Ring Field Integral Domain : Ring , Field and Integral Domain each of them is a group that satisfies some axioms.
Ring :
Field axioms
Example (1)
Prove that \(Q\left( {\sqrt 2 } \right) = \left\{ {a + b\sqrt 2 : a,b \in Q} \right\}\) is a field.
Solution:
Set \(Q\left( {\sqrt 2 } \right) \) need to satisfy following properties.
Let \(a_1 + b_1\sqrt 2 , a_2 + b_2\sqrt 2 , a_3 + b_3\sqrt 2∈Q\left( {\sqrt 2 } \right)\)
(1) \(Q\left( {\sqrt 2 } \right) \) is a commutative group under addition operation
(i) Closure
\[a_1 + b_1\sqrt 2+a_2 + b_2\sqrt 2=a_1+a_2+(b_1+b_2) \sqrt 2∈Q\left( {\sqrt 2 } \right)\]
(ii) Associativity
\[a_1 + b_1\sqrt 2+(a_2 + b_2\sqrt 2+a_3 + b_3\sqrt 2)=(a_1 + b_1\sqrt 2+a_2 + b_2\sqrt 2)+a_3 + b_3\sqrt 2\]
(iii) Commutative
\[a_1 + b_1\sqrt 2+a_2 + b_2\sqrt 2=a_2 + b_2\sqrt 2+a_1 + b_1\sqrt 2\]
(iv) Additive identity
\[a + b\sqrt 2+0+0\sqrt 2 =a + b\sqrt 2\]
(v) Additive inverse
\[a + b\sqrt 2-a-b√2 =0+0\sqrt 2\]
(2) \(Q\left( {\sqrt 2 } \right) \) is a commutative group under multiplication operation
(i) Closure
\[(a_1 + b_1\sqrt 2)(a_2 + b_2\sqrt 2)=(a_1a_2+2b_1b_2)+(a_1b_2+b_1a_2)\sqrt 2∈Q\left( {\sqrt 2 } \right)\]
(ii) Associativity
\[(a_1 + b_1\sqrt 2)*((a_2 + b_2\sqrt 2)*(a_3 + b_3\sqrt 2)) =((a_1 + b_1\sqrt 2) *(a_2 + b_2\sqrt 2))*(a_3 + b_3\sqrt 2)\]
(iii) Commutative
\[(a_1 + b_1\sqrt 2)*(a_2 + b_2\sqrt 2)=(a_2 + b_2\sqrt 2)*(a_1 + b_1\sqrt 2)\]
(iv) Multiplicative identity
\[(a + b\sqrt 2)*(1+0\sqrt 2) =a + b\sqrt 2\]
(v) Multiplicative inverse
Let either \(a_1≠0\) or \(b_1≠0\)
such that \(( a_1 + b_1\sqrt 2)(a_2 + b_2\sqrt 2)=1+0\sqrt 2\)
\[(a_1a_2+2b_1b_2)+(a_1b_2+b_1a_2)\sqrt 2=1+0\sqrt 2\]
comparing \[a_1a_2+2b_1b_2=1,a_1b_2+b_1a_2=0\]
If we write it as a system of matrix equation
\[\left( {\begin{array}{*{20}{c}}{{a_1}}&{2{b_1}}\\{{b_1}}&{{a_1}}\end{array}} \right)\left( \begin{array}{l}{a_2}\\{b_2}\end{array} \right) = \left( \begin{array}{l}1\\0\end{array} \right)\]
Since \(Det\left( {\begin{array}{*{20}{c}}{{a_1}}&{2{b_1}}\\{{b_1}}&{{a_1}}\end{array}} \right) =a_1^2-2b_1^2\)
If \(a_1^2- 2b_1^2 =0 \) then \( a_1=±b_1\sqrt 2\) which is not possible since \(b_1\sqrt 2\) is not a rational contradiction to \( a_1\) is a rational.
So \( a_1^2-2b_1^2≠0\)
Thus solution of the matrix equation exists.
