hcju/proofwikiqa huggingface dataset:
It is shown that $\struct{G, +_G, \circ’}$ satisfies the right module axioms By definition of the opposite ring: :$\forall x, y \in R: x *_R y = y \times_R x$. === $(RM \, 1)$ : Scalar Multiplication (Right) Distributes over Module Addition ===  Let $\lambda \in R$ and $x, y \in G$. {{begin-eqn}} {{eqn | l = \paren{x +_G y} \circ’ \lambda        | r = \lambda \circ \paren{x +_G y}       | c = Definition of $\circ’$ }} {{eqn | r = \lambda \circ x +_G \lambda \circ  y       | c = Left module axiom $(M \, 1)$ on $\struct{G, +_G, \circ}$ }} {{eqn | r = x \circ’ \lambda +_G y \circ’ \lambda        | c = Definition of $\circ’$ }} {{end-eqn}} {{qed|lemma}} === $(RM \, 2)$ : Scalar Multiplication (Left) Distributes over Scalar Addition ===  Let $\lambda, \mu \in R$ and $x \in G$. {{begin-eqn}} {{eqn | l = x \circ’ \paren {\lambda +_S \mu}        | r = \paren {\lambda +_R \mu} \circ x       | c = Definition of $\circ’$ }} {{eqn | r = \lambda \circ x +_G \mu \circ  x       | c = Left module axiom $(M \, 2)$ on $\struct{G, +_G, \circ}$ }} {{eqn | r = x \circ’ \lambda +_G x \circ’ \mu        | c = Definition of $\circ’$ }} {{end-eqn}} {{qed|lemma}} === $(RM \, 3)$ : Associativity of Scalar Multiplication ===  Let $\lambda, \mu \in S$ and $x \in G$. {{begin-eqn}} {{eqn | l = x \circ’ \paren {\lambda *_R \mu}        | r = \paren {\lambda *_R \mu} \circ x       | c = Definition of $\circ’$ }} {{eqn | r = \paren {\mu \times_R \lambda} \circ x       | c = Definition of $*_R$ }} {{eqn | r = \mu \circ \paren {\lambda \circ  x}       | c = Left module axiom $(M \, 3)$ on $\struct{G, +_G, \circ}$ }} {{eqn | r = \mu \circ \paren{x \circ’ \lambda}        | c = Definition of $\circ’$ }} {{eqn | r = \paren{x \circ’ \lambda} \circ’ \mu        | c = Definition of $\circ’$ }} {{end-eqn}} {{qed}}
It is shown that $\struct{G, +_G, \circ’}$ satisfies the left module axioms By definition of the opposite ring: :$\forall x, y \in S: x *_R y = y \times_R x$. === $(M \, 1)$ : Scalar Multiplication (Left) Distributes over Module Addition ===  Let $\lambda \in R$ and $x, y \in G$. {{begin-eqn}} {{eqn | l = \lambda \circ’ \paren{x +_G y}        | r = \paren{x +_G y} \circ \lambda       | c = Definition of $\circ’$ }} {{eqn | r = x \circ \lambda +_G y \circ  \lambda       | c = Right module axiom $(RM \, 1)$ on $\struct{G, +_G, \circ}$ }} {{eqn | r = \lambda \circ’ x +_G \lambda \circ’ y        | c = Definition of $\circ’$ }} {{end-eqn}} {{qed|lemma}} === $(M \, 2)$ : Scalar Multiplication (Right) Distributes over Scalar Addition ===  Let $\lambda, \mu \in R$ and $x \in G$. {{begin-eqn}} {{eqn | l = \paren {\lambda +_R \mu} \circ’ x         | r = x \circ \paren {\lambda +_R \mu}       | c = Definition of $\circ’$ }} {{eqn | r = x \circ \lambda +_G x \circ  \mu       | c = Right module axiom $(RM \, 2)$ on $\struct{G, +_G, \circ}$ }} {{eqn | r = \lambda \circ’ x +_G \mu \circ’ x        | c = Definition of $\circ’$ }} {{end-eqn}} {{qed|lemma}} === $(M \, 3)$ : Associativity of Scalar Multiplication ===  Let $\lambda, \mu \in R$ and $x \in G$. {{begin-eqn}} {{eqn | l = \paren {\lambda *_R \mu} \circ’ x        | r = x \circ \paren {\lambda *_R \mu}        | c = Definition of $\circ’$ }} {{eqn | r = x \circ \paren {\mu \times_R \lambda}       | c = Definition of $*_R$ }} {{eqn | r = \paren {x \circ  \mu} \circ \lambda       | c = Right module axiom $(RM \, 3)$ on $\struct{G, +_G, \circ}$ }} {{eqn | r = \paren {\mu \circ’ x} \circ \lambda       | c = Definition of $\circ’$ }} {{eqn | r = \lambda \circ’ \paren {\mu \circ’ x}        | c = Definition of $\circ’$ }} {{end-eqn}} {{qed}}
