avewright/proofwiki-math huggingface dataset:
Theorem: Extension Theorem for Distributive Operations/Identity

:If $e$ is an identity for $\circ$, then $e$ is also an identity for $\circ'$.

Proof:
We have that $\circ'$ exists and is unique by Extension Theorem for Distributive Operations: Existence and Uniqueness.
Let $e$ be the identity element of $\struct {R, \circ}$.
Then the restrictions to $R$ of the endomorphisms $\lambda_e: x \mapsto e \circ' x$ and $\rho_e: x \mapsto x \circ' e$ of $\struct {T, *}$ are monomorphisms.
But then $\lambda_e$ and $\rho_e$ are monomorphisms by the Extension Theorem for Homomorphisms.
Hence it follows that $e$ is the identity element of $T$.
{{explain|How does the above explain why $e$ is the identity element of $T$?}}
{{qed}}
Theorem: Extension Theorem for Distributive Operations/Cancellability

:Every element of $R$ cancellable for $\circ$ is also cancellable for $\circ'$.

Proof:
Let $a$ be an element of $R$ cancellable for $\circ$.
Then the restrictions to $R$ of the endomorphisms:
:$\lambda_a: x \mapsto a \circ' x$
:$\rho_a: x \mapsto x \circ' a$
of $\struct {T, *}$ are monomorphisms.
But then $\lambda_a$ and $\rho_a$ are monomorphisms by the Extension Theorem for Homomorphisms.
Hence $a$ is cancellable for $\circ'$.
{{qed}}
Theorem: Modulus of Sine of x Less Than or Equal To Absolute Value of x

:$\size {\sin x} \le \size x$

Proof:
Clearly the inequality holds if $x = 0$.
Take $x \ne 0$. 
From the Mean Value Theorem and Derivative of Sine Function, there exists $c \in \R$ such that: 
:$\ds \frac {\sin x - \sin 0} {x - 0} = \cos c$
so:
:$\sin x = x \cos c$
Then we have: 
{{begin-eqn}}
{{eqn	| l = \size {\sin x}
	| r = \size {x \cos c}
}}
{{eqn	| o = \le
	| r = \size x
}}
{{end-eqn}}
as required.
{{qed}}
Category:Inequalities
Category:Sine Function
Category:Absolute Value Function
3et14vyp689938aqv3wxhzwo2xjmkji
Theorem: Order Modulo n of Power of Integer/Corollary

Let $a$ be a primitive root of $n$.
Then:
:$a^k$ is also a primitive root of $n$
{{iff}}:
:$k \perp \map \phi n$
where $\phi$ is the Euler phi function.
Furthermore, if $n$ has a primitive root, it has exactly $\map \phi {\map \phi n}$ of them.

Proof:
Let $a$ be a primitive root of $n$.
Then $R = \set {a, a^2, \ldots, a^{\map \phi n}}$ is a reduced residue system for $n$.
Hence all primitive roots are contained in $R$.
By Order Modulo n of Power of Integer, the multiplicative order $a^k$ modulo $n$ is $\dfrac {\map \phi n} {\gcd \set {\map \phi n, k} }$.
Hence $a^k$ will be a primitive root of $n$ exactly when $\gcd \set {\map \phi n, k} = 1$.
That is, when $\map \phi n \perp k$.
So the primitive roots are the integers $a^k$, where $k \in \set {1, 2, \ldots, \map \phi n}$.
By definition of $\phi$, there are $\map \phi {\map \phi n}$ such $k$.
Hence there are $\map \phi {\map \phi n}$ primitive roots of $n$.
{{qed}}
Category:Number Theory
t6gt5bnljp4h6xmzanccdjc642bbmri
Theorem: Second Chebyshev Function is Increasing

The second Chebyshev function $\psi$ is increasing.

