Prove that $\phi^-1(J)$ is an ideal of $R$.

Hello, I am trying to prove that $\phi^-1(J)$ is an ideal of $R$ given the above conditions but am stuck on my proof. Any feedback would be appreciated.

Note: $\phi^{-1}(J)$ is a sub-ring of $R$.

Absorbs products:

$\forall r \in,R$, and $x \in \phi^-1(J)$ rx $\in$ $\phi^-1(J)$ and xr $\in$ $\phi^-1(J)$ pf. Let r $\in$ $R$ and x $\in$ $\phi^-1(J)$. Thus x =

I don't know if I am in the right direction, can someone point me in the right direction. Thanks!

  • $\begingroup$ Depending on the convention which is being used, it might be the case that ideals are hardly ever subrings. Specifically if "ring" is meant to stand for "ring with $1$" throughout, then "subrings" should always contain $1$. $\endgroup$
    – user239203
    Nov 22 '19 at 6:27

Recall that $\phi^{-1}(J)=\{r\in R:\phi(r)\in J\}\subseteq R$. To show that $\phi^{-1}(J)$ is an ideal in $R$, we need to check that for any $r\in R$ and $r_{1}\in \phi^{-1}(J)$, $rr_{1},r_{1}r\in\phi^{-1}(J)$ and $\phi^{-1}(J)$ is a subgroup of $R$ under $+$, that is, if $a,b\in\phi^{-1}(J)$, $a-b\in\phi^{-1}(J)$. \begin{align*} \phi(rr_{1})=\phi(r)\phi(r_{1})=\phi(r)j\in J \\ \phi(r_{1}r)=\phi(r_{1})\phi(r)=j\phi(r)\in J \end{align*} since $J$ is an ideal in $S$. Moreover, one can check that $\phi^{-1}(J)$ is an additive subgroup of $R$ and so is a subring of $R$ by the above computation.


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