## Algebra - Rings and fields

### 9. Kernel of a homomorphism

• Definition 9.1: Let $f$ be a homomorphism from a ring $R_1$ to $R_2$. Then Kernel of a homomorphism $f$ is defined by $\{a \in R_1/f(a)=0\}$. Kernel of $f$ is denoted by $K_f$ or ker$f$

• Theorems on Kernel of homomorphism

• Theorem 9.1: Every kernel is a subring of a ring.
Proof: Let $f:R_1\rightarrow R_2$ be an ring homomorphism. Then $K_f=\{a \in R_1/ f(a)=0\}$
To prove: $K_f$ is a subring of $R_1$
Let $a,b \in K_f$ $\Rightarrow$ $f(a)=0$ and $f(b)=0$
Consider $f(a-b)=f(a+(-b))=f(a)+f(-b)=f(a)-f(b)=0$
$\Rightarrow$ $a-b \in K_f$
Consider $f(ab)=f(a)f(b)=0$ $\Rightarrow$ $ab \in K_f$
$\therefore$ $K_f$ is a subring of a ring $R_1$ $\blacksquare$

• Theorem 9.2: Every kernel is an ideal of a ring.
Proof: Let $f:R_1\rightarrow R_2$ be an ring homomorphism. Then $K_f=\{a \in R_1/ f(a)=0\}$
To prove: $K_f$ is an ideal.
Let $a,b \in K_f$ $\Rightarrow$ $f(a)=0$ and $f(b)=0$
Consider, $f(a-b)=f(a+(-b))=f(a)+f(-b)=f(a)-f(b)=0$ $\Rightarrow$ $a-b \in K_f$
$\therefore$ $K_f$ is an additive subgroup of $R_1$
Let $r \in R_1$
Consider $f(ra)=f(r)f(a)=f(r)\times 0=0$ $\Rightarrow$ $ra \in K_f$ and also
$f(ar)=f(a)f(r)=0$ $\Rightarrow$ $ar \in K_f$
$\therefore$ $K_f$ is a ideal of $R_1$. $\blacksquare$

• Theorem 9.3: Onto ring homomorphism $f$ is an isomorphism if and only if $K_f=\{0\}$.
Proof: Let $f:R_1 \rightarrow R_2$ be a onto ring homomorphism.
Suppose $f:R_1 \rightarrow R_2$ be an ring isomorphism.
To prove: $K_f=\{0\}$
Let $a \in K_f$ $\Rightarrow$ $f(a)=0$ $\Rightarrow$ $f(a)=f(0)$ $\Rightarrow$ $a=0$ [$\because$ $f$ is $1-1$]
$\therefore$ $K_f={0}$
Conversely, suppose that $K_f=\{0\}$
To prove: $f$ is an isomorphism, it is enough to show that $f$ is $1-1$
Let $f(a)=f(b)$ $\Rightarrow$ $f(a)-f(b)=0$ $\Rightarrow$ $f(a-b)=0$ $\Rightarrow$ $a-b \in K_f$
Since $K_f=\{0\}$ we have $a-b=0$ $\Rightarrow$ $a=b$
$\therefore$ $f$ is $1-1$. $\blacksquare$

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