## Algebra - Rings and fields

### 4. Subfields

• Definition 4.1: Let $F$ be a field. A non empty subset $K$ of $F$ is said to be a subfield of field $F$, if $K$ is a field with respect to the same binary operations defined on $F$.

• Example: $\mathbb{Q}$ is a subfield of $\mathbb{R}$

• Theorem 4.1: (The necessary and sufficient condition) A non empty subset $K$ of field $F$ is said to be a subfield if:
(i) $a,b \in K$ $\Rightarrow$ $a-b \in K$
(ii) $a,b\neq 0 \in K$ $\Rightarrow$ $ab^{-1} \in K$
Proof: Necessary condition: Let $K$ be a subfield of $F$ $\Rightarrow$ $K$ is a field.
For $b \in K$ $\Rightarrow$ $-b \in K$ (closure law in $K$). Let $a,-b \in K$ $\Rightarrow$ $a+(-b) \in K$ (closure law in $K$) $\Rightarrow$ $a-b \in K$
Let $b\neq 0 \in K$ $\Rightarrow$ $b^{-1} \in K$ (inverse law in $K$). So for $a,b^{-1} \in K$ $\Rightarrow$ $ab^{-1} \in K$ (closure law in $K$)
Sufficient condition: Let $K$ be a non empty subset of $F$ such that $a,b \in K$ $\Rightarrow$ $a-b \in K$ and $a,b\neq 0 \in K$ $\Rightarrow$ $ab^{-1} \in K$
To prove: $K$ is a field. Let $a,b,c \in K$
(1.) Associative law: Associative law holds in $S$, since $S \subset R$ and associative law holds in $R$. i.e., $a+(b+c)=(a+b)+c$ $\forall$ $a,b,c \in K$
(2.) Existence of additive identity: For $a,a \in K$ $\Rightarrow$ $a-a \in K$ $\Rightarrow$ $0 \in K$. $0$ is the additive identity in $K$.
(3.) Existence of additive inverse: For $0,a \in K$ $\Rightarrow$ $0-a \in K$ $\Rightarrow$ $-a \in K$. Additive inverse of every element in $K$ exists.
(4.) Commutative law: Commutative law holds in $S$, since $S \subset R$ and commutative law holds in $R$. i.e., $a+b=b+a$ $\forall$ $a,b \in K$
(5.) Closure law: $\forall$ $a \in K$ $\Rightarrow$ $-a \in K$ So for $b,-a \in K$ $\Rightarrow$ $b-(-a) \in K$ $\Rightarrow$ $a+b \in K$
(6.) Existence of multiplicative identity: Let $a,a\neq 0 \in K$ $\Rightarrow$ $aa^{-1} \in K$ $\Rightarrow$ $aa^{-1}=1 \in K$. $1$ is the multiplicative idenity of $K$.
(7.) Existence of multiplicative inverse: For $1,a\neq 0 \in K$ $\Rightarrow$ $a^{-1} \in K$. So for each non zero $a \in K$ $\exists$ $a^{-1} \in K$.
(8.) Associative law: Associative law holds in $S$, since $S \subset R$ and associative law holds in $R$. i.e., $a(bc)=(ab)c$ $\forall$ $a,b,c \in K$
(9.) Distributive law: Distributive law holds in $S$, since $S \subset R$ and distributive law holds in $R$. i.e., $a(b+c)=ab+ac$ and $(a+b)c=ac+bc$ $\forall$ $a,b,c \in K$
(10.) Commutative law: Commutative law holds in $S$, since $S \subset R$ and commutative law holds in $R$. i.e., $ab=ba$ $\forall$ $a,b \in K$
(11.) Closure law: For $a,b^{-1} \in K$ $\Rightarrow$ $a(b^{-1})^{-1} \in K$ $\Rightarrow$ $ab \in K$
$\therefore$ $K$ is a field. Since $K$ is a subset of $F$, it is a subfield of $F$ $\blacksquare$

