Dado
a
≠
0
{\displaystyle a\neq 0}
a
0
=
1
{\displaystyle a^{0}=1}
n
∈
N
,
a
−
n
=
1
a
n
,
ou seja
,
a
−
n
=
(
a
n
)
−
1
.
{\displaystyle n\in \mathbb {N} ,a^{-n}={1 \over a^{n}},{\mbox{ ou seja}},a^{-n}=(a^{n})^{-1}.}
m
,
n
∈
Z
,
{\displaystyle m,n\in \mathbb {Z} ,}
a
m
⋅
a
n
=
a
m
+
n
.
{\displaystyle a^{m}\cdot a^{n}=a^{m+n}.}
(
a
m
)
n
=
a
m
n
{\displaystyle (a^{m})^{n}=a^{mn}}
a
m
⋅
a
n
=
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
⋅
a
⋅
a
⋅
.
.
.
⋅
a
⏟
n
=
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
+
n
=
a
m
+
n
.
{\displaystyle a^{m}\cdot a^{n}=\underbrace {a\cdot a\cdot ...\cdot a} _{m}\cdot \underbrace {a\cdot a\cdot ...\cdot a} _{n}=\underbrace {a\cdot a\cdot ...\cdot a} _{m+n}=a^{m+n}.}
(
a
m
)
n
=
(
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
)
n
=
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
⋅
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
⋅
.
.
.
⋅
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
⏟
n
=
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
+
m
+
.
.
.
+
m
⏟
n
=
a
⋅
a
⋅
.
.
.
⋅
a
⏟
m
n
=
a
m
n
{\displaystyle (a^{m})^{n}=(\underbrace {a\cdot a\cdot ...\cdot a} _{m})^{n}=\underbrace {\underbrace {a\cdot a\cdot ...\cdot a} _{m}\cdot \underbrace {a\cdot a\cdot ...\cdot a} _{m}\cdot ...\cdot \underbrace {a\cdot a\cdot ...\cdot a} _{m}} _{n}=\underbrace {a\cdot a\cdot ...\cdot a} _{\underbrace {m+m+...+m} _{n}}=\underbrace {a\cdot a\cdot ...\cdot a} _{mn}=a^{mn}}
Sejam K, L corpos. Uma função
f
:
K
→
L
{\displaystyle f:K\rightarrow L}
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
{\displaystyle f(x+y)=f(x)+f(y)}
f
(
x
⋅
y
)
=
f
(
x
)
⋅
f
(
y
)
{\displaystyle f(x\cdot y)=f(x)\cdot f(y)}
x
,
y
∈
K
{\displaystyle x,y\in K}
Dado um homomorfismo
f
:
K
→
L
{\displaystyle f:K\rightarrow L}
f
(
0
)
=
0.
{\displaystyle f(0)=0.}
Prove também que, ou
f
(
x
)
=
0
,
∀
x
∈
K
,
{\displaystyle f(x)=0,\forall \;x\in K,}
f
(
1
)
=
1
{\displaystyle f(1)=1}
Pelo homomorfismo de f, dado
x
,
0
∈
K
,
f
(
x
+
0
K
)
=
f
(
x
)
+
f
(
0
K
)
,
mas
x
+
0
K
=
x
(
em
K
)
⇒
f
(
x
+
0
K
)
=
f
(
x
)
=
f
(
x
)
+
f
(
0
K
)
⇒
{\displaystyle x,0\in K,f(x+0_{K})=f(x)+f(0_{K}),{\mbox{ mas }}x+0_{K}=x({\mbox{ em }}K)\Rightarrow f(x+0_{K})=f(x)=f(x)+f(0_{K})\Rightarrow }
⇒
f
(
0
K
)
=
0
L
{\displaystyle \Rightarrow f(0_{K})=0_{L}}
f
(
0
)
=
0
{\displaystyle f(0)=0}
Suponhamos primeiro que
f
(
1
K
)
=
0
L
,
dado
1
K
,
x
∈
K
,
1
k
⋅
x
=
x
{\displaystyle f(1_{K})=0_{L},{\mbox{ dado }}1_{K},x\in K,1_{k}\cdot x=x}
f
(
1
⋅
x
)
=
f
(
x
)
=
f
(
1
)
⋅
f
(
x
)
=
{\displaystyle f(1\cdot x)=f(x)=f(1)\cdot f(x)=}
=
0
L
⋅
f
(
x
)
=
0
L
{\displaystyle =0_{L}\cdot f(x)=0_{L}}
f
(
x
)
=
0
,
∀
x
∈
K
{\displaystyle f(x)=0,\forall \;x\in K}
Vamos tomar agora
f
(
1
K
)
≠
0
L
,
dado
1
K
∈
K
,
1
K
⋅
1
K
=
1
K
{\displaystyle f(1_{K})\neq 0_{L},{\mbox{ dado }}1_{K}\in K,1_{K}\cdot 1_{K}=1_{K}}
f
(
1
)
=
f
(
1
⋅
1
)
=
f
(
1
)
⋅
f
(
1
)
.
