Let $f:A \to B$ be a homomorphism of rings $A,B$.
Then there is a map
$e:\{$ ideals of $A\} \to\{$ ideals of $B\}$,
which sends each ideal of $A$ to $f(A)B$. I cannot understand why $f(A)B$ is an ideal of $B$. For example, I cannot prove that the summation is closed in $f(A)B$, namely, for any $a,a' \in A$ and $b,b' \in B$ there are $ a'' \in A$ and $ b'' \in B$ such that
$f(a)b + f(a')b' =f(a'')b''.$
After comments
This is the book of Undergraduate commutatie algebra Reid Miles A, page 87 section 6.3 Ideals in $A$ and $S^ {-1}A$.
I repeat here.
Quite generally, given a ring homomorphism $\varphi:A\to B$, there is a coresopondence
$e:\{$ ideals of $A\} \to\{$ ideals of $B\}$, given by $e(I)=\varphi(I)B=IB$ called extension and ....
I cannot understand what does the notaion $e(I)=\varphi(I)B=IB$ mean ?
The notation $f(I)B$ doesn't mean $\{f(a)b:a\in I, b\in B\}$, but rather $$ f(I)B = \left\{\sum_{i\in I}f(a_i)b_i:a_i\in I, b_i\in B, I\text{ is finite} \right\}. $$ Indeed, if you only take the set of products, you will not have an ideal. For example, let $k$ be a field and $A=k[x,y]\subset k[x,y,z,w] = B$. The extension of the ideal $(x,y)\subset A$ contains the element $xz+yw$, which is irreducible, so it is not of the form $f(a)b$.