<The previous article in this series | The table of contents of this series |
definition of \(n \times n\) symplectic group
Topics
About:
group
About:
matrices space
The table of contents of this article
Starting Context
Target Context
-
The reader will have a definition of \(n \times n\) symplectic group.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( \mathbb{H}\): \(= \text{ the quaternions division associative algebra }\)
\( \{M\}\): \(= \text{ the } \mathbb{H} \text{ matrices space }\)
\( n\): \(\in \mathbb{N} \setminus \{0\}\)
\(*Sp (n)\): \(= \{M \in \{M\} \vert M \in \{\text{ the } n \times n \text{ matrices such that } M^* = M^{- 1} \}\}\), where \({M^*}^j_l = \overline{M^l_j}\)
//
Conditions:
//
2: Note
Let us see that \(Sp (n)\) is indeed a group.
Let \(M_1, M_2, M_3 \in Sp (n)\) be any.
\((M_1 M_2)^* = {M_2}^* {M_1}^*\), because \({(M_1 M_2)^*}^j_l = \overline{(M_1 M_2)^l_j} = \overline{{M_1}^l_m {M_2}^m_j} = \overline{{M_2}^m_j} \overline{{M_1}^l_m}\), which is described in the definition of quaternions division associative algebra, \(= {{M_2}^*}^j_m {{M_1}^*}^m_l = ({M_2}^* {M_1}^*)^j_l\).
\({{M_1}^*}^* = M_1\), because \({{{M_1}^*}^*}^j_l = \overline{{{M_1}^*}^l_j} = \overline{\overline{{M_1}^j_l}} = {M_1}^j_l\).
\(M_1 M_2 \in Sp (n)\), because \((M_1 M_2)^* = {M_2}^* {M_1}^* = {M_2}^{- 1} {M_1}^{- 1}\), so, \((M_1 M_2)^* M_1 M_2 = {M_2}^{- 1} {M_1}^{- 1} M_1 M_2 = {M_2}^{- 1} I M_2 = {M_2}^{- 1} M_2 = I\) and \(M_1 M_2 (M_1 M_2)^* = M_1 M_2 {M_2}^{- 1} {M_1}^{- 1} = M_1 I {M_1}^{- 1} = M_1 {M_1}^{- 1} = I\), so, \((M_1 M_2)^* = (M_1 M_2)^{- 1}\).
1) \((M_1 \bullet M_2) \bullet M_3 = M_1 \bullet (M_2 \bullet M_3)\): by the proposition that for any ring, the multiplications of any matrices over the ring are associative.
2) \(i \in Sp (n)\) (called 'identity element') such that \(i \bullet M_1 = M_1 \bullet i = M_1\): the identity matrix, \(I\), is in \(Sp (n)\), because \(I^* = I = I^{-1}\), and \(I M_1 = M_1 I = M_1\).
3) \({M_1}^{- 1} \in Sp (n)\) (called 'inverse element of \(M_1\)') such that \({M_1}^{- 1} \bullet M_1 = M_1 \bullet {M_1}^{- 1} = I\): \({M_1}^* = {M_1}^{- 1} \in Sp (n)\), because \({{M_1}^*}^* = M_1\) and \({M_1}^* M_1 = I = M_1 {M_1}^*\), so, \({{M_1}^*}^{- 1} = M_1 = {{M_1}^*}^*\).
So, \(Sp (n)\) is a group.
References
<The previous article in this series | The table of contents of this series |
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for group, nonempty subset is subgroup iff it is closed under operation and inversion
Topics
About:
group
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any group, any nonempty subset is a subgroup if and only if it is closed under the operation and the inversion.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(G\): \(\in \{\text{ the groups }\}\)
\(S\): \(\subseteq G\) such that \(S \neq \emptyset\)
//
Statements:
\(S \in \{\text{ the subgroups of } G\}\)
\(\iff\)
\(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\)
//
2: Note
The point is that only \(s_1 s_2 \in S\) or only \({s_1}^{- 1} \in S\) is not enough for \(S\) to be a subgroup.