Proof:
Let $x \ge y$. 
Then: 
{{begin-eqn}}
{{eqn	| l = \map \psi y
	| r = \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le y} \ln p
	| c = {{Defof|Second Chebyshev Function}}
}}
{{eqn	| r = \sum_{k \mathop \ge 1} \paren {\sum_{p^k \mathop \le x} \ln p + \sum_{x \mathop < p^k \mathop \le y} \ln p}
}}
{{eqn	| r = \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p + \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p
}}
{{end-eqn}}
From Logarithm is Strictly Increasing:
:$\ln p \ge \ln 2 > 0$ 
So, we have: 
:$\ds \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p \ge 0$
so:
{{begin-eqn}}
{{eqn	| l = \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p + \sum_{k \mathop \ge 1} \sum_{x \mathop < p^k \mathop \le y} \ln p
	| o = \ge 
	| r = \sum_{k \mathop \ge 1} \sum_{p^k \mathop \le x} \ln p
}}
{{eqn	| r = \map \psi x
	| c = {{Defof|Second Chebyshev Function}}
}}
{{end-eqn}}
So if $x \le y$, then:
:$\map \psi x \le \map \psi y$
so:
:$\psi$ is increasing.
{{qed}}
Category:Second Chebyshev Function
gdqgtwwlpztncw304uooc0octwnrm0z
Theorem: Limit to Infinity of Power of x by Exponential of -a x

Let $k$ and $a$ be positive real numbers. 
Then: 
:$\ds \lim_{x \mathop \to \infty} x^k e^{-a x} = 0$

Proof:
By Power Series Expansion for Exponential Function, we have: 
:$\ds e^{a x} = \sum_{n \mathop = 0}^\infty \frac {\paren {a x}^n} {n!}$
Since for $x > 0$ each term in this sum is non-negative, we have: 
:$\ds e^{a x} \ge \frac {\paren {a x}^{\floor k + 1} } {\paren {\floor k + 1}!}$
for each $k$.
So, for each $x > 0$ we have: 
:$\ds 0 < e^{-a x} \le \frac {\paren {\floor k + 1}!} {\paren {a x}^{\floor k + 1} }$
So that: 
:$\ds 0 \le x^k e^{-a x} \le \frac 1 {a^{\floor k + 1} } x^{k - \floor k - 1} \paren {\floor k + 1}!$
From the definition of the floor function, we have: 
:$0 \le k - \floor k < 1$
so:
:$k - \floor k - 1 < 0$
Hence by Limit to Infinity of Power:
:$\ds \lim_{x \mathop \to \infty} \frac 1 {a^{\floor k + 1} } x^{k - \floor k - 1} \paren {\floor k + 1}! = 0$
So, by the Squeeze Theorem, we have: 
:$\ds \lim_{x \mathop \to \infty} x^k e^{-a x} = 0$
{{qed}}
Category:Exponential Function
77lbzee7ev9kwvoz9utffp0jcbt0lhv
Theorem: Limit of Power of x by Absolute Value of Power of Logarithm of x

Let $\alpha$ and $\beta$ be positive real numbers. 
Then: 
:$\ds \lim_{x \mathop \to 0^+} x^\alpha \size {\ln x}^\beta = 0$

Proof:
From Order of Natural Logarithm Function, we have: 
:$\ln x = \map \OO {x^{-\frac \alpha {2 \beta} } }$ as $x \to 0^+$
That is, by the definition of big-O notation there exists positive real numbers $x_0$ and $C$ such that: 
:$0 \le \size {\ln x} \le C x^{-\frac \alpha {2 \beta} }$
for $0 < x \le x_0$.
So:
:$0 \le \size {\ln x}^\beta \le C^\beta x^{-\alpha/2}$
for $0 < x \le x_0$. 
That is: 
:$0 \le x^\alpha \size {\ln x}^\beta \le C^\beta x^{\alpha/2}$
We have that: 
:$\ds \lim_{x \mathop \to 0^+} C^\beta x^{\alpha/2} = 0$
so by the Squeeze Theorem for Functions:
:$\ds \lim_{x \mathop \to 0^+} x^\alpha \size {\ln x}^\beta = 0$
{{qed}}
Category:Logarithms
Category:Limits of Real Functions
Category:Limit of Power of x by Absolute Value of Power of Logarithm of x
michvdy298yardw4j9429ktal3abuab
Theorem: X + y + z equals 1 implies xy + yz + zx less than Half

Let $x$, $y$ and $z$ be real numbers such that: 
:$x + y + z = 1$
Then:
:$x y + y z + z x < \dfrac 1 2$