{\displaystyle f(1)=f(1\cdot 1)=f(1)\cdot f(1).}
f
(
x
)
⋅
f
(
x
)
−
1
=
1
L
,
∀
x
∈
K
⇒
f
(
1
)
⋅
f
(
1
)
−
1
=
f
(
1
)
f
(
1
)
⋅
f
(
1
)
−
1
⇒
1
L
=
f
(
1
)
⋅
1
L
⇒
f
(
1
)
=
1
L
.
{\displaystyle {f(x)\cdot f(x)}^{-1}=1_{L},\forall \;x\in K\Rightarrow f(1)\cdot f(1)^{-1}=f(1)f(1)\cdot f(1)^{-1}\Rightarrow 1_{L}=f(1)\cdot 1_{L}\Rightarrow f(1)=1_{L}.}
Vamos provar por indução que
f
(
n
K
)
=
n
L
,
onde
n
K
=
1
K
+
1
K
+
.
.
.
+
1
K
⏟
n
,
n
L
=
1
L
+
1
L
+
.
.
.
+
1
L
⏟
n
{\displaystyle f(n_{K})=n_{L},{\mbox{ onde }}n_{K}=\underbrace {1_{K}+1_{K}+...+1_{K}} _{n},n_{L}=\underbrace {1_{L}+1_{L}+...+1_{L}} _{n}}
Vamos mostrar ser válido para n = 1:
f
(
1
K
)
=
1
L
{\displaystyle f(1_{K})=1_{L}}
Vamos supor válido para n=t+1: Assim
f
(
t
K
)
=
t
L
{\displaystyle f(t_{K})=t_{L}}
f
(
1
K
)
,
temos
f
(
t
K
)
+
f
(
1
K
)
=
t
L
+
f
(
1
K
)
{\displaystyle f(1_{K}),{\mbox{ temos }}f(t_{K})+f(1_{K})=t_{L}+f(1_{K})}
⇒
f
(
t
K
+
1
L
)
=
t
L
+
1
L
⇒
f
(
1
K
+
1
K
+
.
.
.
+
1
K
⏟
t
+
1
K
)
=
1
L
+
1
L
+
.
.
.
+
1
L
⏟
t
+
1
L
⇒
{\displaystyle \Rightarrow f(t_{K}+1_{L})=t_{L}+1_{L}\Rightarrow f(\underbrace {1_{K}+1_{K}+...+1_{K}} _{t}+1_{K})=\underbrace {1_{L}+1_{L}+...+1_{L}} _{t}+1_{L}\Rightarrow }
⇒
f
(
1
K
+
1
K
+
.
.
.
+
1
K
⏟
t
+
1
)
=
1
L
+
1
L
+
.
.
.
