## Algebra - Rings and fields

### 3. subring

• Definition 3.1: Let $R$ be a ring. A non empty subset $S$ of $R$ is said to be a subring of ring $R$ if $S$ is a ring w.r.t the same binary operations defined on $R$.

• Examples: $\mathbb{Z}$ is a subring of $\mathbb{R}$, $\mathbb{Z}[i]$ is a subring of $\mathbb{C}$.

• Theorem 3.1: (The necessary and sufficient condition) A non empty subset $S$ of a ring $R$ is said to be a subring are if:
(i) $a,b \in S$ $\Rightarrow$ $a-b \in S$
(ii) $a,b \in S$ $\Rightarrow$ $ab \in S$
Proof: Necessary condition: Let $S$ be a subring of ring $R$ $\Rightarrow$ $S$ is a ring.
For $b \in S$ $\Rightarrow$ $-b \in S$ (inverse law in $S$). Let $a,-b \in S$ $\Rightarrow$ $a+(-b) \in S$ (closure law in $S$) $\Rightarrow$ $a-b \in S$.
Also if $a,b \in S$ $\Rightarrow$ $ab \in S$ (closure law in $S$)
Sufficient condition: Let $S$ be a non empty subset of ring $R$ with $a,b \in S$ $\Rightarrow$ $a-b \in S$ and $a,b \in S$ $\Rightarrow$ $ab \in S$
To prove: $S$ is a subring. Let $a,b,c \in S$
(1.) Associative law: Associative law holds in $S$, since $S \subseteq R$ and associative law holds in $R$.
(2.) Existence of additive identity: $\forall$ $a,a \in S$ $\Rightarrow$ $a-a \in S$ $\Rightarrow$ $0 \in S$. $0$ is the additive identity.
(3.) Existence of additive inverse: For $0,a \in S$ $\Rightarrow$ $0-a \in S$ $\Rightarrow$ $-a \in S$. Additive inverse of every element in $S$ exists.
(4.) Commutative law: Commutative law holds in $S$, since $S \subseteq R$ and commutative law holds in $R$.
(5.) Closure law: $\forall$ $a \in S$ $\Rightarrow$ $-a \in S$. So, $b,-a \in S$ $\Rightarrow$ $b-(-a) \in S$ $\Rightarrow$ $a+b \in S$
(6.) Closure law: $\forall$ $a,b \in S$, we have $ab \in$ $S$.
(7.) Associative law: Associative law holds in $S$, since $S \subseteq R$ and associative law holds in $R$.
(8.) Distributive law: Distributive law holds in $S$, since $S \subseteq R$ and distributive law holds in $R$.
$\therefore$ $S$ is a subring. $\blacksquare$

• Theorem 3.2: Intercection of two subrings is a subring.
Proof: Let $S_1$ and $S_2$ be two subrings of a ring $R$. We need to prove that $S_1 \cap S_2$ is a subring of $R$. So it is enough to prove that if $a,b \in S_1 \cap S_2$ $\Rightarrow$ $a-b \in S_1 \cap S_2$ and $ab \in S_1 \cap S_2$.
Since $0 \in S_1$ and $0 \in S_2$ $\Rightarrow$ $0 \in S_1 \cap S_2$ $\therefore$ $S_1 \cap S_2$ is non empty and so $S_1 \cap S_2$ is a non empty subset of $R$.
Let $a,b \in S_1 \cap S_2$ $\Rightarrow$ $a \in S_1$ and $a \in S_2$, $b \in S_1$ and $b \in S_2$
So, we have $a,b \in S_1$ and $a,b \in S_2$
Since $S_1$ is a subring of $R$ we have $a-b \in S_1$ and $ab \in S_1$
Since $S_2$ is a subring of $R$ we have $a-b \in S_2$ and $ab \in S_2$
So $a-b \in S_1 \cap S_2$ and $ab \in S_1 \cap S_2$ $\blacksquare$

• Remark: Union of two subrings of a ring may not be a subring.
Example: Let $S_1=\{...,-6,-4,-2,0,2,4,6,...\}$ and $S_2=\{...-9,-6,-3,0,3,6,9,...\}$. Then clearly $S_1$ and $S_2$ are subrings of the ring $\mathbb{Z}$.
But $S_1 \cup S_2=\{...,-6,-4,-3,-2,0,2,3,4,6,...\}$ is not a subring of $\mathbb{Z}$. (Why?)

