In a class on Algebraic geometry, we learnt the following - ${\mathbb P}^1_{(1,2)} \cong {\mathbb P}^1$ over the field ${\mathbb C}$. I'm not sure I followed the entire argument exactly. I'll reproduce my version of the argument against it. I hope a mistake can be pointed out.
${\mathbb P}^1$ is defined by two lines ${\mathbb A}^1_{\mathbb C}$ (parameterized by $s$ and $t$ respectively) with the transition function $s = \frac{1}{t}$.
Let us look at ${\mathbb P}^1_{(1,2)}$. A point can be described in projective coordinates $[x_0 : x_1 ] = [ \lambda x_0 ; \lambda^2 x_1 ] $. In the patch $U_0$ defined by $x_0 \neq 0$, we can describe a point here by $[1 ; s = \frac{x_1}{x_0^2} ] $. $s$ is unrestricted and therefore this patch is identical to ${\mathbb A}_{\mathbb C}^1$.
In the patch $U_1$ defined by $x_1 \neq 0$, we now choose $\lambda$ so that $x_1 = 1$. However, we note that one can make two choices here $\lambda = \pm \frac{1}{\sqrt{x_1}}$. Thus, the same point in this patch has two possible coordinates $[t = \frac{x_0}{\sqrt{x_1}};1]$ or $[-t;1]$. In this patch $t$ is unrestricted except that $t$ and $-t$ correspond to the same point. This patch is therefore identical to ${\mathbb A}^1_{\mathbb C} / {\mathbb Z}_2$.
Here's where I have an issue now. Clearly the patches are different. How can ${\mathbb P}^1_{(1,2)} \cong {\mathbb P}^1$??? Unless....
Is ${\mathbb A}^1_{\mathbb C} / {\mathbb Z}_2 \cong {\mathbb A}^1_{\mathbb C}$?? If it is true, how should I then proceed with my argument to prove ${\mathbb P}^1_{(1,2)} \cong {\mathbb P}^1$?
Anyway, this was my argument. Let me also try and reproduce the argument presented by the professor as I understand it. I hope I can get a clearer explanation.
According my professor - $\frac{x_0}{\sqrt{x_1}}$ is clearly not a good coordinate on this patch since it is not single valued. We can therefore define a single-valued coordinate $t = \frac{x_0^2}{x_1}$. Now, since $t$ is unrestricted, this patch is identical to ${\mathbb A}^1_{\mathbb C}$. Further, we note that the transition functions are also the same as that of ${\mathbb P}^1$. Hence, ${\mathbb P}^1_{(1,2)} \cong {\mathbb P}^1$!
In a similar discussion, we went on to analyze ${\mathbb P}^2_{(1,1,2)}$. A similar situation arose here, but in this case, the professor said that the patch $U_2$ (where $x_2 \neq 0$ is identical to ${\mathbb A}^2_{\mathbb C}/{\mathbb Z}_2$ and therefore has an orbifold singularity. I don't understand how this space is singular if the one discussed above is not.
Thank you.
PS - I'm a student of physics and am not well-versed in math jargon. So any clarification of math words that you used would be greatly appreciated. Thanks.
Your space $\mathbb{P}^n_{(a_0, \ldots, a_n)}$ is the quotient of $\mathbb{A}^{n+1} - 0$ (not an affine space, since its global functions are teh same as $\mathbb{A}^{n+1}$) by a $\mathbb{G}_m$ action with weights $a_1, \ldots, a_n$. Say $\mathbb{A}^{n+1} = k[x_0, \ldots, x_n]$. The $\mathbb{G}_m$ stable open affines are still given by $x_i \ne 0$. So you can do everything explicitly. Note that taking the quotient corresponds to taking invariants on local affines in the rings, and invariants for a $\mathbb{G}_m$ action are the degree 0 pieces (this is an exercise: that a $\mathbb{G}_m$-action on a space is the same as a $\mathbb{Z}$-grading on its ring of functions, and that the invariants are the degree 0 part of the grading).
So, for $\mathbb{P}^1_{(1, 2)}$, one has $k[x,y]$ where $x,y$ have weights $1,2$. When $x \ne 0$ one has $k[y/x^2]$ as the degree zero (i.e. invariant) sections. When $y \ne 0$ one has $k[x^2/y]$. These generators are reciprocals of each other, so it's the same as in $\mathbb{P}^1$: two copies of $\mathbb{A}^1$ glued in the same way.
For $\mathbb{P}^2_{(1,1,2)}$ one has $k[x,y,z]$ with weights $1,1,2$. The open affines are $k[y/x, z/x^2]$, $k[x/y, z/y^2]$ and $k[x^2/z, y^2/z, xy/z^2]$. Notice that the while the first two rings are just $\mathbb{A}^2$, the generators for the last ring satisfy a relation: that is $$(x^2/z) (y^2/z) = (xy/z)^2$$ so it's isomorphic to the ring $$k[A,B,C]/AB - C^2$$ which has a type $A_1$/ordinary double point/rational surface singularity, i.e. obtained from quotienting $\mathbb{A}^2$ by $\mathbb{Z}/2$.
Edit: Also, $\mathbb{A}^1/(\mathbb{Z}/2) = \mathbb{A}^1$. To see this, take invariants: $\mathbb{Z}/2$ sends $x \mapsto -x$ in $k[x]$ and the invariant ring is $k[x^2]$ which is still a polynomial ring on one generator. However, $\mathbb{A}^2/(\mathbb{Z}/2)$ where $x \mapsto -x$ and $y \mapsto -y$ (think of $\mathbb{Z}/2$ as the central subgroup of $SL_2$ acting on $\mathbb{A}^2$) has invariant ring $k[x^2, xy, y^2] = k[A,B,C]/AC-B^2$, which is the same ring as above, and singular.