Rings Isomorphic to each other's Subrings Say $R,S$ are rings, $R$ isomorphic to a subring of $S$, $S$ isomorphic to a subring of $R$, does this imply that $R$ is isomorphic to $S$? If yes, how to show that?
I tried to construct an isomorphism between R and S given the two assumed isomorphisms, but surprisingly for me, it didn't work out.
 A: In fact, the answer is no. Consider the field $R=\mathbb{Q}(x_n:n\in\mathbb{N})$ of rational functions over $\mathbb{Q}$ in infinitely many variables, where $\mathbb{N}$ denotes the non-zero natural numbers. Let $S$ be the subring $\mathbb{Q}(x_n:n>1)[x_1]\subset R$. Then $R\not\cong S$ as rings, since $R$ is a field and $S$ is not. But $R$ is isomorphic to the subring $\mathbb{Q}(x_n:n>1)$ of $S$, via the map taking $x_{n}$ to $x_{n+1}$ for each $n\in\mathbb{N}$.

Edit: In response to your comment, here is the reason that the map $\alpha:R\to\mathbb{Q}(x_n:n>1)$ defined by extending $x_n\mapsto x_{n+1}$ for each $n\in\mathbb{N}$ is an isomorphism. First note that, since $R$ is a field and the codomain of $\alpha$ is a non-zero ring, $\alpha$ is automatically injective. To see that $\alpha$ is surjective, let $h$ be an arbitrary element of $\mathbb{Q}(x_n:n>1)$. By definition of the field $\mathbb{Q}(x_n:n>1)$, we have $$h=\frac{f(x_{m_1},\dots,x_{m_k})}{g(x_{n_1},\dots,x_{n_l})}$$ for some $f,g\in\mathbb{Q}[x_n:n>1]$, with $g$ non-zero. Now $g(x_{n_1-1},\dots, x_{n_l-1})$ is a non-zero element of $\mathbb{Q}[x_n:n\in\mathbb{N}]$ (why?), so $$\tilde{h}:=\frac{f(x_{m_1-1},\dots,x_{m_k-1})}{g(x_{n_1-1},\dots,x_{n_l-1})}$$ is a well-defined element of $R$, and now $\alpha(\tilde{h})=h$ (why?), as desired.