Proof:
We have: 
{{begin-eqn}}
{{eqn	| l = 1
	| r = \paren {x + y + z}^2
}}
{{eqn	| r = \paren {\paren {x + y} + z}^2
}}
{{eqn	| r = \paren {x + y}^2 + 2 z \paren {x + y} + z^2
	| c = Square of Sum
}}
{{eqn	| r = x^2 + 2 x y + y^2 + 2 x z + 2 y z + z^2
	| c = Square of Sum
}}
{{eqn	| r = 2 \paren {x y + y z + z x} + \paren {x^2 + y^2 + z^2}
}}
{{end-eqn}}
So:
:$2 \paren {x y + y z + z x} = 1 - \paren {x^2 + y^2 + z^2}$
We have: 
:$x^2 + y^2 + z^2 \ge 0$
for all real numbers $x$, $y$, $z$ with equality only if:
:$x = y = z = 0$
This cannot be the case since $x + y + z = 1$, so we have: 
:$x^2 + y^2 + z^2 > 0$
so:
:$2 \paren {x y + y z + z x} < 1$
giving:
:$x y + y z + z x < \dfrac 1 2$
{{qed}}
Theorem: Inner Product is Continuous

Let $\struct {V, \innerprod \cdot \cdot}$ be a inner product space. 
Let $\norm \cdot$ be the inner product norm on $V$. 
Let $x, y \in V$.
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be sequences converging in $\struct {V, \norm \cdot}$ to $x$ and $y$ respectively.
Then we have: 
:$\innerprod {x_n} {y_n} \to \innerprod x y$
as $n \to \infty$.

Proof:
Let $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ be sequences converging to $x$ and $y$ respectively.
From Modulus of Limit in Normed Vector Space, we have that: 
:$\norm {x_n} \to \norm x$
and:
:$\norm {y_n} \to \norm y$
We have:
{{begin-eqn}}
{{eqn	| l = \size {\innerprod {x_n} {y_n} - \innerprod x y}
	| r = \size {\innerprod {x_n - x} {y_n} + \innerprod x {y_n - y} }
}}
{{eqn	| o = \le
	| r = \size {\innerprod {x_n - x} {y_n} } + \size {\innerprod x {y_n - y} }
	| c = Triangle Inequality
}}
{{eqn	| o = \le
	| r = \norm {x_n - x} \norm {y_n} + \norm x \norm {y_n - y}
	| c = Cauchy-Bunyakovsky-Schwarz Inequality
}}
{{end-eqn}}
From Convergent Real Sequence is Bounded, we have: 
:$\sequence {\norm {y_n} }$ is bounded.
That is, there exists a positive real number $M$ such that: 
:$\norm {y_n} < M$
Let $\epsilon > 0$. 
From the definition of a convergent sequence in a normed vector space, there exists $N_1 \in \N$ such that: 
:$\norm {x_n - x} < \dfrac \epsilon {2 M}$
for $n > N_1$.
If $x = 0$, we have: 
:$\norm x = 0$
Then:
{{begin-eqn}}
{{eqn | l = \size {\innerprod {x_n} {y_n} - \innerprod x y}
      | o = \le
      | r = \norm {x_n - x} \norm {y_n}
}}
{{eqn | o = <
      | r = M \norm {x_n - x}
}}
{{eqn | o = <
      | r = \frac \epsilon 2
}}
{{eqn | o = <
      | r = \epsilon
}}
{{end-eqn}}
for $n > N_1$, so we are done in this case. 
Now take $x \ne 0$, then: 
:$\norm x \ne 0$
So by the definition of a convergent sequence in a normed vector space, there exists $N_2 \in \N$ such that: 
:$\norm {y_n - y} < \dfrac \epsilon {2 \norm x}$
Let:
:$N = \max \set {N_1, N_2}$
Then, for $n > N$, we have: 
{{begin-eqn}}
{{eqn | l = \size {\innerprod {x_n} {y_n} - \innerprod x y}
      | o = \le
      | r = \norm {x_n - x} \norm {y_n} + \norm x \norm {y_n - y}
}}
{{eqn | o = <
      | r = M \times \frac \epsilon {2 M} + \norm x \times \frac \epsilon {2 \norm x}
}}
{{eqn | r = \frac \epsilon 2 + \frac \epsilon 2 
}}
{{eqn | r = \epsilon
}}
{{end-eqn}}
So we have: 
:$\innerprod {x_n} {y_n} \to \innerprod x y$
as $n \to \infty$, hence the claim.
{{qed}}
Category:Inner Product Spaces
8b76k0j194fkrt1z4jce7r25ullcuw0
Theorem: Convergence of Sequence of Test Functions in Test Function Space implies Convergence in Schwartz Space