• Theorem 4.2: Intersection of two subfields is a subfield.
Proof: Let $K_1$ and $K_2$ be two subfields of a field $F$. We need to prove that $K_1 \cap K_2$ is a subfield of $K$. So it is enough to prove that if $a,b \in K_1 \cap K_2$ $\Rightarrow$ $a-b \in K_1 \cap K_2$ and $ab^{-1} \in K_1 \cap K_2$.
Since $0 \in K_1$ and $0 \in K_2$ $\Rightarrow$ $0 \in K_1 \cap K_2$ $\therefore$ $K_1 \cap K_2$ is non empty and so $K_1\cap K_2$ is a non empty subset of $F$.
Let $a,b \in S_1 \cap S_2$ $\Rightarrow$ $a \in S_1$ and $a \in S_2$, $b \in K_1$ and $b \in K_2$
So, we have $a,b \in K_1$ and $a,b \in K_2$
Since $K_1$ is a subfield of $F$, we have $a-b \in K_1$ and $ab^{-1} \in K_1$
Since $K_2$ is a subfield of $F$, we have $a-b \in K_2$ and $ab^{-1} \in K_2$
So $a-b \in K_1 \cap K_2$ and $ab^{-1} \in K_1 \cap K_2$. $\blacksquare$

Problems 4
1. Show that $\mathbb{Q}$ is a subfield of $\mathbb{R}$
Solution: $\mathbb{Q}= \Big\{\frac{a}{b} \big/ a,b \in \mathbb{Z}$ and $b\neq 0 \Big\}$
Clearly $\mathbb{Q} \subseteq \mathbb{R}$ and $\mathbb{Q}$ is non empty.
Let $x,y \in \mathbb{Q}$ $\Rightarrow$ $x=\frac{a_1}{b_1}$, $y=\frac{a_2}{b_2}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Z}$ and $b_1,b_2 \neq 0$
(1.) Consider $x-y=\frac{a_1}{b_1}-\frac{a_2}{b_2} \in \mathbb{Q}$
(2.) Consider $xy^{-1}=\frac{a_1}{b_1}\frac{b_2}{a_2}=\frac{a_1b_2}{b_1a_2} \in \mathbb{Q}$
$\therefore$ $\mathbb{Q}$ is a subfield of $\mathbb{R}$ $\spadesuit$

2. Show that the set of numbers of the form $a+b\sqrt{2}$ where $a$ and $b$ are rational numbers, is a subfield of $\mathbb{R}$.
Solution: Consider $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}/a,b \in \mathbb{Q}\}$
Clearly $\mathbb{Q}[\sqrt{2}] \subseteq \mathbb{R}$ and $0+0\sqrt{2} \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $0 \in \mathbb{Q}[\sqrt{2}]$. So $\mathbb{Q}[\sqrt{2}]$ is non empty.
Let $x,y \in \mathbb{Q}[\sqrt{2}]$ $\Rightarrow$ $x=a_1+b_1\sqrt{2}$, $y=a_2+b_2\sqrt{2}$ where $a_1,a_2,b_1,b_2 \in \mathbb{Q}$
(1.) Consider, $x-y=(a_1+b_1\sqrt{2})-(a_2+b_2\sqrt{2})=(a_1-a_2)+(b_1-b_2)\sqrt{2} \in \mathbb{Q}[\sqrt{2}]$
(2.) Consider, $xy^{-1}$
$=(a_1+b_1\sqrt{2})(a_2+b_2\sqrt{2})^{-1}$
$=\frac{a_1+b_1\sqrt{2}}{a_2+b_2\sqrt{2}}$
$=\frac{a_1+b_1\sqrt{2}}{a_2+b_2\sqrt{2}} \times \frac{a_2-b_2\sqrt{2}}{a_2-b_2\sqrt{2}}$
$=\frac{(a_1a_2-2b_1b_2)+(a_2b_1-a_1b_2)\sqrt{2}}{a_2^2-2b_2^2}$
$=\frac{(a_1a_2-2b_1b_2)}{a_2^2-2b_2^2}+\frac{(a_2b_1-a_1b_2)}{a_2^2-2b_2^2}\sqrt{2} \in \mathbb{Q}[\sqrt{2}]$
$\therefore$ $\mathbb{Q}[\sqrt{2}]$ is a subfield of $R$. $\spadesuit$

Exercise 4
1. Show that $\mathbb{Q}[\omega]$ is a subfield of complex numbers.

2. Show that $\mathbb{Q}[\sqrt{3}]$ is a subfield of realnumbers.

3. Show that $\mathbb{Z}$ is not a subfield of $\mathbb{R}$.

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