitchell|cat2 = Freyd}}
Theorem: Eigenvalues of Hermitian Operator have Orthogonal Eigenspaces

The eigenvectors of a Hermitian operator have eigenspaces which are orthogonal.

Proof:
Directly follows from Hermitian Operator is Normal and Eigenvalues of Normal Operator have Orthogonal Eigenspaces.
{{qed}}
Category:Linear Algebra
Category:Linear Operators
Category:Linear Transformations on Hilbert Spaces
2m73pteuz0snmf9xolvtxhj5n6jm72n
Theorem: Top is Complete

The category of topological spaces is complete.

Proof:
Let $\II$ be a small category.
Let $D : \II \to \mathbf {Top}$ be a diagram in the category of topological spaces $\mathbf {Top}$.
Let $\family {\lim D, \family {\pi_i}_{i \mathop \in \II}}$ be the limit of topological spaces of $D$.
By Limit of Topological Spaces is Limit, $\family {\lim D, \family {\pi_i}_{i \mathop \in \II}}$ is a categorical limit of $D$.
{{qed}}
Category:Topological Spaces
Category:Category Theory
pd4t2z95dgmnub0zwhvms9067ttvz9t
Theorem: Projective Resolution Exists Iff Enough Projectives

Let $\AA$ be an abelian category.
Then $\AA$ has enough projectives {{iff}} any object in $\AA$ has a projective resolution.

Proof:
Suppose $\AA$ has enough projectives.
Let $X$ be an object in $\AA$.
Then there is an epimorphism $\varepsilon : P_0 \to X$ for some projective object $P_0$.
In particular
: $P_0 \to X \to 0$
is exact at $X$.
Since $\AA$ is abelian it has kernels.
Thus $\varepsilon$ has a kernel $K \to P_0$.
Since $\AA$ has enough projectives, there is an epimorphism $\tilde d_1 : P_1 \to K$.
Define $d_1 : P_1 \to K \to P_0$ by composition.
By Uniqueness of Epi-Mono Factorization and since this is an epi-mono factorization of $d_1$, $K$ is the image of $d_1$.
Hence
:$P_1 \to P_0 \to X \to 0$
is exact at $P_0$.
By induction there is a projective resolution of $X$.
Conversely suppose, that any object $X$ admits a projective resolution.
Then the augmentation map $\varepsilon: P_0 \to X$ is an epimorphism by a projective object $P_0$.
Hence $\AA$ has enough projectives.
{{qed}}
Category:Homological Algebra
a5mdr3eqr3cytmda6cg2v6sepkcn2r9
Theorem: Inner Product with Zero Vector

Let $\struct {V, \innerprod \cdot \cdot}$ be an inner product space.
Let $0_V$ be the zero vector of $V$.
Then: 
:$\innerprod {0_V} x = \innerprod x {0_V} = 0$
for all $x \in V$.

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = \innerprod {0_V} x
	| r = \innerprod {0_V + 0_V} x
	| c = {{Defof|Zero Vector}}
}}
{{eqn	| r = \innerprod {0_V} x + \innerprod {0_V} x
	| c = linearity of inner product in first argument
}}
{{end-eqn}}
so:
:$\innerprod {0_V} x = 0$
From conjugate symmetry, we have: 
:$\innerprod x {0_V} = \overline {\innerprod {0_V} x}$
so:
:$\innerprod x {0_V} = \overline 0 = 0$
{{qed}}
Category:Inner Product Spaces
ib0hqdy8rhebaeblu6g9b9k0ub1l3ul
Theorem: Distributional Solution to x T = 0

Let $T \in \map {\DD'} \R$ be a Schwartz distribution.
Let $\delta$ be the Dirac delta distribution.
Let $\mathbf 0$ be the zero distribution.
Suppose $T$ satisfies the following equation in the distributional sense:
:$x T = \mathbf 0$
Then $T = \alpha \delta$ where $c \in \C$.

