If $f: A \to B$ is a morphism with $f(I) \subset J \trianglelefteq B$, then there is $\tilde{f}: I/I^2 \to J/J^2$.

Question

Let $A, B$ be commutative rings, $I\trianglelefteq A$, $J\trianglelefteq B$ and a ring morphism $f: A \to B$ such that $f(I) \subset J$. I was wondering if this map $f$ induces a map $$\tilde{f}:I/I^2 \to J/J^2.$$ My intuition tells me that it is true but I have absolutly no idea how to show it.. Could one of you help me ?

Answer 1

Since $f(I)\subseteq J$, then $f$ restrict to a morphism $f:I\rightarrow J$ and you can define $\tilde f:I/I^2\rightarrow J/J^2$ directly as $\tilde f(a+I^2)=f(a)+J^2$. Since $a\in I\rightarrow f(a^2)=f(a)^2\in J^2$, therefore $$\{a^2:a\in I\}\subseteq \ker f\rightarrow I^2\subseteq\ker f$$ so $\tilde f$ is well defined