Let $\map \DD \R$ be the test function space.
Let $\map \SS \R$ be the Schwartz space.
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence of test functions in $\map \DD \R$.
Let $\mathbf 0 : \R \to 0$ be the zero mapping.
Suppose $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \DD \R$:
:$\phi_n \stackrel \DD \longrightarrow \mathbf 0$
Then $\sequence {\phi_n}$ converges to $\mathbf 0$ in $\map \SS \R$:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$

Proof:
For all $n \in \N$ let $\phi_n$ be a test function.
By definition, $\phi_n$ has a compact support $I_n \subset \R$:
:$\forall x \notin I_n \implies \map {\phi_n} x = 0$
Let:
:$a \in \R : a > 0 : \forall n \in \N : I_n \subseteq \closedint {-a} a$
Then:
{{begin-eqn}}
{{eqn | l = \forall m,k \in \N : \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x}
      | r = \map {\sup_{\size x \mathop \le a} } {\size x^k \size {\map {\phi^{\paren m}_n} x} }
      | c = Absolute Value Function is Completely Multiplicative
}}
{{eqn | o = \le 
      | r = a^k \sup_{\size x \mathop \le a} \size {\map {\phi^{\paren m}_n} x}
}}
{{end-eqn}}
Suppose:
:$\phi_n \stackrel \DD \longrightarrow \mathbf 0$
Then all derivatives of $\phi_n$ on $\closedint {-a} a$ converge uniformly to $\mathbf 0$.
Hence:
{{begin-eqn}}
{{eqn | l = \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x}
      | o = \le
      | r = a^k \lim_{n \mathop \to \infty} \sup_{\size x \mathop \le a} \size {\map {\phi^{\paren m}_n} x}
}}
{{eqn | r = a^k \sup_{\size x \mathop \le a} \size {\mathbf 0}
      | c = $\phi_n \stackrel \DD \longrightarrow \mathbf 0$
}}
{{eqn | r = a^k \cdot 0
}}
{{eqn | r = 0
}}
{{end-eqn}}
{{Help|passing the limit through supremum}}
In other words:
:$\ds \forall m,k \in \N :  \lim_{n \mathop \to \infty} \sup_{x \mathop \in \R} \size {x^k \map {\phi^{\paren m}_n} x} = 0$
By definition:
:$\phi_n \stackrel \SS \longrightarrow \mathbf 0$
{{qed}}
Theorem: Null Ring is Commutative Ring

Let $R$ be the null ring.
That is, let:
:$R := \struct {\set {0_R}, +, \circ}$
where ring addition and ring product are defined as:
{{begin-eqn}}
{{eqn | l = 0_R + 0_R
      | r = 0_R
      | c = 
}}
{{eqn | l = 0_R \circ 0_R
      | r = 0_R
      | c = 
}}
{{end-eqn}}
Then $R$ is a commutative ring.

Proof:
From Null Ring is Trivial Ring, we have that $R$ is a trivial ring.
The result follows from Trivial Ring is Commutative Ring.
{{qed}}
Theorem: Orthogonal Projection is Linear Transformation

Let $\GF \in \set {\R, \C}$.
{{explain|Does this hold for all subfields of $\C$?}}
Let $H$ be a Hilbert space over $\mathbb F$ with inner product $\innerprod \cdot \cdot$.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then $P_K$ is a linear transformation on $H$.

