Let $k$ be a field. $U_0 = \mathbb{A}^1_k = \operatorname{Spec}(k[T])$ and $U_1=\mathbb{A}^1_k = \operatorname{Spec}(k[S])$.
$U_{01} = D(T) = \mathbb{A}^1_k\backslash \{0\} = \operatorname{Spec}(k[T]_T)$ and
$U_{10} = D(S) = \mathbb{A}^1_k\backslash \{0\} = \operatorname{Spec}(k[S]_S)$.
Then we have two different Isomorphisms between $U_{01}$ and $U_{10}$:
- We glue $U_{01}$ and $U_{10}$ via the isomorphism $\varphi:U_{01}\to U_{10}$ induced by $k[S]_S\to k[T]_T$ with $S\mapsto T$.
We obtain the affine line with a double origin. - We glue with the isomorphism $\psi: U_{01}\to U_{10}$ induced by $k[S]_S\to k[T]_T$ with $S\mapsto T^{-1}$. We obtain the projective line, which we can imagine to be the scheme 'bowed' to a circle, where we use the extra point to glue $-\infty$ and $\infty$.
I am quite new to the whole scheme structure and the gluing. But I also have problems to see how the spectrum works here:
- Why is $\mathbb{A}^1_k\backslash \{0\} = \operatorname{Spec}(k[T]_T)$? I can maybe imagine how $\operatorname{Spec}(k[T]_T) = D(T)$. Is it because in $D(T)$ are the Ideals with $T$ not in it. And the localization $k[T]_T$ also contains no ideals with $T$ in it?
- My next question is, why does the gluing work as described above? I would imagine that gluing $U_0$ and $U_1$ would give a line with no origin, or even two lines with no origin... How does it happen to be the affine line with a double origin?
Also the circle is not clear to me.
I hope someone can help me understand.
All the best, Luca