Proving one form of Hilbert's Nullstellensatz I have been trying to prove the following problem in Atiyah Macdonald concerning one form of Hilbert's Nullstellensatz.
The problem is as follows:

If $X$ is an affine algebraic variety (the set of all points satisfying $f_a(t_1, \ldots, t_n) = 0$, where each polynomial in the collection is in $k[t_1, \ldots, t_n]$, $k$ algebraically closed) consider the ring $$P(X) = k[t_1, \ldots, t_n]/I(X)$$ where $I(X) = \{g \in P(X) : g(x) = 0 \hspace{2mm} \forall x \in X\}.$ If $\mu$ is a map from $X$ to the set of all maximal ideals in $P(X)$ which we call $\operatorname{Max}(P(X))$, defined by $x \mapsto m_x$ then $\mu$ is surjective.
  $$m_x := \{f \in P(X) : f(x) = 0 \}.$$

My approach is as follows: If I can realise every maximal ideal $m \in P(X)$ as some $m_x$ for $x\in X$, I would have proven surjectivity  by definition of $\mu$. In fact it suffices to prove that $m \subseteq m_x$ because by maximality we would have $m = m_x$.
By the lattice theorem we know that maximal ideals in $P(X)$ are in one to one correspondence with maximal ideals in $k[t_1, \ldots,t_n]$ containing the kernel. Also, we know by the weak Nullstellensatz that because $k$ is algebraically closed, every maximal ideal in $k[t_1, \ldots, t_n]$ is of the form 
$$(t_1 - a_1, \ldots, t_n - a_n)$$ where the $a_i's \in k$. Therefore if I know that every maximal ideal $m$ in $P(X)$ is of the form $(\xi_1 - b_1, \ldots \xi_n - b_n)$ where the $\xi_i$ are the coordinate functions on $X$ and $b = (b_1, \ldots b_n)$ a point of $X$ this should be it
For then every function in $m$ vanishes at $b\in X$ so that $m \subseteq m_b$ proving surjectivity.
My problem comes in showing why every maximal ideal $m$ in $P(X)$ must be of the form $(\xi_1 - b_1, \ldots ,\xi_n - b_n)$, $ b= (b_1, \ldots, b_n) \in X$. I guess you could say this is equivalent to asking:

"If $(t_1 - a_1, \ldots, t_n - a_n)$ is a maximal ideal in $k[t_1, \ldots t_n]$ that contains $I(X)$, then why is it that the image of every generator be of the form $\xi_i - b_i$ where the $b_i's$ are as defined above?

I get how the $t_i$ maps to $\xi_i$ but not where the constants map to, so I tried to assume that $b = (b_1, \ldots ,b_n) \notin X$. Then this means that the functions $\xi_i - b_i$ do not vanish at any $x \in X$, meaning for all $x\in X$, $\xi_i - b_i \notin m_x$ so that we always have $(\xi_i - b_i, m_x) = P(X)$. From here, I tried to deduce a contradiction but nothing has worked and I am out of ideas.
Any help is appreciated but please do not give it all away.
Thanks.
 A: By the comment of Azarel above, the quotient map fixes the constants so that we can just speak of the ideal $(\xi_1 - a_1, \ldots, \xi_n - a_n)$ in the quotient  $k[t_1, \ldots t_n]/I(X)$ or $(t_1 - a_1, \ldots, t_n - a_n)$ in the polynomial ring $k[t_1, \ldots t_n]$.
Now we know that given $x \in X$, $f(x) = 0$ for all $f \in I(X)$. The converse is also true because if $f(x) = 0$ for all $f \in I(X)$ but $x \notin X$, recall that $X$ is the set of all points that satisfy $f_\alpha (t_1, \ldots, t_n) = 0$ for $\alpha \in \Lambda$. 
Therefore if $x \notin X$, this means that there exists $\alpha^{\ast} \in \Lambda$ such that $f_{\alpha^{\ast}} (x) \neq 0$. But then $f_{{\alpha}^{\ast}}$ is necessarily contained in $I(X)$ so this is a contradiction.
So before I was having trouble in determining why the point $a = (a_1, \ldots, a_n)$ was in $X$. Suppose it were not, then by the result above $a = (a_1, \ldots, a_n) \notin X$ would mean that there is a polynomial $f(x) \in I(X)$ such that $f(a) \neq 0)$.
But then $f(x) \in I(X) \subset (t_1 - a_1, \ldots, t_n - a_n)$ so that 
$$f(x) = \sum_{i=1}^n g_i(t_i - a_i),$$
for $g_i \in k[t_1, \ldots, t_n]$. The right hand side vanishes at the point $a = (a_1, \ldots, a_n)$ which is a contradiction so that $a \in X$, which finishes the proof!
$\hspace{6in} \square$
A: I might be oversimplifying this question and missing something non-trivial but here goes. 
Ok so you have lifted the maximal ideal $m$ to a maximal ideal $M$ of $K[t_1, ..., t_n]$. 
By the Weak Nullstellensatz this $M$ must have the form $<t_1 - a_1, ..., t_n - a_n>$. Surely the point $x = (a_1, ..., a_n)$ is what you seek? 
The ideal $M$ consists of ALL functions that vanish at $(a_1, ..., a_n)$, so once you move into the quotient you will just get ALL "different" functions that vanish at $(a_1, ..., a_n)$, i.e. $m_x$ . Remember, all this ring is doing is making classes of functions that have the same effect on $X$.
