On Atiyah-Macdonald Exercise 3.26 I am trying to prove the exercise 3.26 on Atiyah-Macdonlad: 

Let $(B_{\alpha},g_{\alpha \beta})$ a direct system of rings and $B$ the direct limit. For each $\alpha$, let $f_{\alpha}:A\rightarrow B_{\alpha}$ be a ring homomorphism such that $g_{\alpha \beta}\circ f_{\alpha}=f_{\beta}$ whenever $\alpha\leq \beta$. Then $f_{\alpha}$ induce $f:A\rightarrow B$. Show that 
  $$f^{\ast}(\mathrm{Spec}(B))=\bigcap f_{\alpha}^{\ast}(\mathrm{Spec}(B_{\alpha}))$$


Following the hint, I figured out 
$$\begin{aligned}\mathfrak{p}\notin f^{\ast}(\mathrm{Spec}(B)) &\Leftrightarrow  \varinjlim(B_{\alpha}\otimes_A k(\mathfrak{p}))=0\end{aligned}$$
and
$$\begin{aligned}\mathfrak{p}\notin \bigcap f_{\alpha}^{\ast}(\mathrm{Spec}(B_{\alpha}))&\Leftrightarrow  B_{\alpha}\otimes_A k(\mathfrak{p})=0 \text{  for some }\alpha\end{aligned}$$
By exercise 2.21 on Atiyah-Macdonald, we have 
$$ \varinjlim(B_{\alpha}\otimes_A k(\mathfrak{p}))=0 \Rightarrow B_{\alpha}\otimes_A k(\mathfrak{p})=0 \text{  for some }\alpha$$
But I have no idea how to prove the converse, which is true according to Atiyah-Macdonald.
Since the direct limit is the direct sum modulo something, consider the direct sum of $B_{\alpha}$. The zero rings will be killed. The quotient part is unknown but intuitively should not be the direct sum of the rest non-zero rings. Does the fact that $B_{\alpha}\otimes_A k(\mathfrak{p})$ is a $k$-module matter here?
Any hint and answers are welcomed!
 A: In general, it is not true that $\varinjlim B_{\alpha}=0$ if and only if $B_{\alpha}=0$ for some $\alpha$.
Consider the trivial counterexample: the direct system consists of $k$ and $0$ and 
$$0 \rightarrow 0 \rightarrow 0\rightarrow \cdots$$
$$
k \xrightarrow{\mathrm{id}} k\xrightarrow{\mathrm{id}}k\xrightarrow{\mathrm{id}}\cdots $$
where $k$ is a field (also an $k$-algebra). However, we have $B_{\alpha}=0$ for some $\alpha$ and $\varinjlim B_{\alpha}=k~\sqcup \{0\}$, which is obviously not a zero ring.
In fact, we can prove the converse easily. Following the hint, we already prove that 
$$f^{\ast}(\mathrm{Spec}(B))\supset\bigcap f_{\alpha}^{\ast}(\mathrm{Spec}(B_{\alpha})).$$
Now, we need to show
$$f^{\ast}(\mathrm{Spec}(B))\subset\bigcap f_{\alpha}^{\ast}(\mathrm{Spec}(B_{\alpha})).$$
Take a $\mathfrak{p}\in f^{\ast}(\mathrm{Spec}(B))$, that is, there exists $\mathfrak{q}$ such that $f^{-1}(\mathfrak{q})=\mathfrak{p}$. Suppose the project morphisms $$\mu_{\alpha}:B_{\alpha}\rightarrow B$$
and we have $f=\mu_{\alpha}\circ f_{\alpha}$ for any $\alpha$. So take the prime ideal $\mathfrak{p}_{\alpha}=\mu_{\alpha}^{-1}(\mathfrak{q})$, then 
$$f^{\ast}_{\alpha}(\mathfrak{q}_{\alpha})=f_{\alpha}^{-1}\mu_{\alpha}^{-1}(\mathfrak{q})=(\mu_{\alpha}\circ f_{\alpha})^{-1}(\mathfrak{q})=f^{-1}(\mathfrak{q})=\mathfrak{p}$$
It follows that
$$\mathfrak{p}\in \bigcap f_{\alpha}^{\ast}(\mathrm{Spec}(B_{\alpha})). $$

So the equivalence the author claimed, that is,
$$\varinjlim(B_{\alpha}\otimes_A k(\mathfrak{p}))=0 \Leftrightarrow B_{\alpha}\otimes_A k(\mathfrak{p})=0 \text{  for some }\alpha$$ 
is true, by the above equivalence. But I don't know how to prove the converse direction directly so far.
A: In general, $\varinjlim B_{\alpha}=0$ if and only if $B_{\alpha}=0$ for some $\alpha$.
Below is a proof of the fact that $\boxed{ \text{if $B_\alpha=0$ for some $\alpha$, then $\varinjlim B_{\alpha}=0$}}:$
Recall by Exercise 2.21, the multiplicative identity $1\in \varinjlim B_{\alpha}$ is the image of $1_\alpha\in B_\alpha$ under the canonical ring homomorphism $g_\alpha: B_\alpha \to \varinjlim B_\alpha$.
Since $B_\alpha=0$, we have that $1_\alpha=0_\alpha$.
So, $1=g_\alpha(1_\alpha)=g_\alpha(0_\alpha)=0$.

Below is an unnecessary calculation showing that everything in $\varinjlim B_\alpha$ equals $0$.
Let $u\in \varinjlim B_\alpha$. Then by Exercise 2.15, there exists $v\in B_\beta$ for some $\beta$ such that $u=g_\beta(v)\in \mathrm{Im}(g_\beta)$.
Since the index set is directed, we can find an index $\gamma\geq \alpha,\beta$.
Thus,
\begin{align*}
u &= 1 \cdot u 
\\&= g_\alpha(1_\alpha) \cdot g_\beta(v) 
\\&= g_\alpha(0_\alpha) \cdot g_\beta(v) 
\\&= g_\gamma(g_{\alpha\gamma}(0_\alpha) \cdot g_{\beta\gamma}(v))
\\&= g_\gamma(0_\gamma \cdot g_{\beta\gamma}(v))
\\&= g_\gamma(0_\gamma) 
\\&= 0
\end{align*}
, which implies that $\varinjlim B_\alpha = 0$.
