Answer to Question #276147 in Abstract Algebra for medyl

Question

Give an example of a subring of a ring, say A, that is not an ideal of A

Answer 1

Consider the ring $(\R,+,\cdot)$ of real numbers. Let $A=\mathbb Q$ be the set of rational numbers. Since for $a,b\in\mathbb Q$ we get that $a-b\in\mathbb Q$ and $a\cdot b\in\mathbb Q,$ we conclude that $(\mathbb Q,+,\cdot)$ is a subring of the ring $(\R,+,\cdot).$ On the other hand, for $\sqrt{2}\in\R$ and $1\in\mathbb Q$ we get that $\sqrt{2}\cdot 1=\sqrt{2}\notin\mathbb Q,$ and hence $(\mathbb Q,+,\cdot)$ is not an ideal of the ring $(\R,+,\cdot).$