## Algebra - Rings and fields

### 6. Ring homomorphism

• Definition 6.1: A mapping $f$ from a ring $R$ to a ring $R'$ is called a homomorphism if
(1) $f(a+b)=f(a)+f(b)$ $\forall$ $a,b \in R$
(2) $f(ab)=f(a)f(b)$ $\forall$ $a,b \in R$

• Definition 6.2: An homomorphism $f$ from a ring $R$ to a ring $R'$ is called a isomorphism if it is $1-1$ and onto, denoted by $R\cong R'$.

• Definition 6.3: An homomorphism $f$ from a ring $R$ to a ring $R$ is called a automorphism if it is $1-1$ and onto.

• Theorems on homomorphism and isomorphism

• Theorem 6.1: Zero elements are mapped to a zero elements under homomorphism.
Proof: Let $f$ be a homomorphism from a ring $R$ into $R'$. i.e., $f:R\rightarrow R'$ be a ring homomorphism. Let $0$ and $0'$ be the zero element in $R$ and $R'$ respectively. To prove: $f(0)=0'$
Since $a+0=a$ $\forall$ $a \in \mathbb{R}$
$\Rightarrow$ $f(a+0)=f(a)$ $\forall$ $a \in \mathbb{R}$
$\Rightarrow$ $f(a)+f(0)=f(a)$ $\forall$ $a \in \mathbb{R}$ [$\because f$ is an homomorphism]
$\Rightarrow$ $f(a)+f(0)=f(a)+0'$
$\Rightarrow$ $f(0)=0'$ [by left cancellation law in $R'$] $\blacksquare$

• Theorem 6.2: Additive inverse elements are mapped to a additive inverse elements under homomorphism.
Proof: Let $f$ be a homomorphism from a ring $R$ into $R'$ i.e., $f:R\rightarrow R'$ be a ring homomorphism defined by $f(a)=a'$ $\forall$ $a \in R$.
To prove: $f(-a)=-f(a)$ $\forall$ $a \in R$
Since $a+(-a)=0$ $\forall$ $a \in R$
$\Rightarrow$ $f[(a)+(-a)]=f(0)$
$\Rightarrow$ $f(a)+f(-a)=0'$ [$\because f$ is an homomorphism and $f(0)=0'$]
$\Rightarrow$ $a'+f(-a)=0'$
$\Rightarrow$ $f(-a)=0'-a'$
$\Rightarrow$ $f(-a)=-a'$
$\Rightarrow$ $f(-a)=-f(a)$ $\forall$ $a \in R$ $\blacksquare$

• Theorem 6.3: Let $f$ be a homomorphism from a ring $R$ to ring $R'$. Then $f(R)$ is a subring of $R'$.
Proof: Let $f:R\rightarrow R'$ is a ring homomorphism. Then $f(R)=\{ f(a) / a\in R \}$ To prove: $f(R)$ is a subring of $R'$
Since $0 \in R$ $\Rightarrow$ $f(0) \in f(R)$ $\Rightarrow$ $0' \in f(R)$
$\therefore$ $f(R)\neq \emptyset$ and clearly $f(R) \subseteq R'$.
Let $a',b' \in f(R)$ $\Rightarrow$ $a'=f(a)$ and $b'=f(b)$ where $a,b \in R$
Consider $a'-b'$
$=f(a)-f(b)$
$=f(a-b)$ [$f$ is an homomorphism]
$=f(a-b) \in f(R)$ [$\because$ $a-b \in R$]
$\therefore$ we have $a',b' \in f(R)$ $\Rightarrow$ $a'-b' \in f(R)$
Consider $a'b'$
$=f(a)f(b)$
$=f(ab)$ [$f$ is an homomorphism]
$=f(ab) \in f(R)$ [$\because$ $ab \in R$]
$\therefore$ we have $a',b' \in f(R)$ $\Rightarrow$ $a'b' \in f(R)$
Hence, $f(R)$ is a subring of $R'$. $\blacksquare$

