a topological question regarding the blowing-up of a line

Question

The context of this question is argument (3) in the blow-up section p.28 in Hartshorne. All necessary details are given.

Let $x_1,\dots,x_n$ be affine coordinates for $\mathbb{A}^n$ and $y_1,\dots,y_n$ homogeneous coordinates of $\mathbb{P}^{n-1}$. Let $O:=(0,\dots,0) \in \mathbb{A}^n$. Then the blowing-up of $\mathbb{A}^n$ at $O$ is defined to be the closed subset $X$ of $\mathbb{A}^n \times \mathbb{P}^{n-1}$ given by the equations $x_i y_j = x_j y_i$. Finally, define $\phi: X \rightarrow \mathbb{A}^n$ to be the restriction to $X$ of the projection on the first factor $\mathbb{A}^n \times \mathbb{P}^{n-1} \rightarrow \mathbb{A}^n$.

Now let $L$ be a line of $\mathbb{A}^n$ passing through the origin, given by the parametric equations $x_i = a_i t$ with $a_1 \neq 0$ and $t \in \mathbb{A}^1$. Then $L':=\phi^{-1}(L-O)$ is given by the equations $x_i = a_i t, y_i = a_i, t \in \mathbb{A}^1 - 0$. Now i can state my

Question: Let $\bar{L'}$ be the closed subset of $X$ defined by the equations $x_i = a_i t, y_i = a_i, t \in \mathbb{A}^1$ (observe now that $t$ can take the value $0$). Why is $\bar{L'}$ the closure of $L'$?

Remark: It is enough to show that $O \times (a_1,\dots,a_n)$ is a limit point of $L'$. This is clear in the classical (euclidean) topology by continuity arguments. How about the Zariski topology?

Answer 1

Because $X\setminus \varphi^{-1}(O)$ is isomorphic to $\mathbb A^n\setminus O$, we know that $L'$ is essentially a line in affine space minus one point, which means $L'$ is not closed in $X$**. The given equations uniquely determine a point of the fibre over $O$, namely $Q = (0,0,\ldots, 0)\times [a_1:a_2:\cdots :a_n]$, so that $L'\cup \{Q\}\subseteq X$ is the closed set defined by the equations. Given that $L'$ and its closure must differ in at least one point, $\overline{L'} = L'\cup \{Q\}$.

** Edit: To see why $L'$ is not closed, note that this blowup is the local picture of the blowup of $\mathbb P^n$ at $O = [1:0:0:\cdots:0]$. Given a line $\ell\subseteq \mathbb P^n$ passing through the centre as before, $\ell' = \varphi^{-1}(\ell\setminus \{O\})$ is not closed: if it were, then $\varphi(\ell') = \ell\setminus \{O\}$ would be closed, since morphisms between projective varieties send closed sets to closed sets, but we know that the latter set is not closed. In particular, we can take $\ell = L\cup \{\infty\}$ to be the projective closure of $L$ in $\mathbb P^n$. If $L'$ were closed, then $\ell' = L'\cup\{\infty\}$ would be closed, a contradiction.