Proof:
Let $\phi \in \map \DD \R$ be a test function.
Let $c \in \C$.
Suppose:
:$T = c \delta$
Then:
{{begin-eqn}}
{{eqn | l = x \map T {\map \phi x}
      | r = x \paren{c \map \delta {\map \phi x} } 
}}
{{eqn | r = \map \delta {c x \map \phi x} 
      | c = {{Defof|Multiplication of Schwartz Distribution by Smooth Function}}
}}
{{eqn | r = 0
      | c = {{Defof|Dirac Delta Distribution}}
}}
{{eqn | r = \map {\mathbf 0} \phi
}}
{{end-eqn}}
Theorem: Restriction of Sheaf to Open Set is Sheaf

Let $X$ be a topological space.
Let $\FF$ be a sheaf of sets on $X$.
Let $U \subset X$ be an open subset.
Then the restriction $\FF {\restriction_U}$ of $\FF$ to $U$ is a sheaf.

Proof:
{{ProofWanted}}
Category:Algebraic Geometry
Category:Sheaf Theory
8zp0igs959sgsu9ajsja8o89lifbtxa
Theorem: Open Neighborhood contains Affine Open Neighborhood

Let $\struct {X, \OO_X}$ be a scheme.
Let $U \subset X$ be an open subset.
Let $x \in U$.
Then there exists an open subset $V \subset U$ with $x \in V$, such that the restriction $\struct {V, \OO_X {\restriction_V}}$ of $\struct {X, \OO_X}$ to $V$ is an affine scheme.

Proof:
{{ProofWanted}}
Category:Algebraic Geometry
Category:Schemes
gd4gmnyt0n0yjt571srduhp8c0n2psa
Theorem: Open Subscheme is Scheme

Let $\struct {X, \OO_X}$ be a scheme.
Let $U \subset X$ be an open subset.
Then the open subscheme $\struct {U, \OO_X {\restriction U}}$ defined by $U$ is a scheme.

Proof:
Let $x \in U$.
By Open Neighborhood contains Affine Open Neighborhood, there is an open subset $V \subset U$ with $x \in V$, such that $\struct {V, \OO_X {\restriction V}}$ is an affine scheme.
By Restriction of Restriction of Functor is Restriction $\OO_X {\restriction U} {\restriction V} = \OO_X {\restriction V}$.
By definition $\struct {U, \OO_X {\restriction U}}$ is a scheme.
{{qed}}
Category:Algebraic Geometry
Category:Schemes
fjv297q3cevaj4k3itzwfcyldizxl9k
Theorem: Vanishing of Quasi-Coherent Sheaf Cohomology of Affine Scheme

Let $X = \Spec A$ be the spectrum of a commutative ring $A$.
Let $\FF$ be a quasi-coherent sheaf on $X$.
Then for all $i \in \Z$ with $i > 0$ the $i$-th sheaf cohomology $\map {H^i} {X, \FF} = 0$.

Proof:
{{ProofWanted|Proof following EGA III (1.3.1)}}
Theorem: Separated Morphism is Quasi-Separated

Let $f$ be a separated morphism of schemes.
Then $f$ is quasi-separated.

Proof:
Let $f$ be a separated morphism of schemes.
By definition, the diagonal morphism $\Delta_f$ is a closed immersion.
By Closed Immersion is Quasi-Compact $\Delta_f$ is quasi-compact.
Thus, by definition, $f$ is quasi-separated.
{{qed}}
Category:Algebraic Geometry
Category:Schemes
tj2m4qmcx0na8v74jggb5uj5np4p8rp
Theorem: Restriction of Ringed Space to Open Set is Ringed Space

Let $\struct {X, \OO_X}$ be a ringed space.
Let $U \subset X$ be an open subset of $X$.
Let $\struct {U, \OO_X {\restriction_U}}$ denote the restriction of $\struct {X, \OO_X}$ to $U$.
Then $\struct {U, \OO_X {\restriction_U}}$ is a ringed space.