By definition of the opposite ring: :$\forall x, y \in R: x * y = y \times x$ === Necessary Condition === Let $\struct {R, +, \times}$ be a commutative ring.  Then: {{begin-eqn}} {{eqn | lo= \forall x, y \in R:       | l = x \times y       | r = y \times x }} {{eqn | ll= \leadsto       | l = x \times y       | r = x * y       | c = {{Defof|Opposite Ring|opposite ring}} }} {{eqn | ll= \leadsto       | l = \struct {R, +, \times}       | r = \struct {R, +, *}       | c = Equality of Algebraic Structures }} {{end-eqn}} {{qed}} === Sufficient Condition === Let $\struct {R, +, \times} = \struct {R, +, *}$. Then: {{begin-eqn}} {{eqn | lo= \forall x, y \in R:       | l = x \times y       | r = x * y       | c = Equality of Algebraic Structures }} {{eqn | ll= \leadsto       | l = x \times y       | r = y \times x       | c = {{Defof|Opposite Ring|opposite ring}} }} {{end-eqn}} Thus by definition $\struct {R, +, \times}$ is a commutative ring. {{qed}}
:400px === Necessary Condition === Let $AL$, $BM$ and $CN$ are concurrent. Let the point of concurrency be $P$. Consider the triangles $\triangle ALB$ and $\triangle ALC$. They have the same altitude from the common base $BC$. Hence: :$\dfrac {\map \Area {ALB} } {\map \Area {ALC} } = \dfrac {BL} {LC}$ Similarly, consider the triangles $\triangle PLB$ and $\triangle PLC$. They also have the same altitude from the common base $BC$. Hence: :$\dfrac {\map \Area {PLB} } {\map \Area {PLC} } = \dfrac {BL} {LC}$ Next we consider the triangles $\triangle APB$ and $\triangle APC$. We have: :$\map \Area {APB} = \map \Area {ALB} - \map \Area {PLB}$ and: :$\map \Area {APC} = \map \Area {ALC} - \map \Area {PLC}$ and so: :$\dfrac {\map \Area {APB} } {\map \Area {APC} } = \dfrac {BL} {LC}$ In the same way, we derive: :$\dfrac {\map \Area {BPC} } {\map \Area {APB} } = \dfrac {CM} {MA}$ and: :$\dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {AN} {NB}$ Thus we have: :$\dfrac {\map \Area {APB} } {\map \Area {APC} } \times \dfrac {\map \Area {BPC} } {\map \Area {APB} } \times \dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB}$ The areas on the {{LHS}} cancel out, leaving us with: :$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$ {{qed|lemma}} === Sufficient Condition === Let: :$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$ Let $P$ be the intersection of $AM$ and $CN$ as in the diagram above. We need to show that $A$, $P$ and $L$ are collinear. From the first part, we have that: :$\dfrac {\map \Area {BPC} } {\map \Area {APB} } = \dfrac {CM} {MA}$ and: :$\dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {AN} {NB}$ Multiplying them: :$\dfrac {\map \Area {BPC} } {\map \Area {APB} } \times \dfrac {\map \Area {APC} } {\map \Area {BPC} } = \dfrac {CM} {MA} \times \dfrac {AN} {NB}$ Simplifying: :$\dfrac {\map \Area {APC} } {\map \Area {APB} } = \dfrac {CM} {MA} \times \dfrac {AN} {NB}$ It is given that: :$\dfrac {BL} {LC} \times \dfrac {CM} {MA} \times \dfrac {AN} {NB} = 1$ and so: :$\dfrac {CM} {MA} \times \dfrac {AN} {NB} = \dfrac {LC} {BL} = \dfrac {\map \Area {APC} } {\map \Area {APB} }$ Extend $BP$ to meet $AC$ at point $Z$, say. By the same construction that we have used throughout, we have: :$\dfrac {\map \Area {APC} } {\map \Area {APB} } = \dfrac {ZC} {BZ}$ But then we have just shown that: :$\dfrac {LC} {BL} = \dfrac {\map \Area {APC} } {\map \Area {APB} }$ So $Z$ coincides with $L$ and the result follows. {{qed}}