Proof:
Let $x, y \in H$. 
Let $\alpha, \beta \in \GF$. 
Let $k \in \KK$. 
Since the inner product is linear in its first argument, we have: 
:$\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_K} x + \beta \map {P_K} y} } k = \alpha \innerprod {x - \map {P_K} x} k + \beta \innerprod {y - \map {P_K} y} k$
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have: 
:$\alpha \innerprod {x - \map {P_K} x} k = 0$
and:
:$\beta \innerprod {y - \map {P_K} y} k = 0$
Since $k$ was arbitrary, we have: 
:$\innerprod {\paren {\alpha x + \beta y} - \paren {\alpha \map {P_K} x + \beta \map {P_K} y} } k = 0$
for all $k \in K$. 
Note that from Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space: 
:$\map {P_K} {\alpha x + \beta y}$ is the unique point in $H$ such that $\innerprod {\paren {\alpha x + \beta y} - \map {P_K} {\alpha x + \beta y} } k = 0$ for each $k \in K$. 
So we therefore have: 
:$\alpha \map {P_K} x + \beta \map {P_K} y = \map {P_K} {\alpha x + \beta y}$
for each $x, y \in H$ and $\alpha, \beta \in \mathbb F$.
So $P_K$ is a linear transformation, as required.
{{qed}}
Theorem: Orthogonal Projection is Bounded

Let $H$ be a Hilbert space with inner product $\innerprod \cdot \cdot$ and inner product norm $\norm \cdot$.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then $P_K$ is bounded. 
That is: 
:$\norm {\map {P_K} h} \le \norm h$
for each $h \in H$.

Proof:
Let $h \in H$. 
Note that we can write: 
:$h = \paren {h - \map {P_K} h} + \map {P_K} h$ 
We have, by the definition of orthogonal projection: 
:$\map {P_K} h \in K$
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have: 
:$h - \map {P_K} h \in K^\bot$ 
so that: 
:$\innerprod {\map {P_K} h} {h - \map {P_K} h} = 0$
By Pythagoras's Theorem (Inner Product Space), we therefore have: 
:${\norm h}^2 = \norm {h - \map {P_K} h}^2 + \norm {\map {P_K} h}^2$ 
We have that: 
:$\norm {h - \map {P_K} h}^2 \ge 0$
so:
:${\norm h}^2 \ge \norm {\map {P_K} h}^2$
giving: 
:$\norm {\map {P_K} h} \le \norm h$ 
{{qed}}
Theorem: Orthogonal Projection is Projection

Let $\HH$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection onto $K$.
Then $P_K$ is a projection.

Proof:
Let $h \in H$.
From the definition of the orthogonal projection, we have:
:$\map {P_K} h \in K$
So, from Fixed Points of Orthogonal Projection, we have: 
:$\map {P_K^2} h = \map {P_K} {\map {P_K} h} = \map {P_K} h$
Since $h$ was arbitrary, we have:
:$P_K^2 = P_K$
So $P_K$ is an idempotent.
Further, from Kernel of Orthogonal Projection, we have:
:$\ker {P_K} = K^\bot$
where $K^\bot$ is the orthocomplement of $K$.
From Range of Orthogonal Projection, we have:
:$K = \Img {P_K}$
Hence we conclude:
:$\ker {P_K} = \paren {\Img {P_K} }^\bot$
Hence $P_K$ is a projection.
{{qed}}
Theorem: Kernel of Orthogonal Projection

Let $H$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then: 
:$\ker P_K = K^\bot$
where:
:$\ker P_K$ denotes the kernel of $P_K$
:$K^\bot$ denotes the orthocomplement of $K$.

Proof:
We first prove that: 
:$\ker P_K \subseteq K^\bot$
Let $h \in \ker P_K$.
Then: 
:$\map {P_K} h = 0$
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, ​we have: 
:$h - \map {P_K} h \in K^\bot$
That is: 
:$h \in K^\bot$
So:
:$\ker P_K \subseteq K^\bot$
We now prove that: 
:$K^\bot \subseteq \ker P_K$
Let $h \in K^\bot$. 
From Unique Point of Minimal Distance to Closed Convex Subset of Hilbert Space, we have that: 
:$\map {P_K} h$ is the unique point such that $h - \map {P_K} h \in K^\bot$ 
So, since $h - 0 \in K^\bot$, we have $\map {P_K} h = 0$.
So, we have $h \in \ker P_K$, and we obtain: 
:$K^\bot = \ker P_K$
{{qed}}
Theorem: Range of Orthogonal Projection

Let $H$ be a Hilbert space.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Then: 
:$P_K \sqbrk H = K$
where $P_K \sqbrk H$ denotes the image of $H$ under $P_K$.

