Problems# It is 6 6 6 i ∈ 0 , 2 , 4 , 6 , 8 i \in \\{ 0,2,4,6,8 \\} i ∈ 0 , 2 , 4 , 6 , 8 6 i m o d 10 = i 6i \mod 10 = i 6 i mod 10 = i
It suffices to find a ring with a subgroup which in turn is not closed under multiplication. Particularly the ring of rationals Q \mathbb{Q} Q S = { x 2 ∣ x ∈ Z } = { x r 2 ∣ x ∈ Z , y = 0 , 1 } S = \{ \dfrac{x}{2} \mid x \in \mathbb{Z} \} = \{ x\ \dfrac{r}{2} \mid x \in \mathbb{Z}, y = 0,1 \} S = { 2 x ∣ x ∈ Z } = { x 2 r ∣ x ∈ Z , y = 0 , 1 } x 0 2 + x 1 2 = x 0 + x 1 2 \dfrac{x_0}{2} + \dfrac{x_1}{2} = \dfrac{x_0 + x_1}{2} 2 x 0 + 2 x 1 = 2 x 0 + x 1 x 0 + x 1 ∈ Z x_0 + x_1 \in \mathbb{Z} x 0 + x 1 ∈ Z x 0 2 \dfrac{x_0}{2} 2 x 0 − x 0 2 \dfrac{-x_0}{2} 2 − x 0 x 0 2 + − x 0 2 = 0 \dfrac{x_0}{2} + \dfrac{-x_0}{2} = 0 2 x 0 + 2 − x 0 = 0 1 2 ⋅ 1 2 = 1 4 ∉ S \dfrac{1}{2} \cdot \dfrac{1}{2} = \dfrac{1}{4} \notin S 2 1 ⋅ 2 1 = 4 1 ∈ / S S S S
P.S.
Any subgroup under addition of a ring, satisfies the ring’s definition, except for being closed under multiplication. Any set S S S a b = a + a + ⋯ + a ⏟ b times ∈ S ab = \underbrace{a + a + \dots + a}_{b \text{ times}} \in S ab = b times a + a + ⋯ + a ∈ S Unity’s uniqueness. Let 1 1 1 1 ′ 1' 1 ′ 1 1 ′ = 1 ′ 1 = 1 ′ 11' = 1'1 = 1' 1 1 ′ = 1 ′ 1 = 1 ′ 1 ′ 1 = 1 1 ′ = 1 1'1 = 11' = 1 1 ′ 1 = 1 1 ′ = 1 1 = 1 ′ 1 = 1' 1 = 1 ′
Multiplicative inverse uniqueness. Fix a 0 a_0 a 0 b 0 b_0 b 0 b 1 b_1 b 1 a 0 a_0 a 0 b 0 a 0 = a 0 b 0 = 1 b_0a_0 = a_0b_0 = 1 b 0 a 0 = a 0 b 0 = 1 b 1 a 0 = a 0 b 1 = 1 b_1a_0 = a_0b_1 = 1 b 1 a 0 = a 0 b 1 = 1
a 0 b 0 = a 0 b 1 b 0 ( a 0 b 0 ) = b 0 ( a 0 b 1 ) ( b 0 a 0 ) b 0 = ( b 0 a 0 ) b 1 b 0 = b 1
\begin{aligned}
a_0b_0 &= a_0b_1 \\
b_0(a_0b_0) &= b_0(a_0b_1) \\
(b_0a_0)b_0 &= (b_0a_0)b_1 \\
b_0 &= b_1
\end{aligned}
a 0 b 0 b 0 ( a 0 b 0 ) ( b 0 a 0 ) b 0 b 0 = a 0 b 1 = b 0 ( a 0 b 1 ) = ( b 0 a 0 ) b 1 = b 1 a. For Z 6 Z_6 Z 6 3 2 = 3 3^2 = 3 3 2 = 3 3 ≠ 0 3 \neq 0 3 = 0 3 ≠ 1 3 \neq 1 3 = 1
b. For Z 4 Z_4 Z 4 3 ⋅ 3 = 0 3 \cdot 3 = 0 3 ⋅ 3 = 0 3 ≠ 0 3 \neq 0 3 = 0
