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I have two questions about understanding localization and basic algebraic geometry.

  1. In class, the professor mentioned that given the basic open set $$D(f) = \mathbb{A}^n - Z(f)$$ where $Z(f)$ is the vanishing set of $f\in k[x_1, \cdots x_n] =:A$, then the polynomials on $D(f)$ forms a ring and is isomorphic to the localization $A_f$.

I know the definition of localization and basic properties, but I can not quite get the picture here. And what does "polynomials on $D(f)$" mean?

  1. Give an ideal $\mathfrak{a} \subset A$, and $f\in I(Z(\mathfrak{a}))$ where $I(V)$ is the ideal of polynomials that vanishes on the algebraic set $V$, the professor said $\mathfrak{a} A_f = A_f$, some how $Z(\mathfrak{a})$ is contained in the zero set of $f$, and $A_f$ is like the complement of the zero set of $f$. I can not get the intuition behind this.

Thank you very much!

Xiao
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1 Answers1

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If $U$ is an open set in $k^n$, define $\mathcal O(U)$ to be the $k$-algebra of functions $F: U \rightarrow k$ of the following form: for each $p \in U$, there exists a smaller open set $p \in V \subseteq U$, and polynomials $f, g \in A$, such that for every $q \in V$, $g(q) \neq 0$ and

$$F(q) = \frac{f(q)}{g(q)}$$

In other words, $\mathcal O(U)$ is the $k$-algebra of functions $U \rightarrow k$ which locally look like a quotient of polynomial functions. So "polynomials on $D(f)$" means $\mathcal O(D(f))$.

Let $a = \frac{h}{f^m}$ be an element of the localization $A_f$. By definition, $f(q) \neq 0$ for all $q \in D(f)$. Thus we have a well defined $k$-algebra homomorphism

$$A_f \rightarrow \mathcal O(D(f))$$

which sends $a$ to the function $D(f) \rightarrow k$ given by $q \mapsto \frac{h(q)}{f(q)^m}$. The result you mentioned in (1) is that this is actually an isomorphism. The hard part is surjectivity. A proof is given in Springer, Linear Algebraic Groups, chapter one.

If $f \in I(Z(\mathfrak a)) = \sqrt{\mathfrak a}$, then $Z(\mathfrak a) \subseteq Z(f)$ (by order reversing correspondence). Consequently, every $g \in \mathfrak a$ satisfies $g(q) \neq 0$ for all $q \in D(f)$ (if $g(q) = 0$, then $q$ must lie in $Z(f)$, the complement of $D(f)$). Hence the image of $g$ in $\mathcal O(D(f))$ is a unit, since it now has a well defined inverse there, $\frac{1}{g}$. But identifying $\mathcal O(D(f))$ with $A_f$, this just says that $\mathfrak a A_f = A_f$.

D_S
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  • One quick question, how can we know $g(q) \neq 0$ for $q\in D(f)$, I can see this holds true for $g=f^n\in \mathfrak{a}$. But this statement $\mathfrak{a}A_f = A_f$ is used to prove $I(Z(\mathfrak{a})) =\sqrt {\mathfrak{a}}$ so we can not assume $f^n \in \mathfrak{a}$. And I think we only need to show the existence of one such $g$ is unit in $A_f$. Thank you again! – Xiao Sep 12 '17 at 01:46
  • You're right, my last paragraph is incorrect. I should just say that there exists a $g \in \mathfrak a$ such that $g(q) \neq 0$ for all $q \in D(f)$. If you assume the nullstellensatz, you know that $f \in \sqrt{\mathfrak a}$, so you can take $g = f^m$ for suitably large $m$. If you are trying to prove the nullstellesatz, I'm afraid I don't know what to do. – D_S Sep 12 '17 at 02:57
  • I think we can use a important result from weak Nullstellensatz, given any ideal $I$ , if $Z(I) = \emptyset$, then $I = (1)$. We can look at $Z(\mathfrak{a} A_f)$ we see that the elements here $\frac{h}{f^n}$ where $h\in \mathfrak{a}$ can not have any common zero on its domain $D(f)$. If all $\frac{h}{f^n}$ vanishes at $p$, we must have $p\in Z(\mathfrak{a}) \subset Z(f)$ which is disjoint from $D(f)$. Is this correct? I know localization is kinda like restricting the domain, but I am not quite sure about how to make it rigorous. – Xiao Sep 12 '17 at 03:11