# Sections on a finite union of principal open subsets in affine $n$-space

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.

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?

• Sorry, could you please be more specific? Jun 16, 2020 at 13:17
• Is your issue with the content of my post, or with how it can be used to deduce the final result? Jun 16, 2020 at 14:26
• How it can be used to deduce the final result? Jun 17, 2020 at 3:06
• 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 $\alpha \in k[x_1,\ldots,x_n]_f$. Jun 17, 2020 at 7:12
• 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. Jun 17, 2020 at 7:55