7
$\begingroup$

I am struggling to understand why is this the case. I need to prove this, but I don't understand how it's true. For example, if every non-zero element of $R$ has a multiplicative inverse, then it's a field. So how does $R=\{0\}$?

Thanks you for your time :)

2 Answers 2

10

We have $0 \cdot x = 1$ for some $x \in R$, so $1 = 0$, and it follows that for $x\in R$, $x = x\cdot 1 = x \cdot 0 = 0$.

  • 0
    Thanks man, this was a really bad question. For some reason I forgot that 1 is the identity and $0*0^{-1}=1$ – 2017-01-24
3

First of all if every non zero element of $R$ has a multiplicative inverse that makes it a division ring and not a field a good example is the quaternions it only becomes a field if the multiplication is commutative. Secondly from the axioms of ring theory $0$ does not have a multiplicative inverse except for the ring {$0$}

  • 1
    The [zero ring](https://en.wikipedia.org/wiki/Zero_ring), which has a single element, satisfies the axioms of a ring. That element is then both the additive identity and the multiplicative identity, so is $0=1$, and thus is its own additive and multiplicative inverse. – 2017-01-24
  • 0
    But I do get what you are saying though. Let me edit my answer :-) – 2017-01-24