"surjectively-free" from Matsumura, Commutative Algebra

Question

One of my friend asked me about this exercise from Matsumura's "Commutative Algebra" (not "Commutative Ring Theory"), and I have no idea about the term "surjectively free" and how to solve this exercise.

Let $A$ be a ring and $M$ an $A$-module. We shall say that $M$ is surjectively-free over $A$ if $A = \sum f(M)$ where sum is taken over $f \in \mathrm{Hom}_A(M,A)$. Thus, free $\Rightarrow$ surjectively free. Prove that if $B$ is a surjectively free $A$-algebra, then for any ideal $I$ of $A$ we have $IB \cap A = I$.

I find this problem difficult since $\mathrm{Hom}_A(B,A)$ consists of "module" homomorphisms instead of "algebra" homomorphisms, so I cannot embed $B$ into a product of copies of A as a ring. Can anyone solve this one?

Answer 1

Let $B$ be surjectively free and put $J:= IB\cap A$. It is obvious that $I \subseteq J$, and so it remains to prove $J \subseteq I$.

For any $A$-linear map $f\colon B\to A$ and any $b\in B$ we observe $$ J\cdot f(b) = f(J\cdot b) \subseteq f(IB) = If(B) \subseteq I. $$ Since $B$ is surjectively free, we find $f_1,\dotsc,f_n \in \operatorname{Hom}_A(B,A)$ such that $\sum_{i=1}^nf_i(b_i) = 1_A \in A$ for certain $b_i\in B$. But then $$ J = J\cdot 1_A \subseteq \sum_{i=1}^nJ\cdot f_i(b_i) \subseteq I. $$ Hence, $I = J = IB\cap A$.