c. For Z 4 Z_4 Z 4 2 ⋅ 1 = 2 = 2 ⋅ 3 2 \cdot 1 = 2 = 2 \cdot 3 2 ⋅ 1 = 2 = 2 ⋅ 3 2 ≠ 0 2 \neq 0 2 = 0 1 ≠ 3 1 \neq 3 1 = 3
( → ) (\rightarrow) ( → ) k k k
b k = c b k ⋅ 1 = b k ⋅ a a − 1 = a b ⋅ k a − 1 =
\begin{aligned}
bk &= c \\
bk \cdot 1 &= \\
bk \cdot aa^{-1} &= \\
ab \cdot ka^{-1} &=
\end{aligned}
bk bk ⋅ 1 bk ⋅ a a − 1 ab ⋅ k a − 1 = c = = = ( ← ) (\leftarrow) ( ← ) k k k
a b ⋅ k = c a ⋅ b k =
\begin{aligned}
ab \cdot k &= c \\
a \cdot bk &=
\end{aligned}
ab ⋅ k a ⋅ bk = c = Consider arbitrary a r 0 a , a r 1 a ∈ S ar_0a, ar_1a \in S a r 0 a , a r 1 a ∈ S
a r 0 a ⋅ a r 1 a = a r 0 a 2 r 1 a = a r 0 r 1 a ∈ S
\begin{aligned}
ar_0a \cdot ar_1a \\
&= ar_0a^2r_1a \\
&= ar_0r_1a \in S
\end{aligned}
a r 0 a ⋅ a r 1 a = a r 0 a 2 r 1 a = a r 0 r 1 a ∈ S As r 0 r 1 ∈ R r_0r_1 \in R r 0 r 1 ∈ R
a r 0 a − a r 1 a = a [ r 0 a − r 1 a ] = a [ ( r 0 − r 1 ) a ] = a ( r 0 − r 1 ) a ∈ S
\begin{aligned}
ar_0a - ar_1a \\
&= a [r_0a - r_1a] \\
&= a \left [ (r_0 - r_1) a \right ] \\
&= a(r_0 - r_1)a \in S
\end{aligned}
a r 0 a − a r 1 a = a [ r 0 a − r 1 a ] = a [ ( r 0 − r 1 ) a ] = a ( r 0 − r 1 ) a ∈ S As r 0 − r 1 ∈ R r_0 - r_1 \in R r 0 − r 1 ∈ R
Since 1 ∈ R 1 \in R 1 ∈ R a 1 a ∈ S a1a \in S a 1 a ∈ S a 1 a = a 2 = 1 a1a = a^2 = 1 a 1 a = a 2 = 1
(1), (2), (3), (5), (6) of a ring’s definition in page 227 are satisfied by the usual properties of matrix algebra and integers.
Note the additive identity is the matrix
[ 0 0 + 0 0 + 0 0 ]
\begin{bmatrix}
0 & 0+0 \\
0+0 & 0
\end{bmatrix}
[ 0 0 + 0 0 + 0 0 ] We show (4). For any matrix M ∈ R M \in R M ∈ R
M = [ a a + b a + b b ]
M = \begin{bmatrix} a & a+b \\ a+b & b \end{bmatrix}
M = [ a a + b a + b b ] The matrix − M -M − M
− M = [ − a − a + ( − b ) − a + ( − b ) − b ] -M = \begin{bmatrix} -a & -a+(-b) \\ -a+(-b) & -b \end{bmatrix}
− M = [ − a − a + ( − b ) − a + ( − b ) − b ] is in M 2 ( Z ) M_2(Z) M 2 ( Z ) − a ∈ Z -a \in Z − a ∈ Z a ∈ Z a \in Z a ∈ Z M − M M - M M − M
2 ∈ 2 Z 2 \in 2Z 2 ∈ 2 Z 3 ∈ 3 Z 3 \in 3Z 3 ∈ 3 Z 2 + 3 = 5 ∉ 2 Z ∪ 3 Z 2+3 = 5 \not\in 2Z \cup 3Z 2 + 3 = 5 ∈ 2 Z ∪ 3 Z 2 Z ∪ 3 Z 2Z \cup 3Z 2 Z ∪ 3 Z