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I was reading in Vakil's Foundations of Algebraic Geometry that one can picture the "traditional points" of Spec($\mathbb{C}[x_1,\dotsc,x_n]/(f_1,\dotsc,f_r))$ as the zero locus of the polynomials $f_1,\dotsc,f_r$. (This is early in the book, and I believe he's referring to the maximal ideals as traditional points.) As he says, this is because the prime ideals of a quotient $A/I$ are in natural bijection with the primes of $A$ containing $I$.

I can see partially why this is true. If a maximal ideal $(x_1-a_1,\dotsc,x_n-a_n)$ contains $(f_1,\dotsc,f_r)$, then for each $i$, we can write $f_i=g_{i1}(x_1-a_1)+\dotsb g_{in}(x_n-a_n)$ so that indeed, $f_i(a_1,\dotsc,a_n)=0$ for all $i$. However, I'm not immediately seeing why the converse is true. That is, how does $f_i$ vanishing at $(a_1,\dotsc,a_n)$ for all $i$ give us that $(x_1-a_1,\dotsc,x_n-a_n) \supset (f_1,\dotsc,f_r)$?

Edit: In light of comments below, we have $(f_1,\dotsc,f_r)\subseteq (x_1-b_1,\dotsc,x_n-b_n)$ for some $b_1,\dotsc,b_n\in \mathbb{C}$, using the fact that every ideal is contained in a maximal ideal and that the maximals ideals have that form, by Nullstellensatz. But then we have that $(f_1,\dotsc,f_r)$ vanishes at both $(a_1,\dotsc,a_n)$ and $(b_1,\dotsc,b_n)$. What allows us to conclude that $a_i=b_i$ for all $i$?

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    $\begingroup$ Hint: By the Nullstellensatz, all maximal ideals are of the form $(x_1-a_1,\cdots, x_n-a_n)$. Also, every ideal is contained in a maximal ideal. $\endgroup$ – RKD Jun 3 '14 at 22:56
  • $\begingroup$ @Jake, thanks. I still have a remaining question in light of your comment. We know $(f_1,\dotsc,f_r)\subseteq (x_1-b_1,\dotsc,x_n-b_n)$ for some $b_1,\dotsc,b_n\in \mathbb{C}$. This means $f_1,\dotsc,f_r$ all vanish at $(b_1,\dotsc,b_n)$. Couldn't $(f_1,\dotsc,f_r)$ vanish at both $(a_1,\dotsc,a_n)$ and $(b_1,\dotsc,b_n)$? And nothing tells us $(f_1,\dotsc,f_r)$ is contained in a unique max ideal, right? $\endgroup$ – Erik Jun 3 '14 at 23:10
  • $\begingroup$ Forget the unique max ideal comment. How do we conclude $a_i=b_i$? $\endgroup$ – Erik Jun 3 '14 at 23:29
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Suppose $f(x_1,\ldots, x_n)$ vanishes at $(a_1,\ldots, a_n).$ Let $I=(x_1- a_1, \ldots, x_n - a_n) .$ Then $$f(x_1,\ldots, x_n) + I = f(a_1, \ldots, a_n)+ I =0+I$$

So $f(x_1,\ldots, x_n) \in I.$

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This is an immediate consequence of the Nullstellensatz.

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  • $\begingroup$ This is correct. I was thinking of another version of the Nullstellensatz when you first wrote this. Indeed, if $f$ vanishes at $(a_1,\dotsc,a_n)=Z(x_1-a_1,\dotsc,x_n-a_n)$, then $f\in \sqrt{(x_1-a_1,\dotsc,x_n-a_n)}$ by Nullstellensatz, which equals $(x_1-a_1,\dotsc,x_n-a_n)$ because it's maximal. $\endgroup$ – Erik Jun 6 '14 at 0:21
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Think about the morphism of $\mathbf C$-algebras $$ \mathbf{C}[x_1,\dots,x_n] \to \mathbf{C},\quad \text{such that }x_i \mapsto a_i. $$ The kernel is precisely the ideal generated by the $x_i-a_i$. I admit that one should check this too, but if you believe that the ideal is maximal then there isn't much more work.

Another way is to observe that a polynomial has a "Taylor expansion" centered at $(a_1,\dots,a_n)$, since the $x_i-a_i$ serve as indeterminates over $\mathbf{C}$ just as well as do the $x_i$. Now the question is whether there is a constant term in this expansion.

To the edit: of course the polynomials could have more than one common zero! I don't think that invalidates anything we've done.

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