# Cannot become ring because distribution law does not hold

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 $$1$$.

• +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
– Jojo
Commented Mar 21, 2022 at 13:07
• 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. Commented Mar 21, 2022 at 17:19

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: $$1\times(1+1)=3\neq1\times 1+1\times 1=4$$.

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

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

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