Lets solve the system we get
\[a_2=\frac{a_1}{a_1^2- 2b_1^2 } ,b_2=-\frac{b_1}{a_1^2 – 2b_1^2 }\]
Thus multiplicative inverse of \(a_1 + b_1\sqrt 2\) is \(\frac{a_1}{a_1^2- 2b_1^2 }-\frac{b_1}{a_1^2 – 2b_1^2 }\sqrt 2\)
(3) Distributive property
\[ (a_1 + b_1\sqrt 2)*(a_2 + b_2\sqrt 2)+(a_3 + b_3\sqrt 2) =(a_1 + b_1\sqrt 2) *(a_2 + b_2\sqrt 2)+(a_1 + b_1\sqrt 2)*(a_3 + b_3\sqrt 2)\]
Since \(Q\left( {\sqrt 2 } \right)\) satisfies all the properties of a field hence \(Q\left( {\sqrt 2 } \right)\) is a field.
Example (2)
Prove that \(Q\left( {i } \right)={a+bi:a,b∈Q}\) is a field.
Solution:
Q(i)={a+bi:a,b∈Q}
Set \(Q\left( {i } \right) \) will be a field it satisfies following conditions.
Let \(a_1 + b_1i , a_2 + b_2i , a_3 + b_3i∈Q\left( {i } \right)\)
(1) \(Q\left( {i } \right) \) is a commutative group under addition operation
(i) Closure
\[a_1 + b_1i+a_2 + b_2i=a_1+a_2+(b_1+b_2) i∈Q\left( {i } \right)\]
(ii) Associativity
\[a_1 + b_1i+(a_2 + b_2i+a_3 + b_3i)=(a_1 + b_1i+a_2 + b_2i)+a_3 + b_3i\]
(iii) Commutative
\[a_1 + b_1i+a_2 + b_2i=a_2 + b_2i+a_1 + b_1i\]
(iv) Additive identity
\[a + bi+0+0i=a + bi\]
(v) Additive inverse
\[a + bi-a-bi =0+0i\]
(2) \(Q\left( {i } \right) \) is a commutative group under multiplication operation
(i) Closure
\[(a_1 + b_1i)(a_2 + b_2i)=(a_1a_2- b_1b_2)+(a_1b_2+b_1a_2)i∈Q\left( {i } \right)\]
(ii) Associativity
\[(a_1 + b_1i)*((a_2 + b_2i)*(a_3 + b_3i)) =((a_1 + b_1i) *(a_2 + b_2i))*(a_3 + b_3i)\]
(iii) Commutative
\[(a_1 + b_1i)*(a_2 + b_2i)=(a_2 + b_2i)*(a_1 + b_1i)\]
(iv) Multiplicative identity
\[(a + bi)*(1+0i) =a + bi\]
(v) Multiplicative inverse
Let either \(a_1≠0\) or \(b_1≠0\)
such that \(( a_1 + b_1i)(a_2 + b_2i)=1+0i\)
Since \(i^2 =- 1\)
\[(a_1a_2-b_1b_2)+(a_1b_2+b_1a_2)i=1+0i\]
comparing \[a_1a_2-b_1b_2=1,a_1b_2+b_1a_2=0\]
If we write it as a system of matrix equation
\[\left( {\begin{array}{*{20}{c}}{{a_1}}&{-{b_1}}\\{{b_1}}&{{a_1}}\end{array}} \right)\left( \begin{array}{l}{a_2}\\{b_2}\end{array} \right) = \left( \begin{array}{l}1\\0\end{array} \right)\]
Since \(Det\left( {\begin{array}{*{20}{c}}{{a_1}}&{-{b_1}}\\{{b_1}}&{{a_1}}\end{array}} \right) =a_1^2+b_1^2\)
Since \(a_1^2+b_1^2 =0 \) is not possible because we have let at-least one of \( a_1 , b_1 \) is non-zero.
So \( a_1^2+b_1^2≠0\)
So solution of the matrix equation can be determined.
If we solve the system we get
\[a_2=\frac{a_1}{a_1^2+b_1^2 } ,b_2=-\frac{b_1}{a_1^2 +b_1^2 }\]
Thus multiplicative inverse of \(a_1 + b_1i\) is \(\frac{a_1}{a_1^2+b_1^2 }-\frac{b_1}{a_1^2 +b_1^2 }i\)
(3) Distributive property
\[ (a_1 + b_1i)*(a_2 + b_2i)+(a_3 + b_3i) =(a_1 + b_1i) *(a_2 + b_2i)+(a_1 + b_1i)*(a_3 + b_3i)\]
Since \(Q\left( {i} \right)\) satisfies all the properties of a field hence \(Q\left( {i } \right)\) is a field.