For example, let \(G = \mathbb{Z}\) as the additive group and \(S = \mathbb{N}\), then, \(S\) is closed under the operation, but \(S\) is not any subgroup, because the inverse of \(1\) is not contained in \(S\), for example; let \(G = \mathbb{Z}\) as the additive group and \(S = \{- 1, 1\}\), then, \(S\) is closed under the inversion, but \(S\) is not any subgroup, because \(1 + 1 \notin S\).
3: Proof
Whole Strategy: Step 1: suppose that \(S\) is a subgroup; Step 2: see that \(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\); Step 3: suppose that \(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\); Step 4: see that \(S\) is a subgroup.
Step 1:
Let us suppose that \(S\) is a subgroup.
Step 2:
\(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\) is obvious, because \(S\) is a group.
Step 3:
Let us suppose that \(\forall s_1, s_2 \in S (s_1 s_2 \in S \land {s_1}^{- 1} \in S)\).
Step 4:
Let us see that \(S\) satisfies the conditions to be a group.
As \(\forall s_1, s_2 \in S (s_1 s_2 \in S)\), the operation induced by the operation on \(G\) is well defined on \(S\).
Let \(s_1, s_2, s_3 \in S\) be any.
1) \((s_1 \bullet s_2) \bullet s_3 = s_1 \bullet (s_2 \bullet s_3)\): because it holds on the ambient \(G\).
2) \(i \in S\) (called 'identity element') such that \(i \bullet s_1 = s_1 \bullet i = s_1\): as \(S \neq \emptyset\), there is an \(s \in S\), but \(s^{- 1} \in S\) and \(s s^{- 1} = i \in S\), and \(i s_1 = s_1 i = s_1\), because it holds on the ambient \(G\).
3) \({s_1}^{- 1} \in S\) (called 'inverse element of \(s_1\)') such that \({s_1}^{- 1} \bullet s_1 = s_1 \bullet {s_1}^{- 1} = i\): \({s_1}^{- 1} \in S\), and \({s_1}^{- 1} s_1 = s_1 {s_1}^{- 1} = i\), because it holds on the ambient \(G\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for 'modules - linear morphisms' isomorphism, linearly independent subset or basis of domain is mapped to linearly independent subset or basis
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any 'modules - linear morphisms' isomorphism, any linearly independent subset or basis of the domain is mapped to a linearly independent subset or basis.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M_1\): \(\in \{\text{ the } R \text{ modules }\}\)
\(M_2\): \(\in \{\text{ the } R \text{ modules }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the 'modules - linear morphisms' isomorphisms }\}\)
//
Statements:
\(\forall S_1 \in \{\text{ the linearly independent subsets of } M_1\} (f (S_1) \in \{\text{ the linearly independent subsets of } M_2\})\)
\(\land\)
\(\forall B_1 \in \{\text{ the bases of } M_1\} (f (B_1) \in \{\text{ the bases of } M_2\})\)
//
2: Proof
Whole Strategy: Step 1: let \(S_1 = \{m_{1, j} \vert j \in J\}\) and take any finite \(J^` \subseteq J\), and see that \(\sum_{j^` \in J^`} r_{j^`} f (m_{1, j^`}) = 0\) implies that \(r_{j^`}\) s are \(0\); Step 2: let \(B_1 = \{{b_1}^j \vert j \in J\}\), and see that for each \(m_2 \in M_2\), there is a finite \(J^` \subseteq J\) such that \(m_2 = \sum_{j' \in J^`} r_{j'} f ({b_1}^{j'})\).
Step 1:
Let \(S_1\) be any linearly independent subset of \(M_1\).
\(S_1 = \{m_{1, j} \in M_1 \vert j \in J\}\) where \(J\) is a possibly uncountable index set.
Let \(J^` \subseteq J\) be any finite subset.
Let \(\sum_{j^` \in J^`} r_{j^`} f (m_{1, j^`}) = 0\).
\(\sum_{j^` \in J^`} r_{j^`} f (m_{1, j^`}) = f (\sum_{j^` \in J^`} r_{j^`} m_{1, j^`})\), because \(f\) is linear.