=== Necessary Condition === Let $\struct {G, +_G, \circ'}$ be a right module over $\struct {R, +_R, \times_R}$. Then: {{begin-eqn}} {{eqn | l = \paren {\lambda \times_R \mu} \circ x       | r = x \circ' \paren {\lambda \times_R \mu}       | c = Definition of $\circ'$ }} {{eqn | r = \paren {x \circ' \lambda} \circ' \mu       | c = Right module axiom $(\text {RM} 3)$: Associativity of Scalar Multiplication }} {{eqn | r = \mu \circ \paren {\lambda \circ x}        | c = Definition of $\circ'$ }} {{eqn | r = \paren {\mu \times_R \lambda} \circ x       | c = Left module axiom $(\text M 3)$: Associativity of Scalar Multiplication }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let the scalar multiplication $\circ$ satisfy: :$\forall \lambda, \mu \in R: \forall x \in G: \paren {\lambda \times_R \mu} \circ x = \paren {\mu \times_R \lambda} \circ x$ It needs to be shown that $\struct {G, +_G, \circ'}$ satisfies the right module axioms. ==== $(\text {RM} 1)$ : Scalar Multiplication (Right) Distributes over Module Addition ==== Let $\lambda, \mu \in R, x \in G$. Then: {{begin-eqn}} {{eqn | l = \paren {x +_G y} \circ' \lambda       | r = \lambda \circ \paren {x +_G y}        | c = Definition of $\circ'$ }} {{eqn | r = \lambda \circ x +_G \lambda \circ y       | c = Left module axiom $(\text M 1)$ on $\circ$ }} {{eqn | r = x \circ' \lambda +_G y \circ' \lambda       | c = Definition of $\circ'$ }} {{end-eqn}} {{qed|lemma}} ==== $(\text {RM} 2)$ : Scalar Multiplication (Left) Distributes over Scalar Addition ==== Let $\lambda \in R, x, y \in G$. Then: {{begin-eqn}} {{eqn | l = x \circ' \paren{\lambda +_R \mu}       | r = \paren {\lambda +_R \mu} \circ x       | c = Definition of $\circ'$ }} {{eqn | r = \lambda \circ x +_G \mu \circ y       | c = Left module axiom $(\text M 2)$ on $\circ$ }} {{eqn | r = x \circ' \lambda +_G x \circ' \mu       | c = Definition of $\circ'$ }} {{end-eqn}} {{qed|lemma}} ==== $(\text {RM} 3)$ : Associativity of Scalar Multiplication ==== Let $\lambda, \mu \in R, x \in G$. Then: {{begin-eqn}} {{eqn | l = x \circ' \paren {\lambda \times_R \mu}       | r = \paren {\lambda \times_R \mu} \circ x       | c = Definition of $\circ'$ }} {{eqn | r = \paren {\mu \times_R \lambda} \circ x       | c = Assumption }} {{eqn | r = \mu \circ \paren {\lambda \circ x}       | c = Left module axiom $(\text M 3)$ on $\circ$ }} {{eqn | r = \paren {x \circ' \lambda} \circ' \mu       | c = Definition of $\circ'$ }} {{end-eqn}} {{qed}}