Proof:
By Restriction of Sheaf to Open Set is Sheaf $\OO_X {\restriction_U}$ is a sheaf of commutative rings on $U$.
It follows, that  $\struct {U, \OO_X {\restriction_U}}$ is a ringed space.
{{qed}}
Category:Ringed Spaces
d0mxvtekfjiupgwl52cpnwug2f8ugkl
Theorem: Regular Representation on Subgroup is Bijection to Coset/Left

Let $y H$ denote the left coset of $H$ by $y$.
The mapping $\lambda_x: H \to x H$, where $\lambda_x$ is the left regular representation of $H$ with respect to $x$, is a bijection from $H$ to $x H$.

Proof:
Let $h \in H$.
Then, by definition of left regular representation:
:$\map {\lambda_x} h = x h \in x H$
Thus:
:$\forall h \in H: \map {\lambda_x} h \in x H$
So $\lambda_x: H \to x H$ is a mapping.
A permutation is {{afortiori}} a bijection.
As Regular Representations in Group are Permutations, it follows that $\lambda_x$ is a bijection.
{{qed}}
Theorem: Sheaf Associated to Injective Module over Noetherian Ring is Flasque

Let $A$ be a Noetherian commutative ring.
Let $I$ be an injective $A$-module.
Then the sheaf $\tilde I$ associated to $I$ on $\Spec A$ is flasque.

Proof:
{{ProofWanted|Hartshorne III.3.4}}
Theorem: Injective iff Projective in Dual Category

Let $\mathbf A$ be an abelian category.
Let $X$ be an object in $\mathbf A$.
Then:
:$X$ is injective in $\mathbf A$
{{iff}}:
:$X$ is projective in the dual category $\mathbf A^{\mathrm{op}}$ of $\mathbf A$.

Proof:
=== Sufficient Condition ===
Let $X$ be injective in $\mathbf A$.
Let $f : B \to A$ be an epimorphism in $\mathbf A^{\mathrm {op} }$.
By Monomorphism iff Epimorphism in Dual Category:
:$f : A \to B$ is a monomorphism in $\mathbf A$.
By definition of injective object, the mapping:
:$\alpha: \map {\mathrm {Hom}_{\mathbf A} } {B, X} \to \map {\mathrm {Hom}_{\mathbf A} } {A, X}, \quad g \mapsto g \circ f$ 
is surjective.
{{explain|The notation $\map {\mathrm {Hom}_{\mathbf A} } {A, X}$ needs to be explained. The objective of {{ProofWiki}} is for people unfamiliar with the subject matter to be able to understand a given page by referring back to pages that define all concepts invoked on that page. It is better to write a line defining exactly what $\alpha$ is in terms of its domain and range.
I added a sentence about the Hom set. Technically it is the Hom functor but it maybe would be overkill to refer to it in the text, since i don't need functoriality, so i added an also see-link. --Wandynsky (talk) 09:44, 28 July 2021 (UTC)}}
In the above, $\map {\mathrm {Hom}_{\mathbf A} } {B, X}$ is the set of morphisms in $\mathbf A$ from $B$ to $X$, and $g$ is an arbitrary element of the domain of $\alpha$.
Further, by definition of dual category:
:$\map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, B} = \map {\mathrm {Hom}_{\mathbf A} } {B, X}$
:$\map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, A} = \map {\mathrm {Hom}_{\mathbf A} } {A, X}$
So:
:$\alpha : \map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, B} \to \map {\mathrm {Hom}_{\mathbf A^{\mathrm {op} } } } {X, A}, \quad g \mapsto g \circ f = f \circ_{\mathbf A^{\mathrm{op}}} g$ 
is surjective.
In the above, $\circ_{\mathbf A^{\mathrm{op}}}$ denotes composition in $\mathbf A^{\mathrm {op} }$.
By definition of projective object, $X$ is projective in $\mathbf A^{\mathrm {op} }$.
{{qed|lemma}}
Theorem: Composition with Zero Morphism is Zero Morphism

Let $\mathbf C$ be a category.
Let $0$ be a zero object in $\mathbf C$.
Let $A, B, C, D$ be objects in $\mathbf C$.
Let $f : A \to B$, $g : B \to C$ and $h : C \to D$ be morphisms in $\mathbf C$.
Let $g$ be a zero morphism.
Then $g \circ f$ and $h \circ g$ are zero morphisms.