• Theorem 6.4: Let $f$ be a homomorphism from a ring $R$ to ring $R'$. If $R$ is commutative then $R'$ is also commutative. (Homomorphic image of a commutative ring is commutative)
Proof: Let $f$ be a ring homomorphism from a commuatative ring $R$ into a ring $R'$ defined by $f(a)=a'$ $\forall$ $a \in R$.
To prove: $R'$ is commutative. i.e., $a'b'=b'a'$
Let $a',b' \in R'$ $\Rightarrow$ $a'=f(a)$ and $b'=f(b)$ where $a,b \in R$
Consider $a'b'$
$=f(a)f(b)$
$=f(ab)$ [$f$ is an homomorphism]
$=f(ba)$ [$R$ is commutative]
$=f(b)f(a)$ [$f$ is an homomorphism]
$=b'a'$
$\therefore$ $a'b'=b'a'$ $\forall$ $a',b' \in R'$ $\blacksquare$

• Theorem 6.5: Isomorphic image of a ring with unity element is also a ring with unity element.
Proof: Let $f$ be a isomorphism from a ring with identity $R$ into a ring with identity $R'$. Let $1$ ad $1'$ be identity elements in $R$ and $R'$ respectively.
To prove: $f(1)=1'$
Let $a' \in R'$ then $\exists$ $a \in R$ such that $f(a)=a'$ [$f$ is onto]
Consider $a'f(1)$
$=f(a)f(1)$
$=f(a.1)$ [$f$ is an homomorphism]
$=f(a)$
$=a'$
Thus, we have $a'f(1)=a'$
$\Rightarrow$ $a'f(1)=a'1'$
$\Rightarrow$ $f(1)=1'$ [by left cancellation law in $R'$] $\blacksquare$

• Theorem 6.6: Isomorphic image of a ring without zero divisors is also a ring without zero divisors.
Proof: Let $f$ be a isomorphism from a ring without zero divisors $R$ into a ring $R'$. Let $0$ ad $0'$ be zero elements in $R$ and $R'$ respectively.
Since $R$ is a ring without zero divisor we have $ab=0$ $\Rightarrow$ $a=0$ or $b=0$
To prove: $R'$ has no zero divisors. i.e., $a'b'=0'$ $\Rightarrow$ $a'=0'$ or $b'=0'$
Let $a',b' \in R'$ then $\exists$ $a,b \in R$ such that $a'=f(a)$ and $b'=f(a)$ [$f$ is onto]
Consider $a'b'=0'$
$\Rightarrow$ $f(a)f(b)=f(0)$
$\Rightarrow$ $f(ab)=f(0)$
$\Rightarrow$ $ab=0$
$\Rightarrow$ $a=0$ or $b=0$
$\Rightarrow$ $f(a)=f(0)$ or $f(b)=f(0)$
$\Rightarrow$ $a'=0'$ or $b'=0'$
$\Rightarrow$ $a'b'=0$ $\Rightarrow$ $a'=0'$ or $b'=0'$
$\therefore$ Zero divisors are absent in $R'$ $\blacksquare$

• Theorem 6.7: The only homomorphism from $\mathbb{Z}$ to $\mathbb{Z}$ are the identity map and zero map.
Proof: Let $f:\mathbb{Z}\rightarrow \mathbb{Z}$ be a homomorphism. Suppose $f$ is a zero map i.e., $f(a)=0$ $\forall$ $a \in \mathbb{Z}$ then nothing to prove.
Suppose $f$ is a non zero map, then $f(a)\neq 0$ $\forall$ $a \in \mathbb{Z}$
Consider, $[f(1)]^2$
$=f(1)f(1)$
$=f(1.1)$ [$f$ is a homomorphism]
$=f(1)$
$\neq 0$
$\therefore$ $f(1)f(1)=f(1)$ $\Rightarrow$ $f(1)=1$
Now Consider $f(n)$ where $n \neq 1$
Case 1: if $n>0$
Consider, $f(n)$
$=f(\smash{\underbrace{1+1+...+1}_\text{$n$times}})$

$=\smash{\underbrace{f(1)+f(1)+...+f(1)}_\text{$n$times}}$ [$f$ is an homomorphism]

$=\smash{\underbrace{1+1+...+1}_\text{$n$times}}$

$=n.1$
$\therefore$ $f(n)=n$ when $n>0$.
Case 2: if $n<0$ Let $n=-m$ where $m>0$
Consider,$f(n)$
$=f(-m)$
$=-f(m)$
$=-m$
$=n$
$\therefore$ $f(n)=n$ when $n<0$.
$\therefore$ $f$ is a identity map. $\blacksquare$
Problems 6
1. Is the mapping $\phi(x)=3x$ $\forall$ $x \in \mathbb{Z}$ a homomorphism?
Solution: Given, $\phi(x)=3x$ $\forall$ $x \in \mathbb{Z}$
Consider, $\phi(x+y)=3(x+y)=3x+3y=\phi(x)\phi(y)$
Now consider, $\phi(xy)=3(xy)=(3x)y=\phi(x)y$
$\therefore$ $\phi(xy)\neq \phi(x)\phi(y)$
$\Rightarrow$ $\phi$ is not a homomorphism. $\spadesuit$