But \(\sum_{j^` \in J^`} r_{j^`} m_{1, j^`} = f^{- 1} \circ f (\sum_{j^` \in J^`} r_{j^`} m_{1, j^`}) = f^{- 1} (0) = 0\), because as \(f\) is a 'modules - linear morphisms' isomorphism, \(f^{- 1}\) is a linear morphism.
That implies that \(r_{j^`}\) s are \(0\), because \(S_1\) is linearly independent.
So, \(f (S_1)\) is linearly independent.
Step 2:
Let \(B_1 \subseteq M_1\) be any basis of \(M_1\).
\(B_1 = \{{b_1}^j \vert j \in J\}\) where \(J\) is a possibly uncountable index set.
While \(B_1\) is linearly independent, \(f (B_1)\) is linearly independent, by Step 1.
Let \(m_2 \in M_2\) be any.
There is a finite \(J^` \subseteq J\) such that \(f^{- 1} (m_2) = \sum_{j^` \in J^`} r_{j^`} {b_1}^{j^`}\), because \(B_1\) is a basis.
\(m_2 = f \circ f^{- 1} (m_2) = f (\sum_{j^` \in J^`} r_{j^`} {b_1}^{j^`}) = \sum_{j^` \in J^`} r_{j^`} f ({b_1}^{j^`})\), because \(f\) is linear, so, \(m_2\) is a linear combination of \(f (B_1)\).
So, \(f (B_1)\) is a basis of \(M_2\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module with finite basis, module is 'modules - linear morphisms' isomorphic to components module
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module with any finite basis, the module is 'modules - linear morphisms' isomorphic to the components module.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\), also as the canonical module mentioned in the proposition that any ring is canonically a module with a \(1\)-element basis
\(J\): \(\in \{\text{ the finite index sets }\}\), such that \(\vert J \vert = d\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\), with any finite basis, \(B = \{b^j \vert j \in J\}\)
\(R^d\): \(= \text{ the product module }\)
\(f\): \(: M \to R^d, m_{J_l} b^{J_l} \mapsto (m_{J_1}, ..., m_{J_d})\)
//
Statements:
\(f \in \{\text{ the 'modules - linear morphisms' isomorphisms }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) is valid; Step 2: see that \(f\) is linear; Step 3: see that \(f\) is a bijection; Step 4: conclude the proposition.
Step 1:
\(f\) is valid, because for each \(m \in M\), \((m_{J_1}, ..., m_{J_d})\) is uniquely determined, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique: when \(m_{J_l} b^{J_l}\) does not contain a basis element, the component is defined to be \(0\).
Step 2:
Let us see that \(f\) is linear.
Let \(m, m' \in M\) and \(r, r' \in R\) be any.
\(m = m_{J_l} b^{J_l}\) and \(m' = m'_{J_l} b^{J_l}\).
\(f (r m + r' m') = f (r m_{J_l} b^{J_l} + r' m'_{J_l} b^{J_l}) = f ((r m_{J_l} + r' m'_{J_l}) b^{J_l}) = (r m_{J_1} + r' m'_{J_1}, ..., r m_{J_d} + r' m'_{J_d}) = r (m_{J_1}, ..., m_{J_d}) + r' (m'_{J_1}, ..., m'_{J_d}) = r f (m) + r' f (m')\).
So, \(f\) is linear.
Step 3:
\(f\) is an injection, because for each \(m = m_{J_l} b^{J_l}, m' = m'_{J_l} b^{J_l} \in M\) such that \(m \neq m'\), \((m_{J_1}, ..., m_{J_d}) \neq (m'_{J_1}, ..., m'_{J_d})\), by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(f\) is a surjection, because for each \((r_{J_1}, ..., r_{J_d}) \in R^d\), \(r_{J_l} b^{J_l} \in M\) and \(f (r_{J_l} b^{J_l}) = (r_{J_1}, ..., r_{J_d})\).
So, \(f\) is a bijection.
