consider Z7 =[0,1,2,3......6+7,*7], show that Z7 is a field.
Answers
Answer:
Explanation:
Image 1 :- Addition Module 7
Image 2:- Multiplication Module 7
Refer Image 1 :- Each element in table is also in Z7. Hence Z is closed in +7.
Addition module is associative.
By referring 1st row in image 1, we can say 0 is additive.
Each element in Z7 has additive inverse. i.e. Inverse of 1 is 6
Z7 is additive with respect to +7.
Image 2:-
Each element is closed with respect to x7.
a,b,c ∈ z7
a x7 (b+7 c) = a x7 b + a x7 c
(a +7 b) x7b = a x7c +7 6x7 c is true for a,b,c ∈ z7
Hence Z7 is ring w.r.t addition modulus 7 and multiplication modulus 7.
Data in image 2 is symmetrical, hence z7 is commutative.
2nd row coincides with 1st row in table.
Hence, 1 is multiplicative identity of z7.
Which means Z7 is commutative ring of unity.
Each non-0 element of of Z7 has multiplicative inverse, hence z7 is a field.
Note:- Units of z7 are those elements which are relative prime of 7. Element prime to z7 - 1 to 6, so units of z7 is 1 to 6
Answer:
Addition module is associative.
By referring 1st row in image 1, we can say 0 is additive.
In general, element in Z7 has additive inverse. i.e. Inverse of 1 is 6 Z7 is additive with respect to +7.
With the reference of Image 2: Each element is closed with respect to x7. a,b,c ∈ z7 a x7 (b+7 c) = a x7 b + a x7 c, (a +7 b) x7b = a x7c +7 6x7 c is true for a,b,c ∈ z7.
Each non-0 element of of Z7 has multiplicative inverse, hence z7 is a field.