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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.


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The quotient map is a morphism of $k$-algebras hence the constants are fix. – azarel Feb 27 '12 at 8:46
@azarel Can you elaborate a bit more? For example I know that if you have a homomorphism from $Z[x]$ into some ring the constants must stay fixed. – user38268 Feb 27 '12 at 9:44
if the ring's characteristic is zero... – user38268 Feb 29 '12 at 17:19
up vote 1 down vote accepted

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!

$\hspace{6in} \square$

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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 $X$.

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The point $x$ may just be a point of $k$. We don't even know if it is a point of $X$. Therefore we can't conclude that the functions $\xi_i - a_i$ vanish at some point of $X$... – user38268 Feb 27 '12 at 9:42
Yes, thought I was missing the point :p – fretty Feb 27 '12 at 9:47
I guess the fact that the ideal contains $I(X)$ must give you the fact that $x\in X$ somehow. – fretty Feb 27 '12 at 9:59
Voted down just because I overlooked a simple pathetic. Now there's a badge effort if ever I saw one. – fretty Feb 27 '12 at 22:22
Let me tell you that I did not downvote your post. There is a large discussion on meta about downvoting and stuff, about whether or not users should leave a downvote with a comment, etc. – user38268 Feb 28 '12 at 1:57

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