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Commutative ring with unit is defined as $(R,+,\times)$, where $(R,+)$ is abelian group and $(R,\times)$ is commutative multiplicative monoid with $1$ and $+$ and $\times$ satisfies distributive law.

Could you give me an example $(R,+,\times)$ cannnot be a ring because $+$ and $\times$ does not satisfy distributive law although $(R,+)$ is abelian group and $(R,\times)$ is commutative multiplicative monoid with $1$.

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    $\begingroup$ +1, this is a great question. I was thinking about this a while back and asked this question which you might be interested in; math.stackexchange.com/questions/3900991/…. Your question really nails what I was trying to think about there $\endgroup$
    – Jojo
    Commented Mar 21, 2022 at 13:07
  • $\begingroup$ I thought, in ring, +and × are dependent(relevant) and should not be independent(irrelevant), the only conditional expression that + and × appear(thus seems to relate) is distributive law. That is the background I asked this question. $\endgroup$ Commented Mar 21, 2022 at 17:19

2 Answers 2

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Here is a "dumb" example. Let $R=\mathbb Z$, and let $\times=+$, i.e., addition and multiplication are the same thing. Now $(R,\times)$ is a commutative monoid, with a $1$ (i.e, $0\in R$). This is clearly not distributive: $1\times(1+1)=3\neq1\times 1+1\times 1=4$.

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Let $(R,+) = \mathbb{Z}/3\mathbb{Z}$, and define $\cdot$ to be the unique commutative operation $R \times R \to R$ such that

$[0] \cdot a = [0]$ for all $a$

$[1] \cdot a = a$ for all $a$

$[2] \cdot [2] = [0]$

You can check that $(R,\cdot)$ is a commutative monoid, but the distributivity law does not hold because

$$[2] \cdot [1] + [2] \cdot [1] = [2] + [2] = [1] \neq [0] = [2] \cdot [2] = [2] \cdot ([1] + [1]).$$

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    $\begingroup$ This is not the smallest possible example since $(\Bbb Z/2\Bbb Z, +, +)$ is smaller.. $\endgroup$
    – J.-E. Pin
    Commented Mar 21, 2022 at 6:00
  • $\begingroup$ Oh, good point! Thank you, I'll correct my answer. $\endgroup$ Commented Mar 21, 2022 at 17:30
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    $\begingroup$ The smallest possible example with $0\neq1$ is $(\mathbb Z/2\mathbb Z,+,+')$ where $a+'b=a+b+1$. $\endgroup$
    – mr_e_man
    Commented Mar 21, 2022 at 23:16

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