=== Necessary Condition === Let $\struct {G, +_G, \circ'}$ be a left module over $\struct {R, +_R, \times_R}$. Then: {{begin-eqn}} {{eqn | l = x \circ \paren {\lambda \times_R \mu}       | r = \paren {\lambda \times_R \mu} \circ' x       | c = Definition of $\circ'$ }} {{eqn | r = \lambda \circ' \paren {\mu \circ' x}       | c = Left module axiom $(\text M 3)$: Associativity of Scalar Multiplication }} {{eqn | r = \paren {x \circ \mu} \circ \lambda       | c = Definition of $\circ'$ }} {{eqn | r = x \circ \paren {\mu \times_R \lambda}       | c = Right module axiom $(\text {RM} 3)$: Associativity of Scalar Multiplication }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let the scalar multiplication $\circ$ satisfy: :$\forall \lambda, \mu \in R: \forall x \in G: x \circ \paren {\lambda \times_R \mu} = x \circ \paren {\mu \times_R \lambda}$  It needs to be shown that $\struct {G, +_G, \circ'}$ satisfies the left module axioms. ==== $(\text M 1)$ : Scalar Multiplication (Left) Distributes over Module Addition ==== Let $\lambda, \mu \in R, x \in G$. Then: {{begin-eqn}} {{eqn | l = \lambda \circ' \paren {x +_G y}       | r = \paren {x +_G y}  \circ \lambda        | c = Definition of $\circ'$ }} {{eqn | r = x \circ \lambda +_G y \circ \lambda       | c = Right module axiom $(\text {RM} 1)$ on $\circ$ }} {{eqn | r = \lambda \circ' x +_G \lambda \circ' y       | c = Definition of $\circ'$ }} {{end-eqn}} {{qed|lemma}} ==== $(\text M 2)$ : Scalar Multiplication (Right) Distributes over Scalar Addition ==== Let $\lambda \in R, x, y \in G$. Then: {{begin-eqn}} {{eqn | l = \paren{\lambda +_R \mu} \circ' x       | r = x \circ \paren{\lambda +_R \mu}       | c = Definition of $\circ'$ }} {{eqn | r = x \circ \lambda +_G x \circ \mu       | c = Right module axiom $(\text {RM} 2)$ on $\circ$ }} {{eqn | r = \lambda \circ' x +_G \mu \circ' x       | c = Definition of $\circ'$ }} {{end-eqn}} {{qed|lemma}} ==== $(\text M 3)$ : Associativity of Scalar Multiplication ==== Let $\lambda, \mu \in R, x \in G$. Then: {{begin-eqn}} {{eqn | l = \paren{\lambda \times_R \mu} \circ' x       | r = x \circ \paren{\lambda \times_R \mu}       | c = Definition of $\circ'$ }} {{eqn | r = x \circ \paren{\mu \times_R \lambda}       | c = Assumption }} {{eqn | r = \paren{x \circ \mu} \circ \lambda       | c = Right module axiom $(\text {RM} 3)$ on $\circ$ }} {{eqn | r = \lambda \circ' \paren {\mu \circ' x}       | c = Definition of $\circ'$ }} {{end-eqn}} {{qed}}
The sum of the distances will be a minimum when the lines $PA$, $PB$ and $PC$ all meet at an angle of $120 \degrees$. This is a consequence of the '''Fermat problem''' being a special case of the Steiner Tree Problem. Consider the circles which circumscribe the $3$ equilateral triangles $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$. Consider quadrilaterals formed by $\triangle ABG$, $\triangle BCE$ and $\triangle ACF$ along with another point on each of those circumscribing circles. Because these are cyclic quadrilaterals, the angle formed with these new points is $120 \degrees$. It follows that $\Box APBG$, $\Box BPCE$ and $\Box APCF$ are those cyclic quadrilaterals. Hence $\angle APC = \angle APB = \angle BPC = 120 \degrees$ and the result follows. {{qed}}
From Ring is Commutative iff Opposite Ring is Itself, $\struct {R, +_R, \times_R}$ is its own opposite ring. From Left Module over Ring Induces Right Module over Opposite Ring, $\struct{G, +_G, \circ’}$ is a right module over $\struct {R, +_R, \times_R}$. {{qed}}