• Problem 3.1: Show that the set of even integers is a subring of ring $\mathbb{Z}$.
Solution: Set of integers, $\mathbb{Z}=\{0,\pm 1,\pm 2,\pm 3,...\}$
Set of even integers, $2\mathbb{Z}=\{0,\pm 2, \pm 4, \pm 6,...\}$
Clearly $2\mathbb{Z} \subseteq \mathbb{Z}$ and $2\mathbb{Z}$ is non empty.
Let $a,b \in 2\mathbb{Z}$ $\Rightarrow$ $a=2m$ and $b=2n$ for some integers $m$ and $n$.
(1.) Consider, $a-b=2m-2n=2(m-n) \in 2\mathbb{Z}$
(2.) Consider, $ab=(2n)(2m)=2(2nm) \in 2\mathbb{Z}$
$\therefore$ $2\mathbb{Z}$ is a subring of $\mathbb{Z}$ $\spadesuit$

• Problem 3.2: Show that the ring of gaussian integers is a subring of $\mathbb{C}$.
Solution: Set of complex numbers, $\mathbb{C}=\{a+ib$ /$a,b \in \mathbb{R}$ and $i=\sqrt{-1}\}$
Set of gaussian integers, $\mathbb{Z}[i]=\{a+ib$ /$a,b \in \mathbb{Z}$ and $i=\sqrt{-1}\}$
Clearly $\mathbb{Z}[i] \subseteq \mathbb{C}$ and $\mathbb{Z}[i]$ is non empty.
Let $x,y \in \mathbb{Z}[i]$ $\Rightarrow$ $x=a_1+ib_1$, $y=a_2+ib_2$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$ and $i=\sqrt{-1}$
(1.) Consider $x-y=(a_1+ib_1)-(a_2+ib_2)=(a_1-a_2)+i(b_1-b_2) \in \mathbb{Z}[i]$
(2.) Consider, $xy =(a_1+ib_1)(a_2+ib_2)$ $=a_1a_2+ib_1a_2+ia_1b_2+i^2b_1b_2$ $=(a_1a_2-b_1b_2)+i(a_1b_2+a_2b_1) \in$ $\mathbb{Z}[i]$
$\therefore$ $\mathbb{Z}[i]$ is a subring of $\mathbb{C}$ $\spadesuit$

• Problem 3.3: Show that $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$ is a subring of a ring of $2 \times 2$ real matrices.
Solution: Consider $\mathbb{M}=\Bigg\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big/ a,b,c,d \in \mathbb{R} \Bigg\}$
Given $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & b \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$
Clearly $S \subseteq R$ and Since $0=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$ $\Rightarrow$ $S$ is non empty
Let $X,Y \in S$ $\Rightarrow$ $X=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}$, $Y=\begin{pmatrix} a_2 & 0 \\ 0 & b_2 \end{pmatrix}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$.
(1.) Consider $X-Y$ $=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}-\begin{pmatrix} a_2 & 0 \\ 0 & b_2 \end{pmatrix}$ $=\begin{pmatrix} a_1-a_2 & 0 \\ 0 & b_1-b_2 \end{pmatrix} \in S$
(2.) Consider $XY$ $=\begin{pmatrix} a_1 & 0 \\ 0 & b_1 \end{pmatrix}\begin{pmatrix} a_2 & 0 \\ 0 & b_2 \end{pmatrix}$ $=\begin{pmatrix} a_1a_2 & 0 \\ 0 & b_1b_2 \end{pmatrix} \in S$
$\therefore$ $S$ is a subring of $\mathbb{M}$. $\spadesuit$

• Problem 3.4: Show that $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \Big/ a \in \mathbb{Z} \Bigg\}$ is a subring of a ring of $2 \times 2$ real matrices.
Solution: Consider $\mathbb{M}=\Bigg\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \Big/ a,b,c,d \in \mathbb{R} \Bigg\}$
Given $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \Big/ a \in \mathbb{Z} \Bigg\}$
Clearly $S \subseteq \mathbb{M}$ and Since $0=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$ $\Rightarrow$ $S$ is non empty.
Let $X,Y \in S$ $\Rightarrow$ $X=\begin{pmatrix} a_1 & 0 \\ 0 & 0 \end{pmatrix}$, $Y=\begin{pmatrix} a_2 & 0 \\ 0 & 0 \end{pmatrix}$ where $a_1,a_2 \in \mathbb{Z}$.
(1.) Consider $X-Y$ $=\begin{pmatrix} a_1 & 0 \\ 0 & 0 \end{pmatrix}-\begin{pmatrix} a_2 & 0 \\ 0 & 0 \end{pmatrix}$ $=\begin{pmatrix} a_1-a_2 & 0 \\ 0 & 0 \end{pmatrix} \in S$
(2.) Consider $XY$ $=\begin{pmatrix} a_1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 0 \\ 0 & 0 \end{pmatrix}$ $=\begin{pmatrix} a_1a_2 & 0 \\ 0 & 0 \end{pmatrix} \in S$
$\therefore$ $S$ is a subring of $\mathbb{M}$. $\spadesuit$

