Example of non surjective ring homomorphisim where im$(\phi)$ is not an ideal

Question

Let $\phi:R\mapsto S$ be a ring homomorphism.

What is an example of a $\phi$ such that the sub-ring im($\phi)=\{x\in S\ : \exists y\in R\ x=\phi(y)\} \subset S$ is not an ideal?

If $\phi$ is surjective then there is a $\phi^{-1}(\alpha)$ for every $\alpha\in R$ so there will always be a pre-image of any multiple of an element of im($\phi$)...

Answer 1

Consider the inclusion homomorphism from a ring $R$ to the ring $R[X]$ of polynomials over $R$. Then, the image of $R$ is a subring which is not an ideal.

Answer 2

If $S$ is unitary, then any proper unitary subring $R\subsetneq S$ together with the inclusion map $$i:R\rightarrow S, \ x \mapsto x$$ will do. If it was an ideal, then we had $$S=(1)\subseteq (Im(i))=Im(i) =R \subsetneq S$$ which is a contradiction.

Concrete examples would be $\mathbb{Z}\subsetneq \mathbb{Q}$, $\mathbb{Q}\subsetneq \mathbb{R}$, or also the example by sqtrat (if $R$ is a unitary ring). You can also take (for a given unitary ring $R$) the ring of rational functions $R\subsetneq R(X)$ or the ring of formal power series $R \subsetneq R [[X]]$.