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MAT 236, Elementary Abstract Algebra
The Quaternions Example
In this example, we consider the subset
H =
α β
β α¯
: α, β ∈ C
as a subset of the 2 × 2 matrices with complex entries, M 2
(C). Here ¯α indicates the complex
conjugate of α Our goal is to prove H is a subring of M 2
(C).
Before we begin, we need to review some facts about the complex number and complex
conjugation. Recall, if α is a complex number, then we can write
α = a 1 + a 2 i for some a 1 , a 2 ∈ R and i
2
= − 1.
Then
α¯ = a 1 − a 2 i.
Then we have the following lemma about the properties of complex conjugation.
Lemma: Let α = a 1 + a 2 i and β = b 1 + b 2 i be two complex numbers. Then
- (α) = α
- α + β = ¯α +
β
- −α = −α¯
- αβ = ¯α
β
- α α¯ = a
2
1
2
2
. So α α¯ = 0 if and only if α = 0.
Proof. This is just a matter of doing the corresponding calculations, which we give below.
(1) Note (α) = a 1
− a 2
i = a 1
− (−a 2
)i = a 1
i = α.
(2) α + β = a 1 + a 2 i + b 1 + b 2 i = (a 1 + b 1 ) + (a 2 + b 2 )i = (a 1 + b 1 ) − (a 2 + b 2 )i = (a 1 − a 2 i) +
(b 1
− b 2
i) = ¯α +
β.
(3) −α = −(a 1 + a 2 i) = −a 1 − a 2 i = −a 1 + a 2 i = −(a 1 − a 2 i) = −α¯.
(4) αβ = (a 1
i)(b 1
i) = a 1
b 1
b 2
b 1
)i + a 2
b 2
i
2 = (a 1
b 1
− a 2
b 2
) + (a 1
b 2
b 1
)i =
(a 1
b 1
− a 2
b 2
) − (a 1
b 2
b 1
)i = (a 1
− a 2
i)(b 1
− b 2
i) = ¯α
β.
(5) α α¯ = (a 1
i)(a 1
− a 2
i) = a
2
1
− a
2
2
i
2 = a
2
1
2
2
Now we prove the following theorem:
Theorem: H is a subring of M 2
(C) that is a ring with identity such that for each 0 6 = A ∈ H
there exists a U ∈ H such that AU = I 2
= U A.
Proof. We use Theorem 3.2 to prove this. First, note that ¯0 = 0 and −0 = 0, so
∈ H.
Next, we want to show closure of addition. So let
α β
β α¯
γ δ
δ γ¯
∈ H. Then
α β
β α¯
γ δ
δ ¯γ
α + γ β + δ
β −
δ α¯ + ¯γ
α + γ β + δ
−(β + δ) α + γ
∈ H.
So H is closed under addition.
To show closure under multiplication, we note
α β
β α¯
γ δ
δ ¯γ
αγ + β(−
δ) αδ + βγ¯
βγ + ¯α(−
δ) −
βδ + ¯αγ¯
αγ − β
δ αδ + βγ¯
−(αδ + β¯γ) αγ − β
δ
∈ H.
So H is also closed under multiplication.
Finally, note that
α β
β α¯
−α −β
β −α
−α −β
−(−β) −α
∈ H
so H is closed under additive inverses.
Therefore, by Theorem 3.2, H is a subring of M 2 (C).
Note that I 2
∈ H, so I 2
becomes the identity element of H. Next let A =
α β
β α¯
∈ H
be nonzero. Then A is also a matrix in M 2
(C) and its determinant is det(A) = α α¯ − β(−
β) =
α α¯ + β
β. By our above Lemma, the only way det(A) can be zero is if both α and β are zero.
Therefore, as a matrix, if A is nonzero, it has an inverse given by
A
− 1
=
α α¯ + β
β
α¯ −β
β α
Note that A
− 1 ∈ H, so we let U = A
− 1 and the rest of the Theorem follows.
