Prove that $\left( I_1I_2 \right)^e=I_1^e I_2^e$

Question

I have a problem when proving a basic exercise about extension in commutative ring:

Let $f:A \rightarrow B$ be a ring homomorphism, $I_1$ and $I_2$ are ideals of $A$. Prove that $\left( I_1I_2 \right)^e=I_1^e I_2^e$. ($I^e$ is the extension of $I$, it is the ideal $Bf(I)$.)

I tried to take an element $x=\sum_i y_if(x_i)\in \left( I_1I_2 \right)^e$ with $y_i\in B$ and $x_i\in I_1 I_2$. So that $x_i=\sum_j a_{ij} b_{ij}$ with $a_{ij}\in I_1$ and $b_{ij}\in I_2$.

Hence I have $x=\sum_i y_if(\sum_j a_{ij} b_{ij})=\sum_i \sum_j y_if(a_{ij} b_{ij})$. (*)

I have a problem when taking (*) to the form $\left(\sum y_i f(a_i) \right)\left(\sum y'_i f(b_i) \right)\in I_1^e I_2^e$. Thank for helping me.

Answer 1

Use the fact that $f$ is a ring homomorphism and write $f(a_{ij}b_{ij}) = f(a_{ij})f(b_{ij})$. Then note that $y_if(a_{ij})f(b_{ij})$ is an element of $I_1^eI_2^e$. Indeed $y_if(a_{ij}) \in I_1^e$ and $f(b_{ij}) \in I_2^e$.

Hence as all elements in the summation are elements of $I_1^eI_2^e$ we have that $x \in I_1^eI_2^e$