• Remark: $\mathbb{M}$ is a non commutative ring and $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \Big/ a \in \mathbb{Z} \Bigg\}$ is a commutative ring (Verify!) which is a subring of $\mathbb{M}$. So, a subring of a non commutative ring can be commutative.
• Problem 3.5: Show that $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \Big/ a \in \mathbb{Z} \Bigg\}$ is a subring of a ring $\mathbb{M}=\Bigg\{ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$
Solution: Given $\mathbb{M}=\Bigg\{ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \Big/ a,b \in \mathbb{R} \Bigg\}$ and $S=\Bigg\{ \begin{pmatrix} a & 0 \\ 0 & 0 \end{pmatrix} \Big/ a \in \mathbb{Z} \Bigg\}$
Clearly $S \subseteq \mathbb{M}$ and Since $0=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$ $\Rightarrow$ $S$ is non empty.
Let $X,Y \in S$ $\Rightarrow$ $X=\begin{pmatrix} a_1 & 0 \\ 0 & 0 \end{pmatrix}$, $Y=\begin{pmatrix} a_2 & 0 \\ 0 & 0 \end{pmatrix}$ where $a_1,a_2 \in \mathbb{Z}$.
(1.) Consider $X-Y$ $=\begin{pmatrix} a_1 & 0 \\ 0 & 0 \end{pmatrix}-\begin{pmatrix} a_2 & 0 \\ 0 & 0 \end{pmatrix}$ $=\begin{pmatrix} a_1-a_2 & 0 \\ 0 & 0 \end{pmatrix} \in S$
(2.) Consider $XY$ $=\begin{pmatrix} a_1 & 0 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} a_2 & 0 \\ 0 & 0 \end{pmatrix}$ $=\begin{pmatrix} a_1a_2 & 0 \\ 0 & 0 \end{pmatrix} \in S$
$\therefore$ $S$ is a subring of $\mathbb{M}$. $\spadesuit$

• Problem 3.6: Show that the set of matrices of the form $\begin{bmatrix} a&a\\a&a \end{bmatrix}$ where $a \in \mathbb{R}$ is a subring of $2 \times 2$ real matrices.
Solution: Consider $\mathbb{M}=\Bigg\{ \begin{pmatrix} a & b \\ 0 & 0 \end{pmatrix} \Big/ a,b \in \mathbb{R} \Bigg\}$
and let and $S=\Bigg\{ \begin{pmatrix} a & a \\ a & a \end{pmatrix} \Big/ a \in \mathbb{R} \Bigg\}$
Clearly $S \subseteq \mathbb{M}$ and Since $0=\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \in S$ $\Rightarrow$ $S$ is non empty.
Let $X,Y \in S$ $\Rightarrow$ $X=\begin{pmatrix} a & a \\ a & a \end{pmatrix}$, $Y=\begin{pmatrix} b & b \\ b & b \end{pmatrix}$ where $a,b \in \mathbb{R}$.
(1.) Consider $X-Y$ $=\begin{pmatrix} a&a\\a&a \end{pmatrix}-\begin{pmatrix} b&b\\b&b \end{pmatrix}$ $=\begin{pmatrix} a-b&a-b\\a-b&a-b \end{pmatrix} \in S$
(2.) Consider $XY$ $=\begin{pmatrix} a&a\\a&a \end{pmatrix}\begin{pmatrix} b&b\\b&b \end{pmatrix}$ $=\begin{pmatrix} 2ab&2ab\\2ab&2ab \end{pmatrix} \in S$
$\therefore$ $S$ is a subring of $\mathbb{M}$. $\spadesuit$

• Remark: The identity of $\mathbb{M}$ is $\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$ and the identity of $S$ is $\begin{bmatrix} \frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&\frac{1}{2} \end{bmatrix}$. So, the identity of a subring may be different from the idenity of a ring.
• Problem 3.7: Let $R$ be a ring and $a$ be a fixed element of $R$. Show that $Ra=\{x \in R/ ax=0\}$ is a subring of $R$.
Solution: Given, $R$ is a ring and $Ra=\{x \in R/ ax=0\}$
Clearly $Ra$ is a non empty subset of $R$.
Let $x,y \in Ra$ $\Rightarrow$ $ax=0$ and $ay=0$
(1.) Consider $a(x-y)=ax-ay=0-0=0$ $\therefore$ $x-y \in Ra$
(2.) Consider $a(xy)=(ax)y=0$ $\therefore$ $xy \in Ra$
$\therefore$ $Ra$ is a subring of $R$. $\spadesuit$

• Exercise 3.1: Show that $S=\Bigg\{ \begin{pmatrix} 0 & a \\ 0 & b \end{pmatrix} \Big/ a,b \in \mathbb{Z} \Bigg\}$ is a subring of a $R$ where $R$ is a ring of $2 \times 2$ matrices.

• Exercise 3.2: Let $R$ be a ring and let $C$ be the center of the ring. Prove that $C$ is a subring of $R$.

• Exercise 3.3: Let $R$ be the ring of integers. Let $a$ be any fixed integer and $S$ be any subset of $R$ defined by $S=\{ax/x \in \mathbb{Z\}}$. Then prove that $S$ is a subring of $R$.

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