# Covering of quasi-projective variety

Proposition Let $$Y$$ be a quasi-projective variety, then $$Y$$ is covered by the open sets $$Y\cap U_i$$, $$i=0,\dots, n$$, which are homeomorphic to quasi-affine varieties via the mapping $$\varphi_i\colon U_i\to \mathbb{A}_k^n$$ defined as: if $$P=(a_0,\dots, a_n)\in U_i$$, then $$\varphi(P)=Q$$, where $$Q$$ is the point with affine coordinates $$\bigg(\frac{a_0}{a_i},\cdots,\frac{a_{i-1}}{a_{i}},\frac{a_{i+1}}{a_{i}},\cdots ,\frac{a_n}{a_i}\bigg)$$

proof. Let $$Y$$ be a quasi-projective variety and consider $$Y\cap U_i$$.

Question. $$Y\cap U_i$$ is open in $$U_i$$?

If the answer to the above question is yes, then since $$\varphi_i\colon U_i\to \mathbb{A}_k^n$$ it is a homeomorphism, $$\varphi_i(Y\cap U_i)$$ is open in $$\mathbb{A}_k^n$$, therefore $$\varphi_i(Y\cap U_i)$$ is irreducible and dense in $$\mathbb{A}_k^n$$, that is $$\overline{\varphi_i(Y\cap U_i)}=\mathbb{A}_k^n,$$ hence $$\varphi_i(Y\cap U_i)\subset\overline{\varphi_i(Y\cap U_i)}$$ is a quasi projective variety.

Finally, since the restriction of a homomorphism is a homeomorphism, we have that $$Y\cap U_i \cong \varphi_i(Y\cap U_i)$$

Edit 1. In fact perhaps, there is no need to observe that $$\varphi(Y\cap U_i)$$ is dense and irreducible in $$\mathbb{A}_k^n$$, because $$\mathbb{A}_k^n$$ is an affine variety and therefore $$\varphi(Y\cap U_i)$$ is an open set in a affine variety, that is quasi-affine variety by definition.

Edit 2 Thanks to KReiser's comment and thanks to him2020's detailed answer I realized that $$Y\cap U_i$$ is not open in $$U_i$$, at this point, however, my proof of the above proposition is unsuccessful. Since I know that the thesis of this proposition is true how can I show it at this point since my procedure is wrong?

Thanks!

• You are very turned around here. Try the example of $Y=\Bbb P^1\subset\Bbb P^2$ as $V(x_2)$ and compute each $Y\cap U_i$. What happens? Commented Jun 27, 2021 at 7:43
• @KReiser $Y\cap U_0=U_0\cap H_3$, $Y\cap U_1=U_1\cap H_3$ and $Y\cap U_2=\emptyset$, right? I would like a formal demonstration though, could you give me some suggestions?
– user805324
Commented Jun 28, 2021 at 5:42
• Edit: put $H_2$ in place of $H_3$
– user805324
Commented Jun 28, 2021 at 5:54
• And are those open in each $U_i$? I think it will be very helpful to you to see what happens in an example before doing anything else. Commented Jun 28, 2021 at 6:02
• $Y\cap U_2=\emptyset$ is open in $U_2$, but $Y\cap U_0$ si closed in $U_0$ and also $Y\cap U_1$ is closed in $U_1$
– user805324
Commented Jun 28, 2021 at 6:13

Question: "Is $$Y∩U_i$$ open in $$U_i$$ for all $$i=0,..,n$$?"

Answer: Let $$Y \subseteq \mathbb{P}^n_k$$ be quasi projective over $$k$$, where $$k$$ is the field of complex numbers. Assume $$Y:=V(I)$$ where $$I \subseteq A[x_0,..,x_n]$$ is a homogeneous prime ideal, and where $$dim(Y):=k < n$$. It follows $$V_i:=U_i \cap Y \subseteq U_i$$ is an open subscheme of $$Y$$ and hence $$dim(V_i) \leq dim(Y)=k < n$$ hence $$V_i$$ cannot be open in $$U_i$$. An open subscheme $$U \subseteq U_i \cong \mathbb{A}^n_k$$ must have $$dim(U)=n=dim(U_i)$$. Hence the answer to your question is: Not in general.

Edit 2: "Thanks to KReiser's comment and thanks to him2020's detailed answer I realized that Y∩Ui is not open in $$U_i$$ , at this point, however, my proof of the above proposition is unsuccessful. Since I know that the thesis of this proposition is true how can I show it at this point since my procedure is wrong? Thanks!"

Your claim: "..therefore $$φ_i(Y∩U_i)$$ is irreducible and dense in $$A^n_k$$,"

Answer: Since $$dim(Y \cap U_i)< n$$ it cannot be dense in $$U_i\cong \mathbb{A}^n_k$$ which has dimension $$n$$.

Comment: "Thanks for the answer, but maybe I didn't make myself clear, how can I prove the above proposition? – Nat. 17 hours ago"

Proposition. "Let Y be a quasi-projective variety, then Y is covered by the open sets $$Y∩U_i$$, i=0,…,n, which are homeomorphic to quasi-affine varieties via the mapping $$φ_i:U_i→A^n_k$$."

Proof: Since $$Y \subseteq \mathbb{P}^n_k$$ is quasi projective, there is a projective variety $$Y' \subseteq \mathbb{P}^n_k$$ with $$Y \subseteq Y'$$ an open subvariety. It follows $$Y' \cap U_i \cong \mathbb{A}^n_k$$ is an affine algebraic variety, and $$Y \cap U_i \subseteq Y' \cap U_i$$ is an open subvariety of an affine variety, hence $$V_i:=Y\cap U_i$$ is a quasi affine variety.QED.

• Thanks for the answer, but maybe I didn't make myself clear, how can I prove the above proposition?
– user805324
Commented Jun 29, 2021 at 16:18