+
1
L
⏟
t
+
1
⇒
f
(
t
+
1
K
)
=
t
+
1
L
{\displaystyle \Rightarrow f(\underbrace {1_{K}+1_{K}+...+1_{K}} _{t+1})=\underbrace {1_{L}+1_{L}+...+1_{L}} _{t+1}\Rightarrow f(t+1_{K})=t+1_{L}}
Vamos mostrar que f é injetiva, tome
f
(
x
K
)
=
f
(
y
K
)
⇒
x
L
=
y
L
logo
f
{\displaystyle f(x_{K})=f(y_{K})\Rightarrow x_{L}=y_{L}{\mbox{ logo }}f}
Seja
f
:
Q
→
Q
{\displaystyle f:\mathbb {Q} \rightarrow \mathbb {Q} }
f
(
x
+
y
)
=
f
(
x
)
+
f
(
y
)
{\displaystyle f(x+y)=f(x)+f(y)}
f
(
x
⋅
y
)
=
f
(
x
)
⋅
f
(
y
)
{\displaystyle f(x\cdot y)=f(x)\cdot f(y)}
x
,
y
∈
K
{\displaystyle x,y\in K}
f
(
x
)
=
0
,
∀
x
∈
Q
,
{\displaystyle f(x)=0,\forall \;x\in \mathbb {Q} ,}
f
(
x
)
=
x
,
∀
x
∈
Q
{\displaystyle f(x)=x,\forall x\in \mathbb {Q} }
Pelo homomorfismo de f, dado
x
,
0
∈
Q
,
f
(
x
+
0
)
=
f
(
x
)
+
f
(
0
)
,
mas
x
+
0
=
x
⇒
f
(
x
+
0
)
=
f
(
x
)
=
f
(
x
)
+
f
(
0
)
⇒
f
(
0
)
=
0.
{\displaystyle x,0\in \mathbb {Q} ,f(x+0)=f(x)+f(0),{\mbox{ mas }}x+0=x\Rightarrow f(x+0)=f(x)=f(x)+f(0)\Rightarrow f(0)=0.}
Suponhamos primeiro que
f
(
1
)
=
0
,
dado
1
,
x
∈
Q
,
1
⋅
x
=
x
{\displaystyle f(1)=0,{\mbox{ dado }}1,x\in \mathbb {Q} ,1\cdot x=x}
f
(
1
⋅
x
)
=
f
(
x
)
=
f
(
1
)
⋅
f
(
x
)
=
{\displaystyle f(1\cdot x)=f(x)=f(1)\cdot f(x)=}
=
0
⋅
f
(
x
)
=
0
{\displaystyle =0\cdot f(x)=0}
f
(
x
)
=
0
,
∀
x
∈
Q
{\displaystyle f(x)=0,\forall \;x\in \mathbb {Q} }
Vamos tomar agora
f
(
1
)
≠
0
,
como
1
⋅
1
=
1
{\displaystyle f(1)\neq 0,{\mbox{ como }}1\cdot 1=1}
f
(
1
)
=
f
(
1
⋅
1
)
=
f
(
1
)
⋅
f
(
1
)
.
{\displaystyle f(1)=f(1\cdot 1)=f(1)\cdot f(1).}
Q
{\displaystyle \mathbb {Q} \;}
f
(
x
)
⋅
f
(
x
)
−
1
=
1
,
∀
x
∈
Q
⇒
f
(
1
)
⋅
f
(
1
)
−
1
=
f
(
1
)
f
(
1
)
⋅
f
(
1
)
−
1
⇒
1
=
f
(
1
)
⋅
1
⇒
f
(
1
)
=
1.
{\displaystyle f(x)\cdot f(x)^{-1}=1,\forall \;x\in \mathbb {Q} \Rightarrow f(1)\cdot f(1)^{-1}=f(1)f(1)\cdot f(1)^{-1}\Rightarrow 1=f(1)\cdot 1\Rightarrow f(1)=1.}
Vamos provar por indução que
f
(
n
)
=
n
,
onde
n
=
1
+
1
+
.
.
.
+
1
⏟
n
{\displaystyle f(n)=n,{\mbox{ onde }}n=\underbrace {1+1+...+1} _{n}}
Vamos mostrar ser válido para n = 1:
f
(
1
)
=
1
{\displaystyle f(1)=1}
Vamos supor válido para n=t+1: Assim
f
(
t
)
=
t
{\displaystyle f(t)=t}
f
(
1
)
,
temos
f
(
t
)
+
f
(
1
)
=
t
+
f
(
1
)
⇒
{\displaystyle f(1),{\mbox{ temos }}f(t)+f(1)=t+f(1)\Rightarrow }
⇒
f
(
1
+
1
+
.
.
.
+
1
⏟
t
+
1
)
=
1
+
1
+
.
.
.
+
1
⏟
t
+
1
⇒
{\displaystyle \Rightarrow f(\underbrace {1+1+...+1} _{t}+1)=\underbrace {1+1+...+1} _{t}+1\Rightarrow }
⇒
f
(
1
+
1
+
.
.
.
+
1
⏟
t
+
1
)
=
1
+
1
+
.
.
.
+
1
⏟
t
+
1
⇒
f
(
t
+
1
)
=
t
+
1
{\displaystyle \Rightarrow f(\underbrace {1+1+...+1} _{t+1})=\underbrace {1+1+...+1} _{t+1}\Rightarrow f(t+1)=t+1}
Verifique as associatividades da adição e da multiplicação em
Z
2
.
{\displaystyle \mathbb {Z} _{2}.\;}
f
:
Z
→
Z
2
,
f
(
n
)
=
{
0
,
se n é par
1
,
se n é impar
{\displaystyle f:\mathbb {Z} \rightarrow \mathbb {Z} _{2},f(n)={\begin{cases}0,{\mbox{ se n é par }}\\1,{\mbox{ se n é impar }}\end{cases}}\;}
m
,
n
∈
Z
{\displaystyle m,n\in \mathbb {Z} }
P
r
o
p
A
:
f
(
m
+
n
)
=
f
(
m
)
+
f
(
n
)
,
P
r
o
p
B
:
f
(
m
⋅
n
)
=
f
(
m
)
⋅
f
(
n
)
.
{\displaystyle PropA:f(m+n)=f(m)+f(n),PropB:f(m\cdot n)=f(m)\cdot f(n).}
Z
{\displaystyle \mathbb {Z} }
Z
2
)
.
{\displaystyle \mathbb {Z} _{2}).}
Primeiro vamos mostrar a associatividade de
Z
2
{\displaystyle \mathbb {Z} _{2}}
f
(
p
)
+
(
f
(
q
)
+
f
(
r
)
)
=
(
f
(
p
)
+
f
(
q
)
)
+
f
(
r
)
,
{\displaystyle f(p)+(f(q)+f(r))=(f(p)+f(q))+f(r),\;}
∀
p
,
q
,
r
∈
Z
{\displaystyle \forall p,q,r\in \mathbb {Z} }
f
(
p
)
+
(
f
(
q
)
+
f
(
r
)
)
=
f
(
p
)
+
f
(
q
+
r
)
,
{\displaystyle f(p)+(f(q)+f(r))=f(p)+f(q+r),}
=
f
(
(
p
+
(
q
+
r
)
)
,
{\displaystyle =f((p+(q+r)),\;}
=
f
(
(
p
+
q
)
+
r
)
{\displaystyle =f((p+q)+r)\;}
Z
,
=
f
(
p
+
q
)
+
f
(
r
)
,
{\displaystyle \mathbb {Z} ,=f(p+q)+f(r),}
=
(
f
(
p
)
+
f
(
q
)
)
+
f
(
r
)
,
{\displaystyle =(f(p)+f(q))+f(r),}
Agora vamos mostrar a associatividade de
Z
2
{\displaystyle \mathbb {Z} _{2}}
f
(
p
)
⋅
(
f
(
q
)
⋅
f
(
r
)
)
=
(
f
(
p
)
⋅
f
(
q
)
)
⋅
f
(
r
)
,
{\displaystyle f(p)\cdot (f(q)\cdot f(r))=(f(p)\cdot f(q))\cdot f(r),\;}
∀
p
,
q
,
r
∈
Z
{\displaystyle \forall p,q,r\in \mathbb {Z} }
f
(
p
)
⋅
(
f
(
q
)
⋅
f
(
r
)
)
=
f
(
p
)
⋅
f
(
q
⋅
r
)
,
{\displaystyle f(p)\cdot (f(q)\cdot f(r))=f(p)\cdot f(q\cdot r),}
=
f
(
(
p
⋅
(
q
⋅
r
)
)
,
{\displaystyle =f((p\cdot (q\cdot r)),\;}
=
f
(
(
p
⋅
q
)
⋅
r
)
{\displaystyle =f((p\cdot q)\cdot r)\;}
Z
,
=
f
(
p
⋅
q
)
⋅
f
(
r
)
,
{\displaystyle \mathbb {Z} ,=f(p\cdot q)\cdot f(r),}
=
(
f
(
p
)
⋅
f
(
q
)
)
⋅
f
(
r
)
,
{\displaystyle =(f(p)\cdot f(q))\cdot f(r),}
Seja p um número natural primo. Para cada inteiro m, indiquemos com
m
¯
{\displaystyle {\overline {m}}}
Z
p
=
{
0
,
1
,
.
.
.
,
p
−
1
}
{\displaystyle \mathbb {Z} _{p}=\{0,1,...,p-1\}}
⊕
{\displaystyle \oplus }
⊗
,
{\displaystyle \otimes ,}
P
A
:
m
⊕
n
=
m
+
n
¯
e
P
B
:
m
⊗
n
=
m
×
n
¯
.
{\displaystyle P_{A}:m\oplus n={\overline {m+n}}{\mbox{ e }}P_{B}:m\otimes n={\overline {m\times n}}.}
Prove que a função
f
:
Z
→
Z
p
{\displaystyle f:\mathbb {Z} \rightarrow \mathbb {Z} _{p}}
P
C
:
f
(
n
)
=
n
¯
,
{\displaystyle P_{C}:f(n)={\overline {n}},}
P
D
:
f
(
m
+
n
)
=
f
(
m
)
⊕
f
(
n
)
e
P
E
:
f
(
m
×
n
)
=
f
(
m
)
⊗
f
(
n
)
.
{\displaystyle P_{D}:f(m+n)=f(m)\oplus f(n){\mbox{ e }}P_{E}:f(m\times n)=f(m)\otimes f(n).}
Conclua que
⊕
e
⊗
{\displaystyle \oplus {\mbox{ e }}\otimes }
Observe que dados
m
,
n
∈
Z
p
,
m
⊗
n
=
0
⇒
m
=
0
ou
n
=
0.
{\displaystyle m,n\in \mathbb {Z} _{p},m\otimes n=0\Rightarrow m=0{\mbox{ ou }}n=0.}
Conclua que
Z
p
{\displaystyle \mathbb {Z} _{p}}
Ao dividirmos m por p, coloquemos o quociente como
q
m
{\displaystyle q_{m}}
m
¯
,
logo
,
m
¯
=
m
−
p
⋅
q
m
{\displaystyle {\overline {m}},{\mbox{ logo }},{\overline {m}}=m-p\cdot q_{m}}
q
n
{\displaystyle q_{n}}
n
¯
,
logo
,
n
¯
=
n
−
p
⋅
q
n
{\displaystyle {\overline {n}},{\mbox{ logo }},{\overline {n}}=n-p\cdot q_{n}}
m
¯
+
n
¯
=
m
−
p
⋅
q
m
+
n
−
p
⋅
q
n
=
m
+
n
−
p
⋅
(
q
m
+
q
n
)
{\displaystyle {\overline {m}}+{\overline {n}}=m-p\cdot q_{m}+n-p\cdot q_{n}=m+n-p\cdot (q_{m}+q_{n})}
m
¯
+
n
¯
{\displaystyle {\overline {m}}+{\overline {n}}}
m
+
n
por
p
{\displaystyle m+n{\mbox{ por }}p}
P
F
:
m
+
n
¯
=
m
¯
+
n
¯
¯
.
{\displaystyle P_{F}:{\overline {m+n}}={\overline {{\overline {m}}+{\overline {n}}}}.}
Analogamente
m
¯
×
n
¯
=
(
m
−
p
⋅
q
m
)
⋅
(
n
−
p
⋅
q
n
)
=
m
⋅
n
+
p
⋅
(
p
⋅
q
m
⋅
q
n
−
m
⋅
q
n
−
n
⋅
q
m
)
.
{\displaystyle {\overline {m}}\times {\overline {n}}=(m-p\cdot q_{m})\cdot (n-p\cdot q_{n})=m\cdot n+p\cdot (p\cdot q_{m}\cdot q_{n}-m\cdot q_{n}-n\cdot q_{m}).}
m
¯
×
n
¯
{\displaystyle {\overline {m}}\times {\overline {n}}}
m
×
n
por
p
{\displaystyle m\times n{\mbox{ por }}p}
P
G
:
m
×
n
¯
=
m
¯
×
n
¯
¯
.
{\displaystyle P_{G}:{\overline {m\times n}}={\overline {{\overline {m}}\times {\overline {n}}}}.}
Temos que
f
(
m
+
n
)
=
m
+
n
¯
,
pela
P
C
,
=
m
¯
+
n
¯
¯
,
pela
P
F
,
=
m
¯
⊕
n
¯
pela
P
A
,
=
f
(
m
)
⊕
f
(
n
)
,
pela
P
C
{\displaystyle f(m+n)={\overline {m+n}},{\mbox{ pela }}P_{C},={\overline {{\overline {m}}+{\overline {n}}}},{\mbox{ pela }}P_{F},={\overline {m}}\oplus {\overline {n}}{\mbox{ pela }}P_{A},=f(m)\oplus f(n),{\mbox{ pela }}P_{C}}
P
D
{\displaystyle P_{D}\;}
Temos que
f
(
m
×
n
)
=
m
×
n
¯
,
pela
P
C
,
=
m
¯
×
n
¯
¯
,
pela
P
G
,
=
m
¯
⊗
n
¯
pela
P
B
,
=
f
(
m
)
⊗
f
(
n
)
,
pela
P
C
{\displaystyle f(m\times n)={\overline {m\times n}},{\mbox{ pela }}P_{C},={\overline {{\overline {m}}\times {\overline {n}}}},{\mbox{ pela }}P_{G},={\overline {m}}\otimes {\overline {n}}{\mbox{ pela }}P_{B},=f(m)\otimes f(n),{\mbox{ pela }}P_{C}}
P
E
{\displaystyle P_{E}\;}
Vamos mostrar que
⊕
e
⊗
{\displaystyle \oplus {\mbox{ e }}\otimes }
Se a = b nos reais então
f
(
a
)
=
f
(
b
)
.
Logo
∀
m
,
n
∈
R
,
m
+
n
=
n
+
m
⇒
f
(
m
+
n
)
=
f
(
n
+
m
)
⇒
{\displaystyle f(a)=f(b).{\mbox{ Logo }}\forall m,n\in \mathbb {R} ,m+n=n+m\Rightarrow f(m+n)=f(n+m)\Rightarrow }
⇒
f
(
m
)
⊕
f
(
n
)
=
f
(
n
)
⊕
f
(
m
)
⇒
⊕
{\displaystyle \Rightarrow f(m)\oplus f(n)=f(n)\oplus f(m)\Rightarrow \oplus }
Analogamente, dados
∀
m
,
n
∈
R
,
m
×
n
=
n
×
m
⇒
f
(
m
×
n
)
=
f
(
n
×
m
)
⇒
f
(
m
)
⊗
f
(
n
)
=
f
(
n
)
⊗
f
(
m
)
⇒
⊗
{\displaystyle \forall m,n\in \mathbb {R} ,m\times n=n\times m\Rightarrow f(m\times n)=f(n\times m)\Rightarrow f(m)\otimes f(n)=f(n)\otimes f(m)\Rightarrow \otimes }
Vamos mostrar que
⊕
e
⊗
{\displaystyle \oplus {\mbox{ e }}\otimes }
Se a = b nos reais então
f
(
a
)
=
f
(
b
)
.
Logo
∀
m
,
n
,
o
∈
R
,
m
+
(
n
+
o
)
=
(
m
+
n
)
+
o
⇒
f
(
m
+
(
n
+
o
)
)
=
f
(
(
m
+
n
)
+
o
)
⇒
{\displaystyle f(a)=f(b).{\mbox{ Logo }}\forall m,n,o\in \mathbb {R} ,m+(n+o)=(m+n)+o\Rightarrow f(m+(n+o))=f((m+n)+o)\Rightarrow }
⇒
f
(
m
)
⊕
f
(
n
+
o
)
=
f
(
m
+
n
)
⊕
f
(
o
)
⇒
f
(
m
)
⊕
[
f
(
n
)
⊕
f
(
o
)
]
=
[
f
(
m
)
⊕
f
(
n
)
]
⊕
f
(
o
)
⇒
⊕
{\displaystyle \Rightarrow f(m)\oplus f(n+o)=f(m+n)\oplus f(o)\Rightarrow f(m)\oplus [f(n)\oplus f(o)]=[f(m)\oplus f(n)]\oplus f(o)\Rightarrow \oplus }
Analogamente, dados
∀
m
,
n
,
o
∈
R
,
m
×
(
n
×
o
)
=
(
m
×
n
)
×
o
⇒
f
(
m
×
(
n
×
o
)
)
=
f
(
(
m
×
n
)
×
o
)
⇒
{\displaystyle \forall m,n,o\in \mathbb {R} ,m\times (n\times o)=(m\times n)\times o\Rightarrow f(m\times (n\times o))=f((m\times n)\times o)\Rightarrow }
⇒
f
(
m
)
⊗
f
(
n
×
o
)
=
f
(
m
×
n
)
⊗
f
(
o
)
⇒
f
(
m
)
⊗
[
f
(
n
)
⊗
f
(
o
)
]
=
[
f
(
m
)
⊗
f
(
n
)
]
⊗
f
(
o
)
⇒
⊗
{\displaystyle \Rightarrow f(m)\otimes f(n\times o)=f(m\times n)\otimes f(o)\Rightarrow f(m)\otimes [f(n)\otimes f(o)]=[f(m)\otimes f(n)]\otimes f(o)\Rightarrow \otimes }
Vamos mostrar que
⊕
e
⊗
{\displaystyle \oplus {\mbox{ e }}\otimes }
Se a = b nos reais então
f
(
a
)
=
f
(
b
)
.
Logo
∀
m
,
n
,
o
∈
R
,
m
×
(
n
+
o
)
=
[
m
×
n
]
+
[
m
×
o
]
⇒
f
(
m
×
(
n
+
o
)
)
=
f
(
[
m
×
n
]
+
[
m
×
o
]
)
⇒
{\displaystyle f(a)=f(b).{\mbox{ Logo }}\forall m,n,o\in \mathbb {R} ,m\times (n+o)=[m\times n]+[m\times o]\Rightarrow f(m\times (n+o))=f([m\times n]+[m\times o])\Rightarrow }
⇒
f
(
m
)
⊗
f
(
n
+
o
)
=
f
(
m
×
n
)
⊕
f
(
m
×
o
)
⇒
f
(
m
)
⊗
[
f
(
n
)
⊕
f
(
o
)
]
=
[
f
(
m
)
⊗
f
(
n
)
]
⊕
[
f
(
m
)
⊗
f
(
o
)
]
⇒
⊕
,
⊗
{\displaystyle \Rightarrow f(m)\otimes f(n+o)=f(m\times n)\oplus f(m\times o)\Rightarrow f(m)\otimes [f(n)\oplus f(o)]=[f(m)\otimes f(n)]\oplus [f(m)\otimes f(o)]\Rightarrow \oplus ,\otimes }
Analogamente, dados
∀
m
,
n
,
o
∈
R
,
(
m
+
n
)
×
o
=
[
m
×
o
]
+
[
n
×
o
]
⇒
f
(
(
m
+
n
)
×
o
)
=
f
(
[
m
×
o
]
+
[
n
×
o
]
)
⇒
{\displaystyle \forall m,n,o\in \mathbb {R} ,(m+n)\times o=[m\times o]+[n\times o]\Rightarrow f((m+n)\times o)=f([m\times o]+[n\times o])\Rightarrow }
⇒
f
(
m
+
n
)
⊗
f
(
o
)
=
f
(
m
×
o
)
⊕
f
(
n
×
o
)
⇒
[
f
(
m
)
⊕
f
(
n
)
]
⊗
f
(
o
)
=
[
f
(
m
)
⊗
f
(
o
)
]
⊕
[
f
(
n
)
⊗
f
(
o
)
]
⇒
⊕
,
⊗
{\displaystyle \Rightarrow f(m+n)\otimes f(o)=f(m\times o)\oplus f(n\times o)\Rightarrow [f(m)\oplus f(n)]\otimes f(o)=[f(m)\otimes f(o)]\oplus [f(n)\otimes f(o)]\Rightarrow \oplus ,\otimes }
Vamos mostrar que
⊕
e
⊗
{\displaystyle \oplus {\mbox{ e }}\otimes }
Se a = b nos reais então
f
(
a
)
=
f
(
b
)
.
Logo, dado
0
∈
R
;
0
+
0
=
0
,
⇒
f
(
0
+
0
)
=
f
(
0
)
⇒
f
(
0
)
⊕
f
(
0
)
=
f
(
0
)
⇒
f
(
0
)
{\displaystyle f(a)=f(b).{\mbox{ Logo, dado }}0\in \mathbb {R} ;0+0=0,\Rightarrow f(0+0)=f(0)\Rightarrow f(0)\oplus f(0)=f(0)\Rightarrow f(0)}
Z
p
{\displaystyle \mathbb {Z} _{p}}
Analogamente, dado
1
∈
R
,
1
×
1
=
1
⇒
f
(
1
×
1
)
=
f
(
1
)
⇒
f
(
1
)
⊗
f
(
1
)
=
f
(
1
)
⇒
f
(
1
)
{\displaystyle 1\in \mathbb {R} ,1\times 1=1\Rightarrow f(1\times 1)=f(1)\Rightarrow f(1)\otimes f(1)=f(1)\Rightarrow f(1)}
Z
p
{\displaystyle \mathbb {Z} _{p}}
Vamos mostrar que dados
m
,
n
∈
Z
p
,
m
⊗
n
=
0
⇒
m
=
0
ou
n
=
0.
{\displaystyle m,n\in \mathbb {Z} _{p},m\otimes n=0\Rightarrow m=0{\mbox{ ou }}n=0.}
∀
m
,
n
∈
Z
p
,
m
⊗
n
=
m
×
n
¯
=
m
×
n
−
p
×
q
=
0
⇒
m
×
n
=
p
×
q
.
{\displaystyle \forall m,n\in \mathbb {Z} _{p},m\otimes n={\overline {m\times n}}=m\times n-p\times q=0\Rightarrow m\times n=p\times q.}
m
,
n
<
p
{\displaystyle m,n<p}
Z
p
{\displaystyle \mathbb {Z} _{p}}
Seja K um conjunto onde são válidos todos os axiomas de corpo, salvo a existência de inverso multiplicativo.
Dado
a
≠
0
{\displaystyle a\neq 0}
f
:
K
→
K
,
{\displaystyle f:K\rightarrow K,}
f
(
x
)
=
a
x
,
{\displaystyle f(x)=ax,}
Mostre que f e injetiva se, e somente se, vale a lei do corte para a.
Conclua que, se K é finito, a lei do corte é equivalente à existência de inverso para cada elemento não-nulo de K. Resolução não está pronta
editar
![{\displaystyle \Rightarrow f(m)\oplus f(n+o)=f(m+n)\oplus f(o)\Rightarrow f(m)\oplus [f(n)\oplus f(o)]=[f(m)\oplus f(n)]\oplus f(o)\Rightarrow \oplus }](https://wikimedia.org/api/rest_v1/media/math/render/svg/44502eb87d5c3b9c418ce6942047da580e3e11f1)
![{\displaystyle \Rightarrow f(m)\otimes f(n\times o)=f(m\times n)\otimes f(o)\Rightarrow f(m)\otimes [f(n)\otimes f(o)]=[f(m)\otimes f(n)]\otimes f(o)\Rightarrow \otimes }](https://wikimedia.org/api/rest_v1/media/math/render/svg/cefaf6b9f7e8bfd78e6f9c737b83a91ed1b39963)
![{\displaystyle f(a)=f(b).{\mbox{ Logo }}\forall m,n,o\in \mathbb {R} ,m\times (n+o)=[m\times n]+[m\times o]\Rightarrow f(m\times (n+o))=f([m\times n]+[m\times o])\Rightarrow }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0082ea540faf5cfe58c8422f00c9216c5b77bcf8)
![{\displaystyle \Rightarrow f(m)\otimes f(n+o)=f(m\times n)\oplus f(m\times o)\Rightarrow f(m)\otimes [f(n)\oplus f(o)]=[f(m)\otimes f(n)]\oplus [f(m)\otimes f(o)]\Rightarrow \oplus ,\otimes }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6754c13088a9d8f1dccd4de72f6999c955456fa6)
![{\displaystyle \forall m,n,o\in \mathbb {R} ,(m+n)\times o=[m\times o]+[n\times o]\Rightarrow f((m+n)\times o)=f([m\times o]+[n\times o])\Rightarrow }](https://wikimedia.org/api/rest_v1/media/math/render/svg/31a4868671dc01d0cb3089c85566798c2c755420)
![{\displaystyle \Rightarrow f(m+n)\otimes f(o)=f(m\times o)\oplus f(n\times o)\Rightarrow [f(m)\oplus f(n)]\otimes f(o)=[f(m)\otimes f(o)]\oplus [f(n)\otimes f(o)]\Rightarrow \oplus ,\otimes }](https://wikimedia.org/api/rest_v1/media/math/render/svg/ef77ec19685c6d376f9472172d8649a2ff150bd9)