Step 4:
By the proposition that any bijective linear map between any modules is a 'modules - linear morphisms' isomorphism, \(f\) is a 'modules - linear morphisms' isomorphism.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that module linearly independent subsets or bases do not necessarily have some properties of vectors space linearly independent subsets or bases
The table of contents of this article
Main Body
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, basis cannot be supplemented with element to keep linearly independent
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module, any basis cannot be supplemented with any element to keep linearly independent.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(J\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(B\): \(= \{b^j \vert j \in J\}\), \(\in \{\text{ the bases of } M\}\)
//
Statements:
\(\forall m \in M \setminus B (B \cup \{m\} \notin \{\text{ the linearly independent subsets of } M\})\)
//
2: Note
This property holds for any module basis: compare with the proposition that module linearly independent subsets or bases do not necessarily have some properties of vectors space linearly independent subsets or bases.
3: Proof
Whole Strategy: Step 1: see that \(m = \sum_{j^` \in J^`} m_{j^`} b^{j^`}\); Step 2: see that \(\sum_{j^` \in J^`} r_{j^`} b^{j^`} + r m = 0\) can be realized with some nonzero coefficients.
Step 1:
There is a finite subset, \(J^` \subseteq J\), such that \(m = \sum_{j^` \in J^`} m_{j^`} b^{j^`}\), where \(m_{j^`} \in R\).
So, \(0 = - m + m = - m + \sum_{j^` \in J^`} m_{j^`} b^{j^`}\).
Step 2:
Let us suppose that \(\sum_{j^` \in J^`} r_{j^`} b^{j^`} + r m = 0\), where \(r_{j^`} \in R\) and \(r \in R\).
Let us take \(r_{j^`} = m_{j^`}\) and \(r = - 1\).
\(\sum_{j^` \in J^`} m_{j^`} b^{j^`} + (- 1) m = \sum_{j^` \in J^`} m_{j^`} b^{j^`} + (- m)\), by the proposition that for any module, the inverse of each element is the element \(- 1\)-scalar multiplied, \(= 0\), by Step 1.
So, \(\sum_{j^` \in J^`} r_{j^`} b^{j^`} + r m = 0\) can be realized by some nonzero coefficients.
So, \(B \cup \{m\}\) is not linearly independent.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module over division ring, nonzero element nonzero-scalar multiplied is nonzero
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module over any division ring, each nonzero element each-nonzero-scalar multiplied is nonzero.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the division rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
\(\forall m \in M \setminus \{0\}, \forall r \in R \setminus \{0\} (r m \neq 0)\)
//
2: Note
Compare with the proposition that for a module, a nonzero element a-nonzero-scalar multiplied is not necessarily nonzero.
3: Proof
Whole Strategy: Step 1: suppose that \(r m = 0\), and find a contradiction.
Step 1:
Let \(m \in M \setminus \{0\}\) and \(r \in R \setminus \{0\}\) be any.
Let us suppose that \(r m = 0\).
As \(R\) is a division ring and \(r \neq 0\), there is \(r^{- 1} \in R\).
\(r^{- 1} (r m) = r^{- 1} 0 = 0\), by the proposition that for any module, \(0\) each-scalar multiplied is \(0\), but the left hand side is \((r^{- 1} r) m = 1 m = m\), so, \(m = 0\), a contradiction against that \(m \neq 0\).
So, \(r m \neq 0\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, nonzero element nonzero-scalar multiplied is not necessarily nonzero
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for a module, a nonzero element a-nonzero-scalar multiplied is not necessarily nonzero.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
not necessarily "\(\forall m \in M \setminus \{0\}, \forall r \in R \setminus \{0\} (r m \neq 0)\)"
//
2: Note
This is not because \(R\) is not commutative but because \(R\) is not any division ring: compare with the proposition that for any module over any division ring, each nonzero element each-nonzero-scalar multiplied is nonzero.
3: Proof
Whole Strategy: Step 1: see an example such that \(r m = 0\).
Step 1:
Let \(R = \mathbb{Z} / 6\), the integers modulo natural number ring, and \(M = R\), which is indeed a module, by the proposition that any ring is canonically a module with a \(1\)-element basis.