From Ring is Commutative iff Opposite Ring is Itself, $\struct {R, +_R, \times_R}$ is its own opposite ring. From Right Module over Ring Induces Left Module over Opposite Ring, $\struct{G, +_G, \circ'}$ is a left module over $\struct {R, +_R, \times_R}$. {{qed}}
From Ring is Ideal of Itself, $R$ is a right ideal. From Right Ideal is Right Module over Ring, $\struct {R, +, \times}$ is a right module over $\struct {R, +, \times}$. {{qed}}
From Ring is Ideal of Itself, $R$ is a left ideal. From Left Ideal is Left Module over Ring, $\struct {R, +, \times}$ is a left module over $\struct {R, +, \times}$. {{qed}}
{{begin-eqn}} {{eqn | l = \int_0^{+\infty} \map f x \rd x       | r = \int_{x \mathop \to 0}^{x \mathop \to +\infty} \map f x \rd x       | c = {{Defof|Improper Integral}} }} {{eqn | r = \int_{\frac 1 x \mathop \to 0}^{\frac 1 x \mathop \to +\infty} \map f {\frac 1 x} \map \rd {\frac 1 x}       | c = Integration by Substitution }} {{eqn | r = \int_{\frac 1 x \mathop \to 0}^{\frac 1 x \mathop \to +\infty} \map f {\frac 1 x} \paren {-\dfrac 1 {x^2} } \rd x       | c = Power Rule for Derivatives }} {{eqn | r = -\int_{x \mathop \to +\infty}^{x \mathop \to 0} \dfrac {\map f {\frac 1 x} } {x^2} \rd x       | c = simplifying }} {{eqn | r = \int_{x \mathop \to 0}^{x \mathop \to +\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x       | c = Reversal of Limits of Definite Integral }} {{eqn | r = \int_0^{+\infty} \dfrac {\map f {\frac 1 x} } {x^2} \rd x       | c = {{Defof|Improper Integral}} }} {{end-eqn}} {{qed}} Category:Definite Integrals 432bl4lpdc26p0270ebe8yqlcnkbvod
Proof by Counterexample: Let $\struct {S, +, \times}$ be a ring with unity Let $\struct {\map {\mathcal M_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$. From Ring of Square Matrices over Ring is Ring, $\struct {\map {\mathcal M_S} 2, +, \times}$ is a ring. Let $G = \set {\begin{bmatrix} x & 0 \\ y & 0 \end{bmatrix} : x, y \in S }$ === Lemma === {{:Left Module Does Not Necessarily Induce Right Module over Ring/Lemma}}{{qed|lemma}} From Left Ideal is Left Module over Ring, $\struct {G, +, \times}$ is a left module over $\struct {\map {\mathcal M_S} 2, +, \times}$. Let $R = \map {\mathcal M_S} 2$ Let $\mathbf A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},  \mathbf B =  \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix},  \mathbf C =  \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}$ Then: {{begin-eqn}} {{eqn | l = \paren {\mathbf A \times \mathbf B} \times \mathbf C       | r = \paren { \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} } \times \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \paren {\mathbf B \times \mathbf A} \times \mathbf C       | r = \paren { \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} } \times \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} }} {{end-eqn}} Hence for $\mathbf A, \mathbf B \in R,  \mathbf C \in G$: :$ \paren {\mathbf A \times \mathbf B} \times \mathbf C \neq \paren {\mathbf B \times \mathbf A} \times \mathbf C$ Let $\circ : G \times R \to R$ be the binary operation defined by: :$\forall \mathbf \Lambda \in R: \forall \mathbf X \in G: \mathbf X \circ \mathbf \Lambda = \mathbf \Lambda \times \mathbf X$ From Left Module induces Right Module over same Ring iff Actions are Commutative, $\struct {G, +, \circ}$ is not a right module over $\struct {R, +, \times}$ {{qed}}
