1
$\begingroup$

This is exercise 2.5.12 of Liu's Algebraic Geometry.

Let $k$ be a field. Let $X = \bigcup_{i=1}^rD(f_i)$ be a finite union of principal open subsets of $\mathbb{A}_k^n$. Show that $\mathcal{O}_{\mathbb{A}_k^n}(X) = k[T_1,\dots,T_n]_f$ where $f = \mathrm{gcd}(f_1,\dots,f_r)$.

Can anyone help me solve this? Thank you.

I have some progress: First, note that $X \subset D(f)$. So we have restriction maps $$ \mathcal{O}_{\mathbb{A}_k^n}(D(f)) \to \mathcal{O}_{\mathbb{A}_k^n}(X) \to \mathcal{O}_{\mathbb{A}_k^n}(D(f_i)) $$ Since $\mathbb{A}_k^n$ is an integral scheme, all the above restriction maps are injective. It suffices to show $\mathcal{O}_{\mathbb{A}_k^n}(D(f)) \to \mathcal{O}_{\mathbb{A}_k^n}(X)$ is surjective. Because elements of $\mathcal{O}_{\mathbb{A}_k^n}(X)$ is in one-to-one correspondence with elements $(a_1,\dots,a_r) \in \prod_{i=1}^r\mathcal{O}_{\mathbb{A}_k^n}(D(f_i))$ verifying $a_i|_{D(f_if_j)} = a_j|_{D(f_if_j)}$ for all $i,j \in [r]$. So, it suffices to find $a \in \mathcal{O}_{\mathbb{A}_k^n}(D(f))$ verifying $a|_D(f_i) = a_i$ for all $i \in [r]$. Suppose $a_i = g_i/f_i^u \in \mathcal{O}_{\mathbb{A}_k^n}(D(f_i)) = k[T_1,\dots,T_n]_{f_i}$. (Because there are finitely many $a_i$, $u$ can be chose independent of $i$). $a_i|_{D(f_if_j)} = a_j|_{D(f_if_j)}$ then means $$ \frac{g_if_j^u}{(f_if_j)^u} = \frac{g_jf_i^u}{(f_if_j)^u}, \quad \mathrm{i.e.,} \quad g_if_j^u = g_jf_i^u $$ All the ring above can be thought of subrings of $k(T_1,\dots,T_n)$. So, in $k(T_1,\dots,T_n)$, we have $$ \frac{g_i}{f_i^u} = \frac{g_j}{f_j^u} $$ Here I got stuck. I cannot find $g/f^l$ to represent $g_i/f_i^u$ simultaneously.

$\endgroup$

1 Answer 1

1
$\begingroup$

The main point that you are not using is that $k[x_1,\ldots,x_n]$ is a UFD.

Thus, any element $\alpha$ of $k(x_1,\ldots,x_n)$ can be written in a unique way as a quotient $\frac{a}{b}$ with $a$ and $b$ having no common irreducible divisor and in $k[x_1,\ldots,x_n]$ (up to scalars).

Then, for any nonzero polynomial $g$, $\alpha \in k[x_1,\ldots,x_n]_g$ iff $b$ divides some power of $g$, that is, if every irreducible factor of the denominator of $\alpha$ occurs in $g$.

Can you finish the proof using that?

$\endgroup$
5
  • $\begingroup$ Sorry, could you please be more specific? $\endgroup$
    – Yu Ning
    Jun 16, 2020 at 13:17
  • $\begingroup$ Is your issue with the content of my post, or with how it can be used to deduce the final result? $\endgroup$
    – Aphelli
    Jun 16, 2020 at 14:26
  • $\begingroup$ How it can be used to deduce the final result? $\endgroup$
    – Yu Ning
    Jun 17, 2020 at 3:06
  • $\begingroup$ By properties of integral schemes, $\mathcal{O}(X)$ is the intersection of the $R_i=\mathcal{O}(D(f_i))$. Now, $R_i$ is the subring of $\alpha \in k(x_1,\ldots,x_n)$ such that its denominator contains only irreducible factors of $f_i$. So $\mathcal{O}(X)$ is the subring of $\alpha \in k(x_1,\ldots,x_n)$ such that all the irreducible factors of its denominator occur in each $f_i$. But this is equivalent (by UFD) to all the irreducible factors of the denominator of $\alpha$ to divide $f$, which is equivalent to $\alpha \in k[x_1,\ldots,x_n]_f$. $\endgroup$
    – Aphelli
    Jun 17, 2020 at 7:12
  • $\begingroup$ Thank you for your clarafication. I think I got it. Also, from your point, if I assume $f_i,f_j$ are distinct irreducible (equivalent to prime in a UFD), then $g_jf_i^u = g_if_j^u$. Hence, $f_i \mid g_i$. So, we may assume $u = 0$, and $g_i/f_i^u$ glues to $g = g_1 = g_2=\dots=g_r$. If $f_i$ are not irreducible, it's indeed difficult to write out their glueing explicitly. $\endgroup$
    – Yu Ning
    Jun 17, 2020 at 7:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .