Homomorphism and Ideal
Thm 1
Let $\phi : R \to R'$ be ring homomorphism. Then
1. $\phi(0) = 0$
2. $\forall a \in R$, $\phi(-a) = -\phi(a)$
3. $S$ is subring of $R$ $\Rightarrow$ $\phi[S]$ is subring of $R'$
4. $S'$ is subring of $R'$ $\Rightarrow$ $\phi^{-1}[S]$ is subring of $R$
5. $\phi(1_{R}) = 1_{\phi[R]}$
( In general, $\phi(1_{R}) \neq 1_{R'}$ )
6. $\ker \phi = \left\{0 \right\}$ $\Leftrightarrow$ $\phi$ is 1-1
7. $\forall a \in R$, $\phi^{-1}[\left\{\phi(a) \right\}] = a + \ker \phi$
pf)
1. 2.
# Prop 1
3.
Let $S$ be a subring of $R$.
So $S$ is subgroup of $R$.
By Prop 1,
$\phi[S]$ is a subgroup of $R'$
Show : closed under $\cdot$
$\forall a', b' \in \phi[S]$
So there exists $a, b \in S$ s.t.
$a' = \phi(a)$, $b' = \phi(b)$
Since $S$ is ring,
$ab \in S$
$\therefore$ $a'b' = \phi(a)\phi(b) = \phi(ab) \in \phi[S]$
end
4.
Let $S'$ be a subring of $R'$.
So $S'$ be a subring of $R'$.
By Prop 1,
$\phi^{-1}[S]$ is subgroup of $R$
Show : closed under $\cdot$
$\forall a, b \in \phi^{-1}[S]$
So $\phi(a), \phi(b) \in S$
Since $S$ is ring,
$\phi(ab) = \phi(a) \phi(b) \in S$
$\therefore$ $ab \in \phi^{-1}[S]$
end
5.
$\forall a' \in \phi[R]$
So there exists $a \in R$ s.t.
$a' = \phi(a)$
Therefore
$a' \phi(1_{R}) = \phi(a)\phi(1_{R}) = \phi(a 1_{R}) = \phi(a) = a'$
$\phi(1_{R}) a' = \phi(1_{R}) \phi(a) = \phi(1_{R} a) = \phi(a) = a'$
$\therefore$ $\phi(1_{R}) = 1_{\phi[R]}$
6. 7.
# Thm 1
Def 1
Let $R$ be a ring and $S$ be subring of $R$.
$S$ is an ideal of $R$ if
$aS \subset S$ and $Sa \subset S$ for all $a \in R$
Cor 1
Let $R$ be a ring. Then
$\left\{0 \right\}$, $R$ are ideal of $R$
Def 2
Let $R$ be a ring and $N$ be ideal of $R$.
1. $N = \left\{0 \right\}$ : trivial ideal
2. $N = R$ : improper ideal
3. $N \neq \left\{0 \right\}$ : nontrivial ideal
4. $N \neq R$ : proper ideal
5. $N \neq \left\{0 \right\}, R$ : nontrivial proper ideal
Thm 2
Let $R$ be ring with $1$ and $N$ be ideal of $R$. Then
$u \in N$ for some unit $u$ in $R$ $\Leftrightarrow$ $N = R$
$\Rightarrow)$
Since $u$ is unit in $R$, there exists $u^{-1} \in R$
Since $N$ is ideal of $R$,
$1 = u^{-1}u \in N$
$\forall r \in R$
Since $N$ is ideal of $R$,
$r = r1 \in N$
$\therefore$ $R \subset N$
$\therefore$ $N = R$
$\Leftarrow)$
Since $N = R$,
$1 \in N$ is unit in $R$
Thm 3
Let $\phi : R \to R'$ be ring homomorphism. Then
$\ker \phi$ is an ideal of $R$
Show 1 : $\ker \phi$ is subring of $R$
It is clear by Thm 1.
end
Show 2 : $a \ker \phi \subset \ker \phi$ for all $a \in R$
$\forall a \in R$, $\forall b \in \ker \phi$
So $\phi(b) = 0'$
Therefore
$\phi (ab) = \phi(a) \phi(b) = \phi(a) 0' = 0'$
$\therefore$ $ab \in \ker \phi$
end
Show 3 : $\ker \phi a \subset \ker \phi$ for all $a \in R$
We can prove similar to Show 2
end
Thm 4
Let $\phi : R \to R'$ be ring homomorphism. Then
1. $S$ is ideal of $R$ $\Rightarrow$ $\phi[S]$ is ideal of $\phi[R]$
2. $S'$ is ideal of $R'$ $\Rightarrow$ $\phi^{-1}[S']$ is ideal of $R$
pf)
Show 1 : $\phi[S]$ is subring of $\phi[R]$
It is clear by Thm 1.
end
Show 2 : $a' \phi[S] \subset \phi[S]$ for all $a' \in \phi[R]$
$\forall a' \in \phi[R]$, $\forall b' \in \phi[S]$
So there exists $a \in R$, $b \in S$ s.t.
$a' = \phi (a)$, $b' = \phi(b)$
Since $S$ is ideal of $R$,
$ab \in S$
$\therefore$ $a'b' = \phi(a) \phi(b) = \phi(ab) \in \phi[S]$
end
Show 3 : $\phi[S] a' \subset \phi[R]$ for all $a' \in \phi[R]$
We can prove similar to Show 2
end
Show 1 : $\phi^{-1}[S']$ is subring of $R$
It is clear by Thm 1.
end
Show 2 : $a \phi^{-1}[S'] \subset \phi^{-1}[S']$ for all $a \in R$
$\forall a \in R$, $\forall b \in \phi^{-1}[S']$
So $\phi(b) \in S'$
Since $S'$ is ideal of $R'$,
$\phi(ab) = \phi(a) \phi(b) \in S'$
$\therefore$ $ab \in \phi^{-1}[S']$
end
Show 3 : $\phi^{-1}[S'] a \subset \phi^{-1}[S']$ for all $a \in R$
We can prove similar to Show 2
end