Proof:
We first show that $P_K \sqbrk H \subseteq K$. 
Let $k \in P_K \sqbrk H$.
Then there exists $h \in H$ such that: 
:$\map {P_K} h = k$
From the definition of the orthogonal projection, we have: 
:$\map {P_K} h \in K$
so:
:$h \in K$
giving:
:$P_K \sqbrk H \subseteq K$ 
{{qed|lemma}}
We now show that: 
:$K \subseteq P_K \sqbrk H$
Let $k \in K$. 
Then, from Fixed Points of Orthogonal Projection: 
:$\map {P_K} k = k$
So:
:$k \in P_K \sqbrk H$
and so: 
:$K \subseteq P_K \sqbrk H$
{{qed|lemma}}
Hence, by definition of set equality:
:$K = P_K \sqbrk H$
{{qed}}
Theorem: Fixed Points of Orthogonal Projection

Let $\struct {H, \innerprod \cdot \cdot}$ be a Hilbert space.
Let $\norm \cdot$ be the inner product norm of $H$.
Let $K$ be a closed linear subspace of $H$.
Let $P_K$ denote the orthogonal projection on $K$.
Let $h \in H$. 
Then: 
:$\map {P_K} h = h$
{{iff}} $h \in K$.

Proof:
Let $d$ be the metric induced by $\norm \cdot$. 
Let $h \in H$.
By the definition of orthogonal projection, we have: 
:$\map d {h, \map {P_K} h} = \map d {h, K}$
Note that by the definition of a metric, we have that: 
:$\map d {h, \map {P_K} h} = 0$
{{iff}}:
:$h = \map {P_K} h$
So, we have: 
:$h = \map {P_K} h$
{{iff}}:
:$\map d {h, K} = 0$
Since $K$ is closed, from Subset of Metric Space is Closed iff contains all Zero Distance Points, this is the case {{iff}}: 
:$h \in K$
{{qed}}
Category:Orthogonal Projections
e2a139tzismqzkme0xxypnayp6q1lcj
Theorem: Sum of Cosets of Ideals is Sum in Quotient Ring

The sum $X +_\PP Y$ in $\powerset R$ is also their sum in the quotient ring $R / J$.

Proof:
As $\struct {R, +, \circ}$ is a ring, it follows that $\struct {R, +}$ is an abelian group.
Thus by Subgroup of Abelian Group is Normal, all subgroups of $\struct {R, +, \circ}$ are normal.
So from the definition of quotient group, it follows directly that $X +_\PP Y$ in $\powerset R$ is also the sum in the quotient ring $R / J$.
{{qed}}
Theorem: Principal Ideal is Smallest Ideal

Let $\struct {R, +, \circ}$ be a ring with unity.
Let $a \in R$.
Let $\ideal a$ be the principal ideal of $R$ generated by $a$.
Let $J$ be an ideal of $R$ such that $a \in J$.
Then $\ideal a \subseteq J$.
That is, $\ideal a$ is the smallest ideal of $R$ to which $a$ belongs.

Proof:
Let $J$ be an ideal of $R$ such that $a \in J$.
By the definition of an ideal:
:$\forall r, s \in R: r \circ a \circ s \in J$
Also, $J$ is a group under $+$.
So every element of $\ideal a$ is in $J$.
Thus $\ideal a \subseteq J$.
{{qed}}
Theorem: Prime Element of Integral Domain is Irreducible

Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $p$ be a prime element of $\struct {D, +, \circ}$.
Then $p$ is an irreducible element of $\struct {D, +, \circ}$.

Proof:
By definition of prime element, $p$ is neither zero nor a unit of $\struct {D, +, \circ}$.
{{AimForCont}}:
:$p = a \circ b$
for some non-units $a, b \in D$.
From Element of Integral Domain is Divisor of Itself:
:$p \divides a \circ b$
By definition of prime element:
:$p \divides a$ or $p \divides b$
{{WLOG}}, suppose $p \divides a$.
That is:
:$a = p \circ c$
for some $c \in D$.
{{begin-eqn}}
{{eqn | l = p \circ 1_D
      | r = p
      | c = {{Defof|Unity of Ring}}
}}
{{eqn | r = a \circ b
}}
{{eqn | r = p \circ c \circ b
      | c = $a = p \circ c$
}}
{{eqn | ll = \leadsto
      | l = 1_D
      | r = c \circ b
      | c = Cancellation Law for Ring Product of Integral Domain
}}
{{end-eqn}}
Thus $b$ is a unit of $\struct {D, +, \circ}$.
This contradicts the assertion that $a$ and $b$ are not units of $\struct {D, +, \circ}$.
Thus by Proof by Contradiction $p$ is an irreducible element of $\struct {D, +, \circ}$.
{{qed}}
Category:Integral Domains
Category:Irreducible Elements of Rings
Category:Prime Elements of Rings
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Theorem: Functor Category is Category

