1. ## Finite field

So I uploaded a picture that is written a claim that I didnt understand. This is the way that is written in my notebook and I didnt understand why its not a field.
So from what I have been told that it is not field because some numbers in Zn their inverse. But how is this proof , proving that exactly?
And I wanted to ask that if this is a contradiction to the uniqueness of 0? I mean in multiplication only when we multiply with zero we get zero but here we didnt multiply with zero is that contradiction to that? thank you

2. ## Re: Finite field

Originally Posted by davidciprut
So I uploaded a picture that is written a claim that I didnt understand. This is the way that is written in my notebook and I didnt understand why its not a field. So from what I have been told that it is not field because some numbers in Zn their inverse. But how is this proof , proving that exactly? And I wanted to ask that if this is a contradiction to the uniqueness of 0? I mean in multiplication only when we multiply with zero we get zero but here we didnt multiply with zero is that contradiction to that? thank you
It simply says that if $n$ is not prime then $\mathbb{Z}_n$ is not a field.
You will have non-zero divisors of zero.

In $\mathbb{Z}_{12}$ we have $2\cdot 6=0$.

BUT in $\mathbb{Z}_{13}$ and $a\ne 0~\&~b\ne 0$ we have $a\cdot b \ne 0$

3. ## Re: Finite field

So the reason is it's a contradiction to the uniqueness of 0?

4. ## Re: Finite field

No, it contradicts the fact that every non-zero member has a multiplicative inverse, which itself implies that there cannot be 'zero divisors', non-zero numbers whose product is 0. In particular if 6 had an inverse in $Z_{12}$, then multiplying both sides of $6*2= 0$ would give $6^{-1}(6*2)= (6^{-1}*6)2= 1*2= 2= 0$ which is false.

5. ## Re: Finite field

Def: Zn is a field if ax=b(modn) has a unique solution (modn) for a≠0.

Theorem: Zn is a field if n =p, p prime.

It follows that: ab=0(modp), a≠0, -> b=0.

Def: If ab=0(modn), a≠0, b≠0, a and b are divisors of zero.

There are no divisors of zero in Zp.