Suppose $S$ is a subring $R$ and R is free of finite rank as a module over $S$. Is there a Ring homomorphism from $R$ to $S$? The reason I ask is to prove that if S has invariable basis number(IBN) then R has IBN.
I know that since $R$ is free of finite rank there exists a module isomorphism $R\cong SxSx..xS$ for finite number of $S$. But I don't see how I can use that module isomorphism to generate a ring homomorphism.
Any ideas?
Edit: The accepted answer shows that there need not be a ring isomorphism. Hence, does anyone know how to prove the IBN statement?
Best Answer
As Dustan's answer suggests, a ring morphism $R\to S$ does not necessarily exist.
However, if $S$ has IBN, so does $R$ under your assumption. Indeed, assume $R\simeq_S S^n$. Now, let $r,s\geq 0$ such that $R^r\simeq_R R^s$. Then $R^r\simeq_S R^s$ (an $R$-linear isomorphism is also $S$-linear), and thus $S^{nr}\simeq_S S^{ns}$. Since $S$ has IBN, $nr=ns$, so $r=s$ and $R$ has IBN.