description/proof of that for module with \(d\)-elements basis and general linear group of module, there is canonical 'groups - homomorphisms' isomorphism w.r.t. basis from general linear group onto group of invertible \(d \times d\) ring matrices
Topics
About: module
About: group
The table of contents of this article
Starting Context
- The reader knows a definition of general linear group of module.
- The reader knows a definition of basis of module.
- The reader knows a definition of %ring name% matrices space.
- The reader knows a definition of %category name% isomorphism.
- The reader admits the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix.
- The reader admits the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
- The reader admits the proposition that any map is a bijection if and only if it has an inverse.
- The reader admits the proposition that for any ring, the multiplications of any matrices over the ring are associative.
- The reader admits the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that for any module with any \(d\)-elements basis and the general linear group of the module, there is the canonical 'groups - homomorphisms' isomorphism with respect to the basis from the general linear group onto the group of the invertible \(d \times d\) ring matrices.
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:
\(R\): \(\in \{\text{ the rings }\}\)
\(M\): \(\in \{\text{ the } R \text{ modules }\}\) with any basis, \(B = \{b^1, ..., b^d\}\)
\(GL (M)\): \(= \text{ the general linear group of } M\)
\(M_d (R)\): \(= \text{ the ring of the } d \times d R \text{ matrices }\)
\(M_d (R)^{\times}\): \(= \text{ the group of the invertible } d \times d R \text{ matrices }\)
\(g\): \(: GL (M) \to M_d (R)^{\times}, f \mapsto A \text{ such that } f (b^j) = A^j_l b^l\)
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Statements:
\(g \in \{\text{ the 'groups - homomorphisms' isomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(M_d (R)^{\times}\) is a group; Step 2: see that \(g\) is a group homomorphism; Step 3: see that \(g\) is a bijection; Step 4: conclude the proposition.
Step 1:
Let us see that \(M_d (R)^{\times}\) is a group with respect to matrix multiplication.
Note that each element of \(M_d (R)^{\times}\) has an inverse in \(M_d (R)\), by the definition of \(M_d (R)^{\times}\), but the inverse is not presupposed to be in \(M_d (R)^{\times}\).
Let \(A_1, A_2, A_3 \in M_d (R)^{\times}\) be any.
We will hereafter use the fact that multiplications in \(M_d (R)\) are associative, which is true, by the proposition that for any ring, the multiplications of any matrices over the ring are associative.
\(A_1 A_2 \in M_d (R)^{\times}\)?
\(A_1\) has an \({A_1}^{- 1} \in M_d (R)\) and \(A_2\) has an \({A_2}^{- 1} \in M_d (R)\), by the definition of \(M_d (R)^{\times}\).
\({A_2}^{- 1} {A_1}^{- 1} A_1 A_2 = {A_2}^{- 1} ({A_1}^{- 1} A_1) A_2 = {A_2}^{- 1} I A_2 = {A_2}^{- 1} A_2 = I\).
\(A_1 A_2 {A_2}^{- 1} {A_1}^{- 1} = A_1 (A_2 {A_2}^{- 1}) {A_1}^{- 1} = A_1 I {A_1}^{- 1} = A_1 {A_1}^{- 1} = I\).
So, \(A_1 A_2\) is invertible and \(A_1 A_2 \in M_d (R)^{\times}\).
\({A_1}^{- 1} \in M_d (R)^{\times}\), because \({A_1}^{- 1} A_1 = I\) and \(A_1 {A_1}^{- 1} = I\) means that \({A_1}^{- 1}\) is invertible.
1) \((A_1 \bullet A_2) \bullet A_3 = A_1 \bullet (A_2 \bullet A_3)\): by the proposition that for any ring, the multiplications of any matrices over the ring are associative.
2) \(i \in M_d (R)^{\times}\) (called 'identity element') such that \(i \bullet A_1 = A_1 \bullet i = A_1\): the identity matrix, \(I\), is in \(M_d (R)^{\times}\), because \(I^{- 1} = I\), because \(I I = I\), and is an identity element, because \((I A_1)^j_m = I^j_l {A_1}^l_m = \delta^j_l {A_1}^l_m = {A_1}^j_m\) and \((A_1 I)^j_m = {A_1}^j_l I^l_m = {A_1}^j_l \delta^l_m = {A_1}^j_m\): the proposition that for any ring, any multiple of 0 is 0
3) \({A_1}^{- 1} \in M_d (R)^{\times}\) (called 'inverse element of \(A_1\)') such that \(A_1^{-1} \bullet A_1 = A_1 \bullet {A_1}^{-1} = i\): the existence of \({A_1}^{- 1} \in M_d (R)\) is by the definition of \(M_d (R)^{\times}\) and \({A_1}^{- 1} \in M_d (R)^{\times}\) has been seen above.
