Subring vs Ideal in $\mathbb Q$

Detailed Solution

First, $\mathbb Z$ is indeed a subring of $\mathbb Q$ because it is closed under addition, subtraction, and multiplication, and contains the multiplicative identity $1$. However, being a **subring** does not imply being an **ideal**. For $\mathbb Z$ to be an ideal of $\mathbb Q$, we would need $$ qz\in\mathbb Z $$ for every $q\in\mathbb Q$ and every $z\in\mathbb Z$. Take $$ q=\frac12\in\mathbb Q \qquad\text{and}\qquad z=1\in\mathbb Z. $$ Then $$ qz=\frac12\notin\mathbb Z. $$ Therefore, $\mathbb Z$ is **not** an ideal of $\mathbb Q$. Hence, $$ \boxed{\mathbb Z\text{ is a subring of }\mathbb Q\text{ but not an ideal of }\mathbb Q.} $$ **Option A:** False. Although $\mathbb Z$ is a subring of $\mathbb Q$, it fails the absorption property required for an ideal. **Option C:** False. $\mathbb Z$ is naturally a subring of $\mathbb Q$. **Option D:** False. A subring of a commutative ring need not be an ideal. The example $$ \mathbb Z\subset\mathbb Q $$ is a direct counterexample. Thus the correct answer is $$ \boxed{\mathrm{B}}. $$ **Key distinction:** A **subring** is closed under the ring operations internally, whereas an **ideal** must additionally satisfy the absorption property $$ rI\subseteq I \qquad\text{for every }r\in R. $$