0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Also the converse holds if $R$ is an ideal. $\endgroup$ – Naive Oct 3 '17 at 7:37
1
$\begingroup$

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 $S$.

$\endgroup$
  • $\begingroup$ I have edited my question, could you please answer my edit if you have time? thanks !! $\endgroup$ – Intuition Oct 5 '17 at 4:23
0
$\begingroup$

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 $R$.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – M. Van Oct 3 '17 at 14:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.