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.