\(R = M = \{[0], ..., [5]\}\)
For \([3] \in M \setminus \{0\}\) and \([2] \in R \setminus \{0\}\), \([2] [3] = [6] = [0] = 0\).
So, for an \(m \in M \setminus \{0\}\) and an \(r \in R \setminus \{0\}\), \(r m \neq 0\) does not necessarily hold.
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, \(0\) scalar multiplied is \(0\)
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module, \(0\) each-scalar multiplied is \(0\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
\(\forall r \in R (r 0 = 0)\)
//
2: Proof
Whole Strategy: Step 1: see that \(r (0 + 0) = r 0 = r 0 + r 0\).
Step 1:
For each \(r \in R\), \(r (0 + 0) = r 0\), because \(0 + 0 = 0\), but the left hand side is \(r 0 + r 0\), so, \(r 0 + r 0 = r 0\).
So, \(- (r 0) + r 0 + r 0 = - (r 0) + r 0 = 0\), but the left hand side is \((- (r 0) + r 0) + r 0 = 0 + r 0 = r 0 + 0 = r 0\), so, \(r 0 = 0\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, inverse of element is element \(- 1\)-scalar multiplied
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module, the inverse of each element is the element \(- 1\)-scalar multiplied.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
\(\forall m \in M (- m = (- 1) m)\)
//
2: Proof
Whole Strategy: Step 1: see that \(m + (- 1) m = 0\).
Step 1:
\(- 1 \in R\) is the additive inverse of \(1\), the multiplicative identity of \(R\).
\(- m\) is the inverse of \(m\).
\(m + (- 1) m = 1 m + (- 1) m = (1 + (- 1)) m = 0 m = 0\), by the proposition that for any module, each element \(0\)-scalar multiplied is \(0\).
So, \((- 1) m\) is an inverse of \(m\), but in fact, it is the inverse of \(m\), by the proposition that for any module, each element has the unique inverse.
So, \((- 1) m = - m\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, each element has unique inverse
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module, each element has the unique inverse.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
\(\forall m \in M (\exists m', m'' \in M (m' + m = 0 \land m'' + m = 0) \implies m' = m'')\)
//
2: Proof
Whole Strategy: Step 1: see that \(m'' + m + m' = m'' = m'\).
Step 1:
\(m + m' = m' + m = 0\).
So, \(m'' + m + m' = m'' + 0 = m''\), but the left hand side is \((m'' + m) + m' = 0 + m' = m' + 0 = m'\).
So, \(m' = m''\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, element \(0\)-scalar multiplied is \(0\)
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module, each element \(0\)-scalar multiplied is \(0\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
//
Statements:
\(\forall m \in M (0 m = 0)\)
//
2: Proof
Whole Strategy: Step 1: see that \((1 + 0) m = m + 0 m = 1 m = m\).
Step 1:
\((1 + 0) m = 1 m + 0 m = m + 0 m\), but the left hand side is \(1 m\), because \(1 + 0 = 1\), \(= m\), so, \(m + 0 m = m\).
So, \((- m) + m + 0 m = (- m) + m = 0\), but the left hand side is \(((- m) + m) + 0 m = 0 + 0 m = 0 m\), so, \(0 m = 0\).
References
<The previous article in this series | The table of contents of this series | The next article in this series>
<The previous article in this series | The table of contents of this series | The next article in this series>
description/proof of that for module, inverse of nonzero element is nonzero element
Topics
About:
module
The table of contents of this article
Starting Context
Target Context
-
The reader will have a description and a proof of the proposition that for any module, the inverse of any nonzero element is a nonzero element.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\)
\(m\): \(\in M\)
//
Statements:
\(m \neq 0\)
\(\implies\)
\(- m \neq 0\)
//
2: Proof
Whole Strategy: Step 1: suppose that \(- m = 0\) and find a contradiction.
Step 1:
\(- m\) is nothing but the inverse element of \(m\).
Let us suppose that \(- m = 0\).
\(m + (- m) = 0\), but the left hand side would be \(m + 0 = m\), so, \(m = 0\), a contradiction against that \(m \neq 0\).
So, \(- m \neq 0\).
References
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