Let $\mathbf C$ and $\mathbf D$ be categories.
Then the functor category $\map {\operatorname {Funct} } {\mathbf C, \mathbf D}$ is a metacategory.

Proof:
We check the metacategory axioms.
Theorem: Equivalence of Definitions of Projective Module

Let $A$ be a ring.
Let $M$ be an $A$-module.
The following are equivalent:
:$(1): \quad$ $M$ is a projective module, that is, $M$ is a projective object in the category of left $A$-modules.
:$(2): \quad$ $M$ is a direct summand of a free module.
:$(3): \quad$ Every short exact sequence of the form:
::<nowiki>$\xymatrix{
0 \ar[r] & X \ar[r]^f & Y \ar[r]^g & M \ar[r] & 0
}$</nowiki>
splits, that is, there is a homomorphism $s : M \to Y$ with $g \circ s = \operatorname {id}_M$.

Proof:
=== $(1)$ implies $(2)$ ===
By Surjection by Free Module there is a free module $Y$ and a surjection $g : Y \to M$.
By Epimorphism of modules iff surjective $g$ is an epimorphism.
By Definition:Projective Object applied to $\operatorname {id}_M$, there is a homomorphism $s : M \to Y$ with $g \circ s = \operatorname {id}_M$.
We have $Y = \map {\operatorname {im} } s \oplus \map \ker g$.
{{qed|lemma}}
Theorem: Tensor Product of Projective Modules is Projective

Let $A$ be a commutative ring with unity.
Let $P$ and $Q$ be projective $A$-modules.
Then the tensor product $P \otimes_A Q$ is a projective $A$-module.

Proof:
By Projective iff Direct Summand of Free Module, there exist $A$-modules $P'$ and $Q'$, such that $P \oplus P'$ and $Q \oplus Q'$ are free.
By Tensor Product Distributes over Direct Sum, there is an isomorphism:
:$\paren {P \oplus P'} \otimes_A \paren {Q \oplus Q'} \cong \paren {P \otimes_A Q} \oplus \paren {P' \otimes_A Q} \oplus \paren {P \otimes_A Q'} \oplus \paren {P' \otimes_A Q'}$
By Tensor Product of Free Modules is Free the {{LHS}} is free.
Hence $P \otimes_A Q$ is a direct summand of a free module.
By Projective iff Direct Summand of Free Module $P \otimes_A Q$ is projective.
{{qed}}
Category:Commutative Algebra
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Theorem: Surjection by Free Module

Let $A$ be a ring.
Let $M$ be a left $A$-module.
Then there exists a free $A$-module $F$ and a surjective $A$-module homomorphism $f : F \to M$.

Proof:
Let $F = A^{\paren M}$ be the free $A$-module on the set $M$.
Let $c : M \to A^{\paren M}$ be the canonical mapping on $F$.
Let $f : F \to M$ be the $A$-module homomorphism induced the by the Universal Property of Free Modules applied to the identity $\operatorname {id}_M$ of $M$.
We have:
:$f \circ c = \operatorname {id}_M$
Thus $f$ is a split epimorphism in the category of sets.
By Split Epimorphism is Epic and Surjection iff Epimorphism in Category of Sets $f$ is surjective.
{{qed}}
Category:Homological Algebra
p1kktwound7fu62agh44c1lv61m5c7d
Theorem: Category of Modules has Enough Projectives

Let $A$ be a ring.
Then the category of left $A$-modules has enough projectives.

Proof:
Let $M$ be an $A$-module.
By Surjection by Free Module there is a free $A$-module $F$ and a surjection $f : F \to M$.
By Epimorphism of Modules Iff Surjection $f$ is an epimorphism.
By Free Module is Projective $F$ is projective.
{{qed}}
Category:Homological Algebra
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