Proof:
By definition of zero morphism, $g$ is the composition of the unique morphism $\alpha : B \to 0$ with the unique morphism $\beta : 0 \to C$.
Since $\alpha \circ f : A \to 0$ is a morphism with codomain $0$ and $0$ is a terminal object, $\alpha \circ f$ is the unique morphism $A \to 0$.
It follows that $g \circ f = \beta \circ \alpha \circ f$ is a zero morphism.
Since $h \circ \beta : 0 \to D$ is a morphism with domain $0$ and $0$ is an initial object, $h \circ \beta$ is the unique morphism $0 \to D$.
It follows that $h \circ g = h \circ \beta \circ \alpha$ is a zero morphism.
{{qed}}
Category:Morphisms
ofpwex1ufb5al3zql7ksphtfaqckuai
Theorem: Differentiability of Function with Translation Property

Let $f$ be a real function.
Let $f$ have the translation property.
Let $c$ be a real number.
Let $\map {f'} c$ exist.
Then:
:$f'$ exists
:$f'$ is a constant function

Proof:
Let $x$ be a real number.
We have:
{{begin-eqn}}
{{eqn | l = \map {f'} x
      | r = \lim_{y \mathop \to x} \frac {\map f y - \map f x} {y - x}
      | c = {{Defof|Differentiable Mapping|subdef = Real Function/Point|index = 1}}
}}
{{eqn | r = \lim_{y \mathop \to x} \frac {\map f {y - x + x - c + c} - \map f {x - c + c} } {y - x}
}}
{{eqn | r = \lim_{y \mathop \to x} \frac {\map f {y - x + c + x - c} - \map f {c + x - c} } {y - x}
}}
{{eqn | r = \lim_{y \mathop \to x} \frac {\map f {y - x + c + t} - \map f {c + t} } {y - x}
      | c = $t = x - c$
}}
{{eqn | r = \lim_{z \mathop + x \mathop \to x} \frac {\map f {z + c + t} - \map f {c + t} } z
      | c = $z = y - x$
}}
{{eqn | r = \lim_{z \mathop \to 0} \frac {\map f {z + c + t} - \map f {c + t} } z
}}
{{eqn | r = \lim_{z \mathop \to 0} \frac {\map f {z + c} - \map f c} z
      | c = {{Defof|Translation Property}}
}}
{{eqn | r = \map {f'} c
      | c = {{Defof|Differentiable Mapping|subdef = Real Function/Point|index = 2}}
}}
{{end-eqn}}
We conclude that $\map {f'} x$ exists as $\map {f'} c$ exists.
Also, $\map {f'} x$ exists everywhere as $x$ is arbitrary.
In other words, $f'$ exists.
Also, $f'$ is a constant function as $\map {f'} x$ equals $\map {f'} c$ for every real number $x$.
{{qed}}
Category:Differential Calculus
hmkrk3numo56ecyw342udvs3sb9fvtq
Theorem: Linearity of Function with Translation Property

Let $f$ be a real function.
Let $c$ be a real number.
Then:
:$f$ has the translation property
{{improve|Obviously, what needs to happen here is that rather than use the suboptimal page translation property, we need to make it so that it directly references the page which defines what a translation is. That is to say "$f$ is a translation", or some such. I hadn't dreamed that this would be a difficult concept to grasp, for which I apologise.}}
:$\map {f'} c$ exists
{{iff}}:
:$f$ is linear