2. Show that the correspondence $x+iy$ $\longleftrightarrow$ $x-iy$ is an automorphism of field of complex numbers.
Solution: Given, $x+iy$ $\longleftrightarrow$ $x-iy$ where $x,y \in \mathbb{C}$
Define a map $f:\mathbb{C} \rightarrow \mathbb{C}$ by $f(x+iy)=x-iy$
Let $x,y \in \mathbb{C}$ $\Rightarrow$ $x=a_1+ib_1$ and $y=a_2+ib_2$ where $a_1,a_2,b_1,b_2 \in \mathbb{R}$
Consider, $f(x+y)$
$=f[(a_1+ib_1)+(a_2+ib_2)]$
$=f[(a_1+a_2)+i(b_1+b_2)]$
$=(a_1+a_2)-i(b_1+b_2)$
$=(a_1-ib_1)+(a_2-ib_2)$
$=f(x)+f(y)$
$\therefore$ $f(x+y)=f(x)+f(y)$
Now consider, $f(xy)$
$=f[(a_1+ib_1)(a_2+ib_2)]$
$=f[(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1)]$
$=(a_1a_2-b_1b_2)-i(a_1b_2+a_2b_1)$
$=a_1a_2-ia_1b_2-b_1b_2-ia_2b_1$
$=a_2(a_1-ib_1)-ib_2(a_1-ib_1)$
$=(a_1-ib_1)(a_2-ib_2)$
$=f(x)f(y)$
$\therefore$ $f(xy)=f(x)f(y)$
$\therefore$ $f$ is an homomorphism.
Consider $f(x)=f(y)$
$\Rightarrow$ $a_1+ib_1=a_2+ib_2$
$\Rightarrow$ $a_1=a_2$ and $b_1=b_2$
$\Rightarrow$ $x=a_1+ib_1=a_2+ib_2=y$
$\therefore$ $f(x)=f(y)$ $\Rightarrow$ $x=y$
$\therefore$ $f$ is $1-1$
The pre image of $a_1-ib_1$ is $a_1+ib_1$ $\Rightarrow$ $a_2-ib_2=f(a_1+ib_1)$
$\therefore$ $f$ is onto
Hence $f$ is an automorphism from $\mathbb{C}$ to $\mathbb{C}$. $\spadesuit$

3. Prove that the transformation $x+y\sqrt{3}$ $\rightarrow$ $x-y\sqrt{3}$ is an automorphism of the domain $\mathbb{Z}[\sqrt{3}]$.
Solution: Given, $x+y\sqrt{3}$ $\longleftrightarrow$ $x-y\sqrt{3}$ where $x,y \in \mathbb{Z}$
Define a map $f:\mathbb{Z}[\sqrt{3}] \rightarrow \mathbb{Z}[\sqrt{3}]$ by $f(x+y\sqrt{3})=x-y\sqrt{3}$
Let $x,y \in \mathbb{C}$ $\Rightarrow$ $x=a_1+b_1\sqrt{3}$ and $y=a_2+b_2\sqrt{3}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
Consider, $f(x+y)$
$=f[(a_1+b_1\sqrt{3})+(a_2+b_2\sqrt{3})]$
$=f[(a_1+a_2)+(b_1+b_2)\sqrt{3}]$
$=(a_1+a_2)-(b_1+b_2)\sqrt{3}$
$=(a_1-b_1\sqrt{3})+(a_2-b_2\sqrt{3})$
$=f(x)+f(y)$
$\therefore$ $f(x+y)=f(x)+f(y)$
Now consider, $f(xy)$
$=f[(a_1+b_1\sqrt{3})(a_2+b_2\sqrt{3})]$
$=f[(a_1a_2+3b_1b_2)+(a_1b_2+a_2b_1)\sqrt{3}]$
$=(a_1a_2+3b_1b_2)-(a_1b_2+a_2b_1)\sqrt{3}$
$=a_1a_2-a_1b_2\sqrt{3}+3b_1b_2-a_2b_1\sqrt{3}$
$=a_2(a_1-b_1\sqrt{3})-b_2\sqrt{3}(a_1-b_1\sqrt{3})$
$=(a_1-b_1\sqrt{3})(a_2-b_2\sqrt{3})$
$=f(x)f(y)$
$\therefore$ $f(xy)=f(x)f(y)$
$\therefore$ $f$ is an homomorphism.
Consider $f(x)=f(y)$
$\Rightarrow$ $a_1+b_1\sqrt{3}=a_2+b_2\sqrt{3}$
$\Rightarrow$ $a_1=a_2$ and $b_1=b_2$
$\Rightarrow$ $x=a_1+b_1\sqrt{3}=a_2+b_2\sqrt{3}=y$
$\therefore$ $f(x)=f(y)$ $\Rightarrow$ $x=y$
$\therefore$ $f$ is $1-1$
The pre image of $a_1-b_1\sqrt{3}$ is $a_1+b_1\sqrt{3}$ $\Rightarrow$ $a_2-b_2\sqrt{3}=f(a_1+b_1\sqrt{3})$
$\therefore$ $f$ is onto
Hence $f$ is an automorphism from $\mathbb{Z}[\sqrt{3}]$ to $\mathbb{Z}[\sqrt{3}]$.$\spadesuit$

