description/proof of that quaternions division associative algebra is 'rings - homomorphisms' isomorphic to subring of \(2 \times 2\) complex matrices ring
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of quaternions division associative algebra.
- 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 ring, the multiplications of any matrices over the ring are associative.
- The reader admits the proposition that for any ring, the multiplications of any matrices over the ring are distributive with respect to addition.
- The reader admits the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism.
Target Context
- The reader will have the proposition that the quaternions division associative algebra is 'rings - homomorphisms' isomorphic to a subring of the \(2 \times 2\) complex matrices ring.
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:
\(\mathbb{H}\): \(= \text{ the quaternions division associative algebra }\)
\(R\): \(= \{\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} \vert c_1, c_2 \in \mathbb{C}\}\) with the canonical addition and multiplication
\(f\): \(: \mathbb{H} \to R, a + b i + c j + d k \mapsto \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix}\)
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Statements:
\(f \in \{\text{ the 'rings - homomorphisms' isomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(R\) is a ring; Step 2: see that \(f\) is a ring homomorphism; Step 3: see that \(f\) is a bijection; Step 4: conclude the proposition.
Step 1:
Let us see that \(R\) satisfies the conditions to be a ring.
Let us see that \(R\) is an Abelian group under addition.
\(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} + \begin{pmatrix} c'_1 & c'_2 \\ \overline{- c'_2} & \overline{c'_1} \end{pmatrix} = \begin{pmatrix} c_1 + c'_1 & c_2 + c'_2 \\ \overline{- c_2} + \overline{- c'_2} & \overline{c_1} + \overline{c'_1} \end{pmatrix}\), but \(\overline{c_1 + c'_1} = \overline{c_1} + \overline{c'_1}\) and \(\overline{- (c_2 + c'_2)} = \overline{- c_2 - c'_2} = \overline{- c_2} + \overline{- c'_2}\), so, \(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} + \begin{pmatrix} c'_1 & c'_2 \\ \overline{- c'_2} & \overline{c'_1} \end{pmatrix} \in R\).
Additions are associative, because \(\mathbb{C}\) is associative in addition.
\(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in R\) is the identity element in addition.
For each \(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix}\), \(\begin{pmatrix} - c_1 & - c_2 \\ \overline{- (- c_2)} & \overline{- c_1} \end{pmatrix} \in R\) is the inverse element, because \(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} + \begin{pmatrix} - c_1 & - c_2 \\ \overline{- (- c_2)} & \overline{- c_1} \end{pmatrix} = \begin{pmatrix} c_1 - c_1 & c_2 - c_2 \\ \overline{- c_2} + \overline{- (- c_2)} & \overline{c_1} + \overline{- c_1} \end{pmatrix} = 0\) and \(\begin{pmatrix} - c_1 & - c_2 \\ \overline{- (- c_2)} & \overline{- c_1} \end{pmatrix} + \begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} = \begin{pmatrix} - c_1 + c_1 & - c_2 + c_2 \\ \overline{- (- c_2)} + \overline{- c_2} & \overline{- c_1} + \overline{c_1} \end{pmatrix} = 0\).
Addition is commutative, because \(\mathbb{C}\) is commutative in addition.
So, \(R\) is an Abelian group under addition.
Let us see that \(R\) is a monoid under multiplication.
\(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} \begin{pmatrix} c'_1 & c'_2 \\ \overline{- c'_2} & \overline{c'_1} \end{pmatrix} \in R\), because \(= \begin{pmatrix} c_1 c'_1 + c_2 \overline{- c'_2} & c_1 c'_2 + c_2 \overline{c'_1} \\ \overline{- c_2} c'_1 + \overline{c_1} \overline{- c'_2} & \overline{- c_2} c'_2 + \overline{c_1} \overline{c'_1} \end{pmatrix}\), but \(\overline{c_1 c'_1 + c_2 \overline{- c'_2}} = \overline{c_1 c'_1} + \overline{c_2 \overline{- c'_2}} = \overline{c_1} \overline{c'_1} + \overline{c_2} (- c'_2) = \overline{- c_2} c'_2 + \overline{c_1} \overline{c'_1}\) and \(\overline{- (c_1 c'_2 + c_2 \overline{c'_1})} = \overline{- c_1 c'_2 - c_2 \overline{c'_1}} = \overline{c_1} \overline{- c'_2} + \overline{- c_2} c'_1 = \overline{- c_2} c'_1 + \overline{c_1} \overline{- c'_2}\).
