Intuition of projective line via glueing

Question

I am reading the construction of projective line (4.4.6) following Prof. Vakil's Algebraic geometry notes FOAG. However, I fail to understand few lines during the discussion.
Let $X=\operatorname{Spec} k[t]$ and $Y=\operatorname{Spec} k[u]$. We glue the open subsets $D(t)\cong \operatorname{Spec}k[t]_t$ and $D(u)\cong \operatorname{Spec}k[u]_u$ along the isomorphism sending $t$ to $\frac{1}{u}$.
Now we let $k$ be algebraically closed to build intuition.
Generic Points: We have generic point $[(0)]\in \operatorname{Spec}k[t]_t$, which is also sent to generic point of $\operatorname{Spec}k[u]_u$ by the isomorphism, hence they are glued together.
Closed Points (also my question): First of all, we see that for $a\in k$, prime ideas of the form $[(t-a)]\in \operatorname{Spec}k[t]$ are the closed points (also of $\operatorname{Spec}k[t]_t$). Then the isomorphism should send it to $[(\frac{1}{u}-a)]$. However, in the text, it is claimed that "the a on the t-line is glued to $\frac{1}{a}$ on the u-line" (if we think of each $\operatorname{Spec}k[t]$ as a line). If I am not misunderstanding him, he claimed that the two points $[(t-a)]$ and $[(u-\frac{1}{a})]$ are glued together. I am really confused.
I appreciate any help! Thanks in advance.

Answer 1

In the ring $k[u]_u$, for $a\neq0$ we have an equality of ideals $(u-1/a)=(1/u-a)$: to go from left to right, multiply the generator by $a/u$, and to go from right to left, multiply the generator by $u/a$.