4. Show that the transformation $a+b\sqrt{2}$ $\rightarrow$ $a+b\sqrt{3}$ is not an isomorphism from $\mathbb{Z}[\sqrt{2}]$ to $\mathbb{Z}[\sqrt{3}]$
Solution: Given, $a+b\sqrt{2}$ $\longleftrightarrow$ $a+b\sqrt{3}$ where $a,b \in \mathbb{Z}$
Define a map $f:\mathbb{Z}[\sqrt{2}] \rightarrow \mathbb{Z}[\sqrt{3}]$ by $f(a+b\sqrt{2})=a+b\sqrt{3}$
Let $a,b \in \mathbb{Z}[\sqrt{2}]$ $\Rightarrow$ $a=a_1+b_1\sqrt{2}$ and $b=a_2+b_2\sqrt{3}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$
Consider, $f(a+b)$
$=f[(a_1+b_1\sqrt{2})+(a_2+b_2\sqrt{2})]$
$=f[(a_1+a_2)+(b_1+b_2)\sqrt{2}]$
$=(a_1+a_2)+(b_1+b_2)\sqrt{3}$
$=(a_1+b_1\sqrt{3})+(a_2+b_2\sqrt{3})$
$=f(a)+f(b)$
$\therefore$ $f(a+b)=f(a)+f(b)$
Consider, $f(ab)$
$=f[(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})]$
$=f[(a_1a_2+2b_1b_2)+(a_1b_2+b_1a_2)\sqrt{2}]$
$=(a_1a_2+2b_1b_2)+(a_1b_2+b_1a_2)\sqrt{3}$ $..........(1)$
Consider, $f(a)f(b)$
$=(a_1+b_1\sqrt{3})(a_2+b_2\sqrt{3})$
$=(a_1a_2+3b_1b_2)+(a_1b_2+b_1a_2)\sqrt{3}$ $..........(2)$
From $(1)$ and $(2)$, we have $f(ab)\neq f(a) f(b)$
$\therefore$ $f$ is not an homomorphism $\Rightarrow$ $f$ is not an isomorphism.$\spadesuit$

5. If $\phi:R\rightarrow R'$ is an homomorphism, then show that $\phi(a-b)=\phi(a)-\phi(b)$.
Solution: Consider, $\phi(a-b)$
$=\phi[a+(-b)]$
$=\phi(a)+\phi(-b)$ [$\because$ $\phi$ is an homomorpshim]
$=\phi(a)-\phi(b)$ $\spadesuit$

Exercise 6

1. Find whether the transformation $f:\mathbb{Z} \rightarrow 2\mathbb{Z}$ defined by $f(x)=2x$ is a homomorphism or not.

2. If $\phi$ ia homomorphism of a ring $R$ into $R'$, then prove that $\phi(-a)=-\phi(a)$

3. Prove that the correspondence $a+b\sqrt{7}$ $\leftrightarrow$ $a-b\sqrt{7}$ is an automorphism in the domain $\mathbb{Z}[\sqrt{7}]$.

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