# A difficulty in understanding a paragraph in Hungerford Algebra.

Could anyone please explain to me why not vice versa in this paragraph, and how are the rings $R[x_1,\dots,x_m]$ and $R[[x]]$ clarify this example? which is a subring from the other?How can I prove that they are $R-modules$? :

If $S$ is a ring and $R$ is a subring, then $S$ is an $R$-module (but not vice versa!) with
$ra$ $(r\in R, a\in S$) being multiplication in $S$. In particular, the rings $R[x_1,\dots,x_m]$ and $R[[x]]$ are $R$-modules.

• Also the converse holds if $R$ is an ideal. – Naive Oct 3 '17 at 7:37

Since $R$ is smaller than $S$, one cannot be sure that multiplying an element $r \in R$ by an element $s \in S$ will produce an element in $R$. In general, $R$ is only closed under multiplication in $R$, not multiplication by any possible element in $S$.
If $R$ is a proper subring, then there exists an element $s\in S\backslash R$, so that $R$ cannot be an $S$-module, because when you multiply the identity of $R$ by $s$ you do not get an element of $R$.
• R can only consist of the identity if $1=0$ in $S$, since $0 \in R$, so $S=0$. So in the case you describe, $R=S=0$ and both are modules over one-another in the obvious way. – M. Van Oct 3 '17 at 14:25