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Let $\Delta$ be a fan, consisting of cones $\sigma_0=conv(e_1,e_1+e_2)$ and $\sigma_2=conv(e_1+e_2,e_2)$ and $\tau=\sigma_0\cap\sigma_1=conv(e_1+e_2)$. The dual cones are $\sigma_0^\vee= conv(e_2,e_1\!-\!e_2)$ and $\sigma_1^\vee= conv(e_2\!-\!e_1,e_1)$ and $\tau^\vee= conv(e_2\!-\!e_1,e_1\!-\!e_2,e_1,e_2)= conv(e_2\!-\!e_1,e_1\!-\!e_2,e_1)= conv(e_2\!-\!e_1,e_1\!-\!e_2,e_2)$.

The corresponding semigroup algebras are $S_{\sigma_0}= \mathbb{C}[\mathbf{x}^{e_2},\mathbf{x}^{e_1\!-\!e_2}] =\mathbb{C}[y,xy^{-1}]$ and $S_{\sigma_1}= \mathbb{C}[\mathbf{x}^{e_2\!-\!e_1},\mathbf{x}^{e_1}]= \mathbb{C}[x^{-1}y,x]$ and $S_\tau= \mathbb{C}[\mathbf{x}^{e_2\!-\!e_1}, \mathbf{x}^{e_1\!-\!e_2}, \mathbf{x}^{e_1},\mathbf{x}^{e_2}]$ $=$ $\mathbb{C}[x^{-1}y,y^{-1}x,x,y]$ $=$ $\mathbb{C}[x^{-1}y,y^{-1}x,x]= \mathbb{C}[x^{-1}y,y^{-1}x,y]$ inside $\mathbb{C}[x^{\pm1},y^{\pm1}]$. The three associated affine schemes $U_{\sigma_0}, U_{\sigma_1}, U_\tau$ are their spectrums.

The homomorphism of $\mathbb{C}$-algebras $\beta_0\!: \mathbb{C}[X,Y]\rightarrow\mathbb{C}[x/y,y]$ that sends $X\!\mapsto\!x/y,Y\!\mapsto\!y$ is an isomorphism. The homomorphism of $\mathbb{C}$-algebras $\beta_1\!: \mathbb{C}[X,Y]\rightarrow\mathbb{C}[x,y/x]$ that sends $X\!\mapsto\!x,Y\!\mapsto\!y/x$ is an isomorphism. Thus $U_{\sigma_0}\!=\!\mathbb{C}^2\!=\!U_{\sigma_1}$ via $\overset{\beta_0^\ast}{\longleftarrow}\!\ldots\!\overset{\beta_1^\ast}{\longrightarrow}$. Furthermore, $\mathbb{C}[X,x]_X\!=\!\mathbb{C}[X,x,X^{-1}]\!\cong\!\mathbb{C}[x/y,x,y/x] \!=\! \mathbb{C}[x/y,y,y/x]\!\cong\!\mathbb{C}[Y^{-1}\!,y,Y]\!=\!\mathbb{C}[y,Y]_Y$, so $\mathbb{C}^\ast\!\!\times\!\mathbb{C} \!=\!U_\tau\!=\! \mathbb{C}\!\times\!\mathbb{C}^\ast$ and $X\!=\!Y^{-1}$.

We have inclusions $\sigma_0\!\hookleftarrow\!\tau\!\hookrightarrow\!\sigma_1$, hence inclusions $\sigma_0^\vee\!\hookrightarrow\!\tau^\vee\!\hookleftarrow\!\sigma_1^\vee$, thus morphisms $\mathbb{C}[\sigma_0^\vee]\!\overset{\iota_0}{\hookrightarrow}\!\mathbb{C}[\tau^\vee]\!\overset{\iota_1}{\hookleftarrow}\!\mathbb{C}[\sigma_1^\vee]$, hence morphisms $Spec\,\mathbb{C}[\sigma_0^\vee]\!\overset{\iota_0^\ast}{\leftarrow}\!Spec\,\mathbb{C}[\tau^\vee]\!\overset{\iota_1^\ast}{\rightarrow}\!Spec\,\mathbb{C}[\sigma_1^\vee]$. Our morphisms $\mathbb{C}[x/y,y]\!\overset{\iota_0}{\hookrightarrow}\!\mathbb{C}[x/y,x,y/x]\!=\!\mathbb{C}[x/y,y,y/x]\!\overset{\iota_1}{\hookleftarrow}\!\mathbb{C}[x,y/x]$ send $(x/y,y)\!\rightarrow\!(x/y,y\!=\!x\frac{y}{x})$ and $(y\frac{x}{y}\!=\!x,y/x)\!\leftarrow\!(x,y/x)$, i.e. morphisms $\mathbb{C}[X,Y]\!\overset{\iota_0}{\hookrightarrow}\!\mathbb{C}[X,x,X^{-1}]\!=\!\mathbb{C}[Y^{-1},y,Y]\!\overset{\iota_1}{\hookleftarrow}\!\mathbb{C}[X,Y]$ send $(X,Y)\!\rightarrow\!(X,x/X)$ and $(y/Y,Y)\!\leftarrow\!(X,Y)$. On ideals, they map $\langle X\!-\!u,Y\!-\!v\rangle \!\overset{\iota_0}{\rightarrow}\! \langle X\!-\!u,x/X\!-\!v\rangle \!=\! \langle X\!-\!u,x\!-\!vX\rangle$ and $\langle y\!-\!uY,Y\!-\!v\rangle \!=\! \langle y/Y\!-\!u,Y\!-\!v\rangle\!\overset{\iota_0}{\leftarrow}\!\langle X\!-\!u,Y\!-\!v\rangle$, so preimage morphisms $\mathbb{C}^2\!\overset{\iota_0^\ast}{\leftarrow}\!\mathbb{C}^\ast\!\!\times\!\mathbb{C}\!=\!\mathbb{C}\!\times\!\mathbb{C}^\ast\!\overset{\iota_1^\ast}{\rightarrow}\!\mathbb{C}^2$ send $\langle X\!-\!u,Y\!-\!???\rangle\!\leftarrow\!\langle X\!-\!u,x\!-\!v\rangle$ and $\langle y\!-\!u,Y\!-\!v\rangle\!\rightarrow\!\langle X\!-\!???,Y\!-\!v\rangle$, i.e. $(u,???)\!\leftarrow\!(u,v)$ and $(u,v)\!\rightarrow\!(???,v)$.