So, \(M_d (R)^{\times}\) is a group.
Step 2:
1st, let us think of \(g': GL (M) \to M_d (R), f \mapsto A \text{ such that } f (b^j) = A^j_l b^l\), the codomain extension of \(g\).
\(g'\) is well-defined, as it is the one mentioned in the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix.
Let \(f_1, f_2 \in GL (M)\) be any.
Let \(A_1 := g' (f_1)\), \(A_2 := g' (f_2)\), and \(A_{2 1} := g' (f_2 \circ f_1)\).
\(f_1 (b^j) = {A_1}^j_m b^m\), \(f_2 (b^m) = {A_2}^m_l b^l\), and \(f_2 \circ f_1 (b^j) = {A_{2 1}}^j_l b^l\).
\({A_{2 1}}^j_l b^l = f_2 \circ f_1 (b^j) = f_2 (f_1 (b^j)) = f_2 ({A_1}^j_m b^m) = {A_1}^j_m f_2 (b^m)\), because \(f_2\) is linear, \(= {A_1}^j_m {A_2}^m_l b^l\).
That implies that \({A_{2 1}}^j_l = {A_1}^j_m {A_2}^m_l = (A_2 A_1)^j_l\), so, \(A_{2 1} = A_2 A_1\).
So, \(g' (f_2 \circ f_1) = g' (f_2) g (f_1)\).
Note that this far, the bijective-nesses of \(f_1\) and \(f_2\) are not used, so even when the domain of \(g'\) is extended to the set of the module endomorphisms on \(M\), \(g' (f_2 \circ f_1) = g' (f_2) g (f_1)\) holds, which will be used later.
\({f_1}^{- 1} \circ {f_1} (b^j) = id_{M} (b^j) = b^j = I^j_l b^l\), so, \(g' ({f_1}^{- 1} \circ {f_1}) = I\), but \(g' ({f_1}^{- 1} \circ {f_1}) = g' ({f_1}^{- 1}) g' (f_1)\), so, \(g' ({f_1}^{- 1}) g' (f_1) = I\)..
\({f_1} \circ {f_1}^{- 1} (b^j) = id_{M} (b^j) = b^j = I^j_l b^l\), so, \(g' ({f_1} \circ {f_1}^{- 1}) = I\), but \(g' ({f_1} \circ {f_1}^{- 1}) = g' (f_1) g' ({f_1}^{- 1})\), so, \(g' (f_1) g' ({f_1}^{- 1}) = I\)..
So, \(g' (f_1)\) is invertible with the inverse, \(g' ({f_1}^{- 1})\).
So, \(g'\) is into \(M_d (R)^{\times}\).
So, \(g\) is valid.
\(g\) is a group homomorphism, by the proposition that any map between any groups that maps the product of any 2 elements to the product of the images of the elements is a group homomorphism.
Step 3:
\(g\) is an injection, because for each \(f_1, f_2 \in GL (M)\) such that \(f_1 \neq f_2\), \(g (f_1) \neq g (f_2)\), because if \(g (f_1) = g (f_2)\), \(f_1 = f_2\), because \(g (f_1)\) determines \(f_1\), by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix.
\(g\) is a surjection, because for each \(A \in M_d (R)^{\times}\), there is an \(A^{- 1} \in M_d (R)^{\times}\), and \(A\) and \(A^{- 1}\) determine the linear maps, \(f\) and \(f'\), by the proposition that for any map from any module with any \(d_1\)-elements basis into any module with any \(d_2\)-elements basis over any same ring, the map is linear if and only if the map is represented by the \(d_1 \times d_2\) ring matrix, and \(f' \circ f\) is represented by \(A^{- 1} A\) as before, \(= I\), so, \(f' \circ f = id_M\); \(f \circ f'\) is represented by \(A A^{- 1}\) as before, \(= I\), so, \(f \circ f' = id_M\), so, \(f\) has an inverse, \(f'\), so, \(f\) is a bijection, by the proposition that any map is a bijection if and only if it has an inverse, so, \(f \in GL (M)\), and \(g (f) = A\).
So, \(g\) is a bijection.
Step 4:
\(g\) is a 'groups - homomorphisms' isomorphism, by the proposition that any bijective group homomorphism is a 'groups - homomorphisms' isomorphism.
