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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?

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  • $\begingroup$ What is your definition of multiplicative inverse? $\endgroup$ – hardmath Jan 26 '16 at 12:29
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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.

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$0.r+0.r=(0+0).r=0.r$ 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.

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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.
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