Show $f^{-1}(I)$ Ideal in $R$?

Question

I'm trying to show that if $f: R \rightarrow S$ is a ring homomorphism, and $I \subset S$ is ideal, then $f^{-1}(I)$ is Ideal in $R$

I'm trying to show that is satisfies the conditions of an ideal, but I'm not getting anywhere.

Answer 1

a different approach which is maybe simpler is to use the Homomorphism Theorem, and consider the composite morphism $f\circ g: R \rightarrow S \rightarrow S/I$, and then we have that $f^{-1}I$ is the kernel of this, therefore an ideal in $R$.

Answer 2

A direct proof :

Note that $f^{-1}(I)$ is non-empty as it contains 0 (infact, the whole of ker f).

Suppose $x,y \in f^-1(I)$ and r in R.

Then, $f(x),f(y) \in I$ implies that $f(x+y) = f(x)+f(y) \in I, f(rx) = f(r)f(x) \in I, f(xr) = f(x)f(r) \in I$ as $I$ is an ideal of S. Hence, $x+y, rx, xr \in f^{-1}(I)$.

So, $f^{-1}(I)$ is an ideal in R.