Define $\mathbb{Z}_2=\{0,1\}$ (Modulo 2). In addition, Define $+$ and $\dot{}$ as addition and multiplication respectively.

Is $\mathbb{Z}_2,+,\dot{}$ a commutative ring ($\dot{}$ commutative)?

The requirements for a commutative group are 1) $\mathbb{Z}_2\neq \phi$, 2)$\mathbb{Z}_2,+$ is an Abelian group. 3) $\mathbb{Z}_2,\dot{}$ is a commutative semi-group with no inverses. 4)Distributive laws.

To verify 1), Clearly $\mathbb{Z}_2 \neq \phi$. To verify 3) $\mathbb{Z}_2,\dot{}$ has neutral element $1_{\mathbb{Z}_2}=1$, and it is associative and commutative. To verify 4) Clearly it follows the distributive laws.

However, To verify 2), $\mathbb{Z}_2,+$ has a neutral element $0_{\mathbb{Z}_2}=0$, and $\mathbb{Z}_2$ is both associative and commutative. However, I can't see how it is closed on + or on the inverse of +. For example, 1 has no inverse in $\mathbb{Z}_2$ And $1+1$ is not in $\mathbb{Z}_2$. Why does my book claim that $\mathbb{Z}_2$ is a Commutative ring when it doesn't satisfy the properties?

  • 1
    $\begingroup$ Actually, $\mathbb{Z}_2$ under multiplication is not required to have no inverses, but it is not required to have inverses for every element. $1$ always has an inverse. Also, any field is a commutative ring and all the non-zero elements have inverses. $\endgroup$
    – user357980
    Sep 18, 2016 at 2:50
  • $\begingroup$ Aha, I see, thanks. $\endgroup$ Sep 18, 2016 at 2:53
  • $\begingroup$ Also, why the down vote? $\endgroup$ Sep 18, 2016 at 2:53

1 Answer 1


As a group, $\mathbb Z_2$ means addition, mod 2. The same is true now. We have the two elements, and we will add them, and multiply them, but do so mod 2. This means that $1$ is its own inverse, as $1 \cdot 1 = 1$, and $1+1 = 0$, as $ 2= 0 \mod 2$.

  • $\begingroup$ I see, thanks for clearing it up $\endgroup$ Sep 18, 2016 at 2:54

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