# Let $\phi: R \to S$ be a homomorphism of rings, $I$ an ideal of $R$, $J$ an ideal of $S$. [duplicate]

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!

• 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 $1$.
– 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.