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Prove that a nonzero ring R is not a group under multiplication. [Hint: what is the inverse of 0?]

I know that if R is a nonzero ring the for x,y in R xy=0 means either x=0 or y=0 and when x is not = 0 there is an x^-1 so that xx^-1=1. Also, with the hint I know that 0r doesn't have an inverse but I'm not sure how to correctly show this.

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    $\begingroup$ «I know that if R is a nonzero ring the for x,y in R xy=0 means either x=0 or y=0» you don't really know that, because it is not true! $\endgroup$ – Mariano Suárez-Álvarez Apr 10 '14 at 4:34
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    $\begingroup$ Assume 0 has an inverse. What does that mean? Write down an equation. Get a contradiction. (Also, when you say $xy=0$ implies one of $x,y$ is zero and that every nonzero element has an inverse, you seem to be describing a division ring, which is a special type of ring. Not all rings are division rings.) $\endgroup$ – anon Apr 10 '14 at 4:34
  • $\begingroup$ @anon, (he is describing an integral domain, which need not be a division ring) $\endgroup$ – Mariano Suárez-Álvarez Apr 10 '14 at 4:36
  • $\begingroup$ @MarianoSuárez-Alvarez A nontrivial ring with no zero divisors («for x,y in R xy=0 means either x=0 or y=0») whose nonzero elements admit right inverses («when x is not = 0 there is an x^-1 so that xx^-1=1») is necessarily a division ring. (I assume you skimmed over the second part.) $\endgroup$ – anon Apr 10 '14 at 4:50
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Assuming the ring has two distinct elements, try to show absorption and cancellation cannot coexist.

  • Absorption is when there is an absorbing element $0$, where $0a=0$ for all $a$.
  • Cancellation is when $ab=ac\Rightarrow b=c$ (left) or $ba=ca\Rightarrow b=c$ (right).

Rings have absorption and groups have cancellation.

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Let $(R, +, \cdot)$ be ring, with additive identity $0$ and multiplicative identity $1$. Suppose $(R, \cdot)$ is a group. Then $0$ has a multiplicative inverse, which we will denote by $0^{-1}$. Let $x \in R$. We know $x \cdot 0 = 0$ for any $x \in R$, so \begin{align*} 0 \cdot 0 = 0 &\implies 0^{-1} \cdot (0 \cdot 0) = 0^{-1} \cdot 0\\ &\implies (0^{-1} \cdot 0) \cdot 0 = 0^{-1} \cdot 0\\ &\implies 1 \cdot 0 = 1\\ &\implies 0 = 1 \end{align*} which implies that $R$ is trivial since for any $x \in R$, $$x = 1 \cdot x = 0 \cdot x = 0.$$

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