Proof:
=== Sufficient Condition ===
Let:
:$f$ have the translation property
:$\map {f'} c$ exist
We need to show that:
:$f$ is linear
The fact that $f$ has the translation property means:
:$\forall x_1, x_2, t \in \R: \map f {x_1 + t} - \map f {x_2 + t} = \map f {x_1} - \map f {x_2}$
$f$ being linear means:
:$\forall x \in \R: \map f x = ax + b$
where $a$ and $b$ are real numbers.
We have that $f'$ exists by Differentiability of Function with Translation Property.
Also, $f'$ is a constant function by the same theorem.
Accordingly:
:$\map {f'} x = \map {f'} c$ for every real number $x$
Let $x$ be a real number.
We have:
{{begin-eqn}}
{{eqn | l = \map f x
      | r = \map f x - \map f 0 + \map f 0
}}
{{eqn | r = \int_0^x \map {f'} y \rd y  + \map f 0
      | c = $f'$ is integrable by Integral of Constant
}}
{{eqn | r = \int_0^x \map {f'} c \rd y  + \map f 0
      | c = $\map {f'} y = \map {f'} c$
}}
{{eqn | r = \map {f'} c x  + \map f 0
      | c = Integral of Constant
}}
{{end-eqn}}
So, $f$ is linear.
{{qed|lemma}}
Theorem: Linearity of Function defined using Function with Translation Property

Let $f$ be a real function.
Let $f$ have the translation property.
Let $x$ and $l$ be real numbers.
Define:
:$\map {f_x} l = \map f {x + l} - \map f x$
Then:
:$\forall q \in \Q: \map {f_x} {q l} = q \map {f_x} l$

Proof:
=== Lemma ===
{{:Linearity of Function defined using Function with Translation Property/Lemma}} {{qed|lemma}}
Let $q$ be a rational number.
Choose integers $n$, $m$ such that:
:$\dfrac n m = q$
We need to prove that:
:$\map {f_x} {q l} = q \map {f_x} l$
We have:
{{begin-eqn}}
{{eqn | l = \map {f_x} {q l}
      | r = \map {f_x} {\frac n m l}
      | c = Definition of $n$, $m$
}}
{{eqn | r = n \map {f_x} {\frac l m}
      | c = Lemma
}}
{{eqn | r = \frac n m m \map {f_x} {\frac l m}
}}
{{eqn | r = \frac n m \map {f_x} {\frac {m l} m}
      | c = Lemma
}}
{{eqn | r = q \map {f_x} l
      | c = Definition of $n$, $m$
}}
{{end-eqn}}
{{qed}}
Category:Linear Real Functions
83z5uuvult6b9nyz7a9ur9okd79czl0
Theorem: Baer's Criterion

Let $R$ be a ring with unity.
Let $M$ be a left $R$-module.
Then $M$ is injective {{iff}} the following condition holds:
:For all left ideals $I$ of $R$ with inclusion map $\iota : I \to R$, and for all $R$-module homomorphisms $f : I \to M$, there exists an $R$-module homomorphism $\tilde f : R \to M$ such that:
::$\tilde f \circ \iota = f$

Proof:
{{ProofWanted}}
{{Namedfor|Reinhold Baer|cat = Baer}}
Category:Homological Algebra
Category:Module Theory
Category:Ring Theory
9q1r8gxmef4xghp9rp3oif3hzsg0762
Theorem: Injective Module over Dedekind Domain

Let $D$ be a Dedekind domain.
Let $M$ be a $D$-module.
Then $M$ is injective {{iff}} it is divisible.

Proof:
{{ProofWanted}}
Category:Commutative Algebra
Category:Homological Algebra
Category:Module Theory
7ejvfxep5vvl24spbd569d1yiwvwttq
Theorem: Injective Module over Principal Ideal Domain

Let $D$ be a principal ideal domain.
Let $M$ be a $D$-module.
Then $M$ is injective {{iff}} it is divisible.

Proof:
By Principal Ideal Domain is Dedekind Domain $D$ is a Dedekind domain.
By Injective Module over Dedekind Domain $M$ is injective {{iff}} it is divisible.
{{qed}}
Category:Principal Ideal Domains
Category:Commutative Algebra
Category:Homological Algebra
Category:Module Theory
s91awp7825oi1x931w87o1txnfzyqls
Theorem: Principal Ideal Domain is Dedekind Domain

Let $D$ be a principal ideal domain which is specifically not a field.
Then $D$ is a Dedekind domain.

