Distinguishing $\mathbb Z_2\times\mathbb Z_2$ from $\mathbb Z_4$

Detailed Solution

We compare the characteristics of the two rings. Step 1: Determine the characteristic of $R$. The ring $$ R=\mathbb Z_2\times\mathbb Z_2 $$ has componentwise addition. Its multiplicative identity is $$ 1_R=(1,1). $$ Adding the identity to itself gives $$ 1_R+1_R=(1+1,1+1)=(0,0). $$ Thus $$ 2\cdot1_R=0_R. $$ Since $1_R\neq0_R$, the characteristic of $R$ is $$ \boxed{\operatorname{char}(R)=2}. $$ Step 2: Determine the characteristic of $S$. In $S=\mathbb Z_4$, the multiplicative identity is $1$. We have $$ 1+1=2\neq0, $$ but $$ 1+1+1+1=4\equiv0\pmod4. $$ Therefore, $$ \boxed{\operatorname{char}(S)=4}. $$ Step 3: Compare the characteristics. A ring isomorphism preserves the multiplicative identity and the additive structure. Consequently, isomorphic rings with unity must have the same characteristic. Since $$ \operatorname{char}(R)=2 \quad\text{and}\quad \operatorname{char}(S)=4, $$ the rings cannot be isomorphic. Hence $$ \boxed{R\not\cong S}. $$ Therefore, option B is correct. Why the other options are incorrect: Option A: False. Although both rings contain four elements, they have different characteristics and therefore cannot be isomorphic. Option C: False. An element $a$ is idempotent if $a^2=a$. In $\mathbb Z_4$, $$ 2^2=4\equiv0\pmod4, $$ so $2^2\neq2$. Thus $2$ is not idempotent. In fact, the only idempotents in $\mathbb Z_4$ are $0$ and $1$. Option D: False. The characteristics are different: $R$ has characteristic $2$, whereas $S$ has characteristic $4$. An additional distinction is that $R$ has nontrivial idempotents: $$ (1,0)^2=(1,0),\qquad (0,1)^2=(0,1). $$ Neither of these equals $0_R=(0,0)$ or $1_R=(1,1)$. By contrast, $\mathbb Z_4$ has no nontrivial idempotents. Thus the correct option is $$ \boxed{\mathrm{B}}. $$