A definition of division ring says that its every element has an inverse under multiplication, except $0_R$, where $0_R$ is the additive identity. Why can't $0_R$ have such an inverse too?

  • $\begingroup$ What is your definition of multiplicative inverse? $\endgroup$
    – hardmath
    Jan 26, 2016 at 12:29

3 Answers 3


Suppose it has, let's call it $0^{-1}$. Then we have $1= 0*0^{-1}$, but from the usual axioms for rings we also have $0*a=0$ for all $a$ in the ring, therefore $0*0^{-1}=0$. That implies $1=0$ in the ring. Now, if $a$ is any element in the ring, $a = a*1 = a*0 = 0$, so our ring is the zero ring. Therefore, the zero ring is the only ring where 0 (the additive identity) has a multiplicative inverse.


$0.r+0.r=(0+0).r=0.r$ impies that $0.r=0$. This for every $r$.

So $0$ has a multiplicative inverse if and only if $0$ and $1$ coincide (i.e. if the ring is trivial).

A division ring is not trivial.


For every $x \in R$ we have $$ 0 \cdot x = (0 + 0) \cdot x = 0 \cdot x + 0 \cdot x $$ and thus $0 \cdot x = 0$. So for $0$ to have an inverse $a \in R$ we must have $$ 1 = 0 \cdot a = 0. $$ So for every $x \in R$ we then have $$ x = 1 \cdot x = 0 \cdot x = 0. $$ So the only ring $R$ in which $0$ is invertible is the zero ring $R = \{0\}$, often simply written as $0$ itself.

To exclude this from the definition of a division ring is a convention which is made because

  1. the zero ring $0$ often behaves differently from non-zero division rings, so we often would need to explicitely exclude $0$ when we make a statement about division ring, and
  2. we don’t need much theory to understand the zero ring or modules over the zero ring, and it is a very unuseful tool when it comes to understanding other situations.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.