Proof:
By definition of principal ideal domain $D$ is an integral domain.
By Principal Ideal Domain is Noetherian $D$ is noetherian.
By Principal Ideal Domain is Integrally Closed $D$ is integrally closed.
By Prime Ideal of Principal Ideal Domain is Maximal $D$ has Krull dimension $\le 1$.
By Integral Domain has Dimension Zero iff Field and since $D$ is not a field, the Krull dimension of $D$ is $1$.
Hence $D$ is a Dedekind domain by definition.
{{qed}}
Category:Principal Ideal Domains
Category:Dedekind Domains
n95ndgen5b1n5snqjwg0anmm8tt9tjf
Theorem: Principal Ideal Domain is Integrally Closed

Let $A$ be a principal ideal domain.
Then $A$ is integrally closed.

Proof:
By Principal Ideal Domain is Unique Factorization Domain $A$ is a unique factorization domain.
By Unique Factorization Domain is Integrally Closed $A$ is integrally closed.
{{qed}}
Category:Principal Ideal Domains
onnfqze63bvk4dhb9ga9bj3vngezdtc
Theorem: Field of Quotients is Divisible Module

Let $D$ be an integral domain.
Let $\map {\operatorname {Quot} } D$ be the field of quotients of $D$.
Then $\map {\operatorname {Quot} } D$ is a divisible $D$-module.

Proof:
Let $a \in D$ be a non zero divisor.
Let $x, y \in D$ such that $y \ne 0$.
Then $\dfrac x y \in \map {\operatorname {Quot} } D$.
By definition of integral domain:
:$a \ne 0$
Thus $\dfrac x {a y}$ is defined in $\map {\operatorname {Quot} } D$.
It follows that:
:$a \cdot \dfrac x {a y} = \dfrac x y$
Thus $\map {\operatorname {Quot} } D$ is a divisible $D$-module.
{{qed}}
Category:Integral Domains
Category:Module Theory
Category:Homological Algebra
009da50c647peur6apgya6jdsoagppn
Theorem: Quotient of Divisible Module is Divisible

Let $R$ be a ring with unity.
Let $M$ be a divisible left $R$-module.
Let $N \subseteq M$ be an $R$-submodule.
Then the quotient module $M / N$ is divisible.

Proof:
Let $r \in R$ be a regular element of $R$.
Hence by definition $r$ is not a zero divisor of $R$.
Let $\eqclass m {} \in M / N$ be an arbitrary element represented by $m \in M$.
Since $M$ is divisible, there exists some $m' \in M$ such that $m = r m'$.
By definition of scalar multiplication on the quotient module $M / N$:
:$r \eqclass m {} = \eqclass {m'} {}$
It follows, that $M / N$ is divisible.
{{qed}}
Category:Homological Algebra
Category:Module Theory
6j8tc3rl5x9qf5w6afivhg0iv51xs9m
Theorem: Quotient of Rationals by Integers is Injective

Let $\struct {\Q, +}$ be the abelian group of rational numbers.
Let $\struct {\Z, +}$ be the abelian group of integers, considered as a subgroup of $\struct {\Q, +}$.
Then the quotient group $\Q / \Z$ is an injective object in the category of abelian groups.

Proof:
By definition, $\struct {\Q, +, \times}$ is the field of quotients of the ring of integers $\struct {\Z, +, \times}$.
By Field of Quotients is Divisible Module $\Q$ is a divisible $\Z$-module.
By Quotient of Divisible Module is Divisible $\Q / \Z$ is a divisible $\Z$-module.
By Ring of Integers is Principal Ideal Domain $\struct {\Z, +, \times}$ is a principal ideal domain.
By Injective Module over Principal Ideal Domain $\Q / \Z$ is an injective $\Z$-module.
Since the category of abelian groups is equivalent to the category of $\Z$-modules, $\Q / \Z$ is an injective object in the category of abelian groups.
{{qed}}
Category:Homological Algebra
Category:Abelian Groups
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