##\mathbb{Z_2}## a field?

[romsofia]

I was helping someone with a problem today, and it was about showing that $\mathbb{Z_2}$ was a field. It's been a while since I've done abstract algebra, but to my knowledge, this means that there has to exist a multiplication inverse for 0? But I don't see how it would be allowed in this set.
$0*a=1 \rightarrow 0=a^{-1}$ but $0*0=0$. Is there something I'm overlooking? Potentially my math is off too, any insights would be helpful.

 

[fresh_42]

I was helping someone with a problem today, and it was about showing that $\mathbb{Z_2}$ was a field. It's been a while since I've done abstract algebra, but to my knowledge, this means that there has to exist a multiplication inverse for 0? But I don't see how it would be allowed in this set.
$0*a=1 \rightarrow 0=a^{-1}$ but $0*0=0$. Is there something I'm overlooking? Potentially my math is off too, any insights would be helpful.

No, there cannot be a multiplicative inverse to $0$, it does not belong to the multiplicative group, which in this case is simply $\{\,1\,\}$.
Multiplication by zero merley comes from the distributive law: $a\cdot 0 := a\cdot (1+(-1)) = a -a = 0$. You see, that it isn't even defined as a multiplication by zero, only by the requirement, that distribution holds for all elements between the two groups. So how there can be an inversion, if there isn't even a multiplication? It only drops out of another rule.

 

[romsofia]

No, there cannot be a multiplicative inverse to $0$, it does not belong to the multiplicative group, which in this case is simply $\{\,1\,\}$.
Multiplication by zero merley comes from the distributive law: $a\cdot 0 := a\cdot (1+(-1)) = a -a = 0$. You see, that it isn't even defined as a multiplication by zero, only by the requirement, that distribution holds for all elements between the two groups. So how there can be an inversion, if there isn't even a multiplication? It only drops out of another rule.

So, at the time, I told the student to ignore the 0 case and talk to their teacher, and prove it for the 1 case. Should that be the advice I have in the future for these kinds of questions?

 

[fresh_42]

So, at the time, I told the student to ignore the 0 case and talk to their teacher, and prove it for the 1 case. Should that be the advice I have in the future for these kinds of questions?

If you like.

$0$ is simply not part of the multiplicative system. We start with an additive structure of all elements that has a $1$ so that we can count. In cases like $\mathbb{Z}_2$ where $1+1=0$ it is counting as well, only that in this case skipping the light switch twice brings us back to where we started, but we counted. And of course we want $1 \cdot a = a$. So up to now we only can add, count and add (subtract) numbers to themselves. To make sense of it, we find the distribution law. In the last step, we ask, if we could reverse those multiplications. This requires a multiplication with division, but it does not require to have this for all elements. Namely $0$ cannot be part of the multiplication, except for what distribution allows us to do. Assuming a division by zero will instantly create contradictions: $1 = 0 / 0 = 0 \cdot [0]^{-1} = (1 -1) \cdot [0]^{-1}= 1 \cdot [0]^{-1} - 1 \cdot [0]^{-1} = [0]^{-1} - [0]^{-1} = 0$ and we cannot distinguish the dot operation from the plus operation anymore.