# I'd like to verify that $\{[0], [1]\}$ is an abelian group under multiplication

Since $[0][1] = [0] \in \mathbb Z_2$, this ring is closed under multiplication.

Let $[1]$ be identity: $[0][1] = [0]$ and $[1][1] = [1]$.

$[0][1] = [0 \cdot 1] = [1 \cdot 0] = [1][0]$ so commutativity holds.

If we assume $\mathbb Z_2$ is a ring, associativity is given.

I am having difficulty showing there's inverse element in $\mathbb Z_2$. What could possibly be the inverse of $[0]$? Please, elaborate on this. Thanks.

• $[0][0]=[0]$ and $[0][1]=[0]$. Since $\mathbb{Z}_2=\{[0],[1]\}$, we conclude that $\forall x\in\mathbb{Z}_2,\ [0]x=[0]$. Since $[0]\neq[1]$, we conclude that $\forall x\in\mathbb{Z}_2,\ [0]x\neq[1]$. Doesn't this show that $[0]$ has no inverse? – gniourf_gniourf Feb 16 '16 at 18:18
• The additive zero does not have a multiplicative inverse in a ring. – John Smith Feb 16 '16 at 18:19
• In a field, the non-zero elements form a group under multiplication. (The zero element in a ring only has an inverse in the trivial ring with only one element.) – Rob Arthan Feb 16 '16 at 18:22
• You need to reread the definition of a field. – Tobias Kildetoft Feb 16 '16 at 18:23
• $(\mathbb{Z_2},+)$ isn't the same as $(\mathbb{Z_2^{*}},(*))$ – user111750 Feb 16 '16 at 18:25