Proof by Counterexample: Let $\struct {S, +, \times}$ be a ring with unity Let $\struct {\map {\mathcal M_S} 2, +, \times}$ denote the ring of square matrices of order $2$ over $S$. From Ring of Square Matrices over Ring is Ring, $\struct {\map {\mathcal M_S} 2, +, \times}$ is a ring. Let $G = \set {\begin{bmatrix} x & y \\ 0 & 0 \end{bmatrix} : x, y \in S }$ === Lemma === {{:Right Module Does Not Necessarily Induce Left Module over Ring/Lemma}}{{qed|lemma}} From Right Ideal is Right Module over Ring, $\struct {G, +, \times}$ is a right module over $\struct {\map {\mathcal M_S} 2, +, \times}$. Let $R = \map {\mathcal M_S} 2$. Let $\mathbf A = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix},  \mathbf B =  \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix},  \mathbf C =  \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$ Then: {{begin-eqn}} {{eqn| l = \mathbf A \times \paren {\mathbf B \times \mathbf C}      | r = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \paren {\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} } }} {{eqn | r = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} }} {{end-eqn}} And: {{begin-eqn}} {{eqn| l = \mathbf A \times \paren {\mathbf C \times \mathbf B}      | r = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \paren {\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} } }} {{eqn | r = \begin{bmatrix} 1 & 1 \\ 0 & 0 \end{bmatrix} \times \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} }} {{eqn | r = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} }} {{end-eqn}} Hence for $\mathbf A, \mathbf B \in R,  \mathbf C \in G$: :$ \mathbf A \times \paren{\mathbf B \times \mathbf C} \neq \mathbf A \times \paren {\mathbf C \times \mathbf B}$ Let $\circ : G \times R \to R$ be the binary operation defined by: :$\forall \mathbf \Lambda \in R: \forall \mathbf X \in G: \mathbf \Lambda \circ \mathbf X = \mathbf X \times \mathbf \Lambda$ From Right Module induces Left Module over same Ring iff Actions are Commutative, $\struct {G, +, \circ}$ is not a left module over $\struct {R, +, \times}$ {{qed}}
From Test for Left Ideal, the following need to be proved: :$(1): \quad G \ne \O$ :$(2): \quad \forall \mathop {\mathbf X}, \mathop {\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$ :$(3): \quad \forall \mathop{\mathbf J} \in G, \mathop {\mathbf R} \in \map {\MM_S} 2: \mathbf R \times \mathbf J \in G$ === Condition $(1): \quad G \ne \O$ === By definition of $G$: :$\quad \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \in G$ {{qed|lemma}} === Condition $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$ === Let: :$\quad \mathbf X = \begin{bmatrix} x_1 & 0 \\ x_2 & 0 \end{bmatrix}, \quad \mathbf Y  = \begin{bmatrix} y_1 & 0 \\ y_2 & 0 \end{bmatrix} \in G$ Then: :$\quad \mathbf X - \mathbf Y = \begin{bmatrix} x_1 - y_1 & 0 \\ x_2 - y_2 & 0 \end{bmatrix} \in G$ {{qed|lemma}} === Condition $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf R \times \mathbf J \in G$ === Let: :$\quad \mathbf J = \begin{bmatrix} j_1 & 0 \\ j_2 & 0 \end{bmatrix} \in G, \quad \mathbf R = \begin{bmatrix} r_{1 1} & r_{2 1} \\ r_{1 2} & r_{2 2} \end{bmatrix} \in \map {\MM_S} 2$ Then: :$\quad \mathbf R \times \mathbf J = \begin{bmatrix} r_{1 1} \times j_1 + r_{2 1} \times j_2 & 0 \\ r_{1 2} \times j_1 + r_{2 2} \times j_2 & 0 \end{bmatrix} \in G$ {{qed}} Category:Left Module Does Not Necessarily Induce Right Module over Ring sm252o3vjhqgi1ob3en3cqc03cxnxv6
