why isnt Z mod 6 a field?
this is an extra credit problem in my pre-calc class...
I know that it must abide to the field axioms(am i right?)
but mostly i dont understand the form Z mod 6
Okay, the basic answer is that it is not a power of a prime. But that is a more advanced answer.
By you can do it by checking whether it is a field or not.
The set is,
It froms a group under addition.
Furthermore, it is a ring (just check the definitions for a ring).
Furthermore, it is a commutative ring.
Furthermore, it has unity (1).
Thus, it is a commutative ring with unity.
Now, if every non-zero element has an inverse the proof is complete.
Thus, we need to show every element has a multiplicative inverse.
The element 2, has no inverse!
Just check each one and convince thyself.
It is not a fair question for a pre-calculus class.
Just write on your paper, that not all elements (numbers in Z6) has an inverse, for example 2 is in Z6 and it got no inverse.
Inverse means we need to find a number 'x' (in Z6) such that,
2x=1
Now, just check all the possibilities,
2(0)=0
2(1)=2
2(2)=4
2(3)=6=0 (because we take the mod of 6)
2(4)=8=2 (same reason)
2(5)=10=4
Note, we went through all possibilities because,
Z6= { 0,1,2,3,4,5 }
And none of them produce 1.