surjectivity condition for induced map of prime ideals Let $A,B$ be commutative rings and $f: A \rightarrow B$ a flat ring homomorphism.
I would like to show that:
if or all $\mathfrak{m} \in$ MaxSpec($A$) one has ($f(\mathfrak{m})$)=:$\mathfrak{m}^e \neq (1)$ then $\phi:$ Spec($B$) $\rightarrow$ Spec($A$), $\mathfrak{q} \mapsto \mathfrak{q}^c:=f^{-1}\mathfrak{q}$ is surjective 
First I need to show that MaxSpec($A$) $\subset$ im($\phi$). But how do I do that? 
And what would be the next steps to show the assertion?
Thank you!
 A: 
For every $A$-module $M\neq\{0\}$ we have $M\otimes_AB\neq\{0\}$

For let $0\neq x\in M$.
Then $Ax\cong A/\mathfrak a$ for some ideal $\mathfrak a\neq A$.
Let $\mathfrak m\subseteq A$ be a maximal ideal containing $\mathfrak a$.
Then $\mathfrak a^e\subseteq\mathfrak m^e\subset B$ and $(Ax)\otimes_AB\cong B/\mathfrak a^e\neq\{0\}$.
Since $A\to B$ is flat, $(Ax)\otimes_AB$ is isomorphic to a submodule of $M\otimes_AB$.

For every $A$-module $M$, the canonical mapping $M\to M\otimes_AB$ is injective.

For let $\{0\}\to N\to M\to M\otimes_AB\to\{0\}$ be exact.
Then $\{0\}\to N\otimes_AB\to M\otimes_AB\to (M\otimes_AB)\otimes_AB\to\{0\}$ is exact as well.
Since $M\otimes_AB\to (M\otimes_AB)\otimes_AB$ is (split) injective, we get $N\otimes_AB=\{0\}$, hence $N=\{0\}$ by previous remark.

For every prime ideal $\mathfrak p\subseteq A$ we have $\mathfrak p^{ec}=\mathfrak p$.

Let $\mathfrak p\subseteq A$ be a prime ideal.
By previous remark, we have a commutative diagram where the vertical arrows are injective
$\require{AMScd}$
\begin{CD}
A@>>> A/\mathfrak p\\
@VVV@VVV\\
B@>>> B/\mathfrak p^e
\end{CD}
Since $\mathfrak p^{ec}$ is the kernel of the composition $A\to B\to B/\mathfrak p^e$, we get $\mathfrak p^{ec}=\mathfrak p$.

$\varphi:\operatorname{Spec}(B)\to\operatorname{Spec}(A)$ is surjective.

Let $\mathfrak p\subseteq A$ be a prime ideal and $S=A\setminus\mathfrak p$.
Since $\mathfrak p=\mathfrak p^{ec}$, the ideal $(S^{-1}B)\mathfrak p$ is proper, hence contained in a maximal ideal $(S^{-1}B)\mathfrak p\subseteq\mathfrak n\subset S^{-1}B$.
Let $\mathfrak q$ be the contraction of $\mathfrak n$ in $B$.
Then $\mathfrak p=\mathfrak p^{ec}\subseteq\mathfrak q^c$ and since $S\cap\mathfrak q^c=\varnothing$, we get $\mathfrak p=\mathfrak q^c$.
A: Let $f: A\rightarrow B$ is a flat ring homomorphism and $m^e\neq B$ for all $m\in \text{Max-Spec }B \implies B$ is a faithfully flat $A$-algebra.
Let $M$ be a finitely generated $A$-module such that $M\otimes_A B=0\implies M/mM\otimes_{A/m} B/mB=0$ after tensoring with $A/m$.
Since $B/mB \neq 0$ we get $M/mM=0\implies M_m=0$ 
Now $m\in \text{Max-Spec }A$ was arbitrary. So we get $M_m=0 \ \forall \ m\in \text{Max-Spec }A\implies M=0 $ as $M$ is finitely generated.      
Now $p^{ec}\supset p $ for $p\in \text{Spec }A$ 
Moreover we have 
$A/p\otimes B=B/p^e$ 
$A/p^{ec}\otimes B=B/p^e$ 
Then we have $A/p^{ec}=A/p$ as $A$-modules and hence $p^{ec}=p$ 
If the last line seems dubious, think of it this way $$A/p\rightarrow A/p^{ec}\rightarrow 0$$ is exact
Since $B$ is flat, we get $$0\rightarrow A/p \otimes B\rightarrow A/p^{ec}\otimes B\rightarrow 0$$ is exact and then by faithful flatness we get $$0\rightarrow A/p\rightarrow A/p^{ec}\rightarrow 0$$ is exact. This forces $p^{ec}=p$