Multiplications are associative, because matrices multiplications are associative, by the proposition that for any ring, the multiplications of any matrices over the ring are associative.
\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \in R\) is the identity element in multiplication.
So, \(R\) is a monoid under multiplication.
Multiplication is distributive with respect to addition, by the proposition that for any ring, the multiplications of any matrices over the ring are distributive with respect to addition.
So, \(R\) is a ring.
Step 2
Let us see that \(f\) is indeed valid.
\(\overline{a + b i} = a - b i\) and \(\overline{- (c + d i)} = \overline{- c - d i} = - c + d i\), so, \(f\) is into \(R\).
So, \(f\) is valid.
\(f (0) = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} = 0\).
\(f (1) = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 1\).
\(f ((a + b i + c j + d k) + (a' + b' i + c' j + d' k)) = f (a + a' + (b + b') i + (c + c') j + (d + d') k) = \begin{pmatrix} (a + a') + (b + b') i & (c + c') + (d + d') i \\ - (c + c') + (d + d') i & (a + a') - (b + b') i \end{pmatrix} = \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix} + \begin{pmatrix} a' + b' i & c' + d' i \\ - c' + d' i & a' - b' i \end{pmatrix} = f (a + b i + c j + d k) + f (a' + b' i + c' j + d' k)\).
\(f ((a + b i + c j + d k) (a' + b' i + c' j + d' k)) = f ((a a' - b b' - c c' - d d') + (a b' + b a' + c d' - d c') i + (a c' - b d' + c a' + d b') j + (a d' + b c' - c b' + d a') k) = \begin{pmatrix} (a a' - b b' - c c' - d d') + (a b' + b a' + c d' - d c') i & (a c' - b d' + c a' + d b') + (a d' + b c' - c b' + d a') i \\ - (a c' - b d' + c a' + d b') + (a d' + b c' - c b' + d a') i & (a a' - b b' - c c' - d d') - (a b' + b a' + c d' - d c') i \end{pmatrix}\), while \(f (a + b i + c j + d k) f (a' + b' i + c' j + d' k) = \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix} \begin{pmatrix} a' + b' i & c' + d' i \\ - c' + d' i & a' - b' i \end{pmatrix} = \begin{pmatrix} (a + b i) (a' + b' i) + (c + d i) (- c' + d' i) & (a + b i) (c' + d' i) + (c + d i) (a' - b' i) \\ (- c + d i) (a' + b' i) + (a - b i) (- c' + d' i) & (- c + d i) (c' + d' i) + (a - b i) (a' - b' i) \end{pmatrix} = \begin{pmatrix} a a' - b b' - c c' - d d' + (a b' + b a' + c d' - d c') i & a c' - b d' + c a' + d b' + (a d' + b c' - c b' + d a') i \\ - c a' - d b' - a c' + b d' + (- c b' + d a' + a d' + b c') i & - c c' - d d' + a a' - b b' + (- c d' + d c' - a b' - b a') i \end{pmatrix} = f ((a + b i + c j + d k) (a' + b' i + c' j + d' k))\).
So, \(f\) is a ring homomorphism.
Step 3:
\(f\) is an injection, because for each \(a + b i + c j + d k \neq a' + b' i + c' j + d' k\), \(a \neq a'\), \(b \neq b'\), \(c \neq c'\), or \(d \neq d'\), and anyway, \(f (a + b i + c j + d k) = \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix} \neq \begin{pmatrix} a' + b' i & c' + d' i \\ - c' + d' i & a' - b' i \end{pmatrix} = f (a' + b' i + c' j + d' k)\).
\(f\) is a surjection, because for each \(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix}\), let \(c_1 = a + b i\) and \(c_2 = c + d i\), then, \(\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} = \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix} = f (a + b i + c j + d k)\).
So, \(f\) is a bijection.
Step 4:
\(f\) is a 'rings - homomorphisms' isomorphism, by the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism.