From Test for Right Ideal, the following need to be proved: : $(1): \quad G \ne \O$ : $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$ : $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$ === Condition $(1): \quad G \ne \O$ === By definition of $G$: :$\quad \begin {bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \in G$ {{qed|lemma}} === Condition $(2): \quad \forall \mathop {\mathbf X}, \mathop{\mathbf Y} \in G: \mathbf X + \paren {-\mathbf Y} \in G$ === Let: :$\quad \mathbf X = \begin{bmatrix} x_1 & x_2 \\ 0 & 0 \end{bmatrix}, \quad \mathbf Y  = \begin{bmatrix} y_1 & y_2 \\ 0 & 0 \end{bmatrix} \in G$ Then: :$\quad \mathbf X - \mathbf Y = \begin {bmatrix} x_1 - y_1 & x_2 - y_2 \\ 0 & 0 \end{bmatrix} \in G$ {{qed|lemma}} === Condition $(3): \quad \forall \mathop{\mathbf J} \in G, \mathop{\mathbf R} \in \map {\MM_S} 2: \mathbf J \times \mathbf R \in G$ === Let: :$\quad \mathbf J = \begin{bmatrix} j_1 & j_2 \\ 0 & 0 \end{bmatrix} \in G, \quad \mathbf R = \begin{bmatrix} r_{1 1} & r_{2 1} \\ r_{1 2} & r_{2 2} \end{bmatrix} \in \map {\MM_S} 2$ Then: :$\quad \mathbf J \times \mathbf R = \begin{bmatrix} j_1 \times r_{1 1} + j_2 \times r_{1 2} & j_1 \times r_{2 1} + j_2 \times r_{2 2} \\ 0 & 0 \end{bmatrix} \in G$ {{qed}} Category:Right Module Does Not Necessarily Induce Left Module over Ring 6rd5qz1edmk5oex5kjo6387gtlyxct0
By definition of an ideal, $J$ is both a left ideal and a right ideal. From Left Ideal is Left Module over Ring then $\struct {J, +, \circ_l}$ is a left module. From Right Ideal is Right Module over Ring then $\struct {J, +, \circ_r}$ is a right module. Then: {{begin-eqn}} {{eqn | lo= \forall x, y \in R \land \forall j \in J:       | l = \paren {x \circ_l j} \circ_r y       | r = \paren {x \times j} \times y       | c = Definition of $\circ_l$ and $\circ_r$ }} {{eqn | r = x \times \paren c{j \times y}       | c = Ring Axiom $(\text M 2)$: Associativity of Product }} {{eqn | r = x \circ_l \paren {j \circ_r y}       | c = Definition of $\circ_l$ and $\circ_r$ }} {{end-eqn}} Hence $\struct {J, +, \circ_l, \circ_r}$ is a bimodule over $\struct {R, +, \times}$ by definition. {{qed}}
From Ring is Ideal of Itself and Ideal is Bimodule over Ring, $\struct {R, +, \times, \times}$ is a bimodule over $\struct {R, +, \times}$. {{qed}}
From Left Module over Commutative Ring induces Right Module, $\struct{G, +_G, \circ’}$ is a right module. Let $\lambda, \mu \in R$ and $x \in G$. Then: {{begin-eqn}} {{eqn | l = \lambda \circ \paren{x \circ’ \mu}       | r = \lambda \circ \paren{\mu \circ x}       | c = Definition of $\circ’$ }} {{eqn | r = \paren {\lambda \circ \mu} \circ x       | c = Left module axiom $(M \,3)$ : Associativity of Scalar Multiplication }} {{eqn | r = \paren {\mu \circ \lambda} \circ x       | c = Ring product $\circ$ is commutative }} {{eqn | r = \mu \circ \paren{\lambda \circ x}       | c = Left module axiom $(M \,3)$ : Associativity of Scalar Multiplication }} {{eqn | r = \paren{\lambda \circ x} \circ’ \mu       | c = Definition of $\circ’$ }} {{end-eqn}} Hence $\struct{G, +_G, \circ, \circ’}$ is a bimodule over $\struct {R, +_R, \times_R}$ by definition. {{qed}}