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Prove or disprove Let $\langle R,+,\cdot\rangle $ be a ring and $a,b \in R$.

  1. If $a\cdot b=0$ then $a=0$ or $b=0$
  2. If $a\cdot a=a$ then $a=1$ or $a= 0$ if $(1\in R)$

I think that '1.' is a false statement.

Counterexample: $\langle Z,+,.\rangle$ is a ring with $a\cdot b=a+b,\,\,\forall a ,b \in R$, where $a=1$ and $b=-1$ then $a\cdot b=1+(-1)=0$. But $a$ and $b$ don't equal zero.

Is that true? And what about the second one?

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    $\begingroup$ I took the liberty of reformatting your problem statement. Can you please edit and insert the work you've done on the problem so far? $\endgroup$ – Matthew Leingang Sep 30 '17 at 17:02
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Both are false. Take as $R$ the ring of $2\times2$ matrices with integer coefficients.

  1. $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right).\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)=\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$, but $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)\neq\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right)$.
  2. $\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right).\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$, but $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)\notin\left\{\left(\begin{smallmatrix}0&0\\0&0\end{smallmatrix}\right),\left(\begin{smallmatrix}1&0\\0&1\end{smallmatrix}\right)\right\}$.
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  • $\begingroup$ That's what I need thank you (+1) $\endgroup$ – Tasneem Sep 30 '17 at 17:36
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One nice counterexample is the ring $R=\mathbb{Z}/6\mathbb{Z}$. In that ring, we have $[2]\cdot [3]=[0]$ and $[3]\cdot [3]=[3]$.

I'm puzzled by the counterexample you provided in your question. It's not true that $\langle\mathbb{Z},+,\cdot\rangle$ is a ring if you redefine multiplication to be addition. Multiplication has to be the operation that distributes over the other operation.

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  1. The first is asking about zero divisors. In particular it asks if all rings are integral domains. This is of course not the case. See here.
  2. The second is asking about idempotents. In an integral domain, the only idempotents are the additive identity $0$ and the multiplicative identity $1$. Outside of an integral domain, this is not necessarily the case.

Indeed, a non-trivial idempotent (nontrivial not equal to $0$ or $1$) is a zero divisor. Let $a\in R$ be a nontrivial idempotent, so that $a^2=a$. Then $$a(a-1)=a^2-a=a-a=0.$$ So $a^2$ and $a-1$ are zero divisors (unless $a=0$ or $a=1$).

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