Question 1: Is everything so far correct?

Question 2: The usual identification is that $Max\,\mathbb{C}[x,y]=\{\langle x\!-\!u,y\!-\!v\rangle;\, u,v\!\in\!\mathbb{C}\}\equiv\mathbb{C}^2$. However, in the literature, $Spec\,\mathbb{C}[x,y]$ is identified with $\mathbb{C}^2$ (I'm not sure this is the right thing to do). I would like to glue together $U_{\sigma_0}$ and $U_{\sigma_1}$ via $U_\tau$ to obtain a scheme. How can I obtain a gluing isomorphism? The problem is that for $\iota_0,\iota_1$, the preimage of a maximal ideal is not a maximal ideal. How does this gluing morphism map points $\mathbb{C}^\ast\!\!\times\!\mathbb{C}\rightarrow\mathbb{C}\!\times\!\mathbb{C}^\ast$? I'd like to see that is sends $(u,v)\mapsto(uv,u^{-1})$. Are we gluing two copies of $\mathbb{C}^2$ along $\mathbb{C}\!\times\!\mathbb{C}^\ast$ or along $\mathbb{C}^\ast\!\times\!\mathbb{C}^\ast$? How are $\mathbb{C}^\ast\!\!\times\!\mathbb{C}$ and $\mathbb{C}\!\times\!\mathbb{C}^\ast$ via $\iota_0^\ast$ and $\iota_1^\ast$ embedded in $\mathbb{C}^2$?

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Your arguments look correct. You identify $S_{\tau} = \mathbb C[x/y,y/x,x,y]$ which is a subring of $\mathbb C(x,y)$ containing $S_{\sigma_0}$ and $S_{\sigma_1}.$ Also, you are correct that $S_{\tau}$ is the localization of $S_{\sigma_0}$ at $x/y$ and of $S_{\sigma_1}$ at $y/x.$ Thus, $U_{\tau}\subseteq U_{\sigma_0},U_{\sigma_1}$ is an open subscheme of each of the larger affine spaces.

Once we know that $U_{\tau}$ is a common open subscheme of $U_{\sigma_0}$ and $U_{\sigma_1},$ we can glue along it. We use the map $\mathbb C[y,x/y,y/x]\to\mathbb C[x,y/x,x/y]$ taking $x/y$ and $y/x$ to their counterparts, and $y\mapsto x\cdot y/x.$ Note that this is an isomorphism, since the inverse will take $x\mapsto y\cdot x/y$ and composing the two we get $y\mapsto(y\cdot x/y)\cdot y/x=y$ and coversely $x\mapsto x.$

In the literature $\mathbb C$ (and $\mathbb C^*$) has a double meaning. Often it means what you think: complex numbers. Sometimes it means the scheme $\operatorname{Spec}(\mathbb C[t])$ however, and similarly for $\mathbb C^2$ and $\operatorname{Spec}(\mathbb C[x,y])$, etc. The closed points of $\operatorname{Spec}(\mathbb C[t])$ correspond to complex numbers, but there is also the generic point $(0)$ lurking about. In $\operatorname{Spec}(\mathbb C[x,y])$ we have all the points $(x-a,y-b)$ which correspond to pairs of complex numbers $(a,b)\in\mathbb C^2$, but there are also points correponding to non-maximal primes like $(x^2+y^2-1)$ and $(0)$ and so on.

The fact that $U_{\tau}$ is a common open subscheme of $U_{\sigma_0}$ and $U_{\sigma_1}$ only depends on the ring homomorphisms we identified. In fact, one of the (many) reasons we work with prime spectra rather than maximal spectra is what you point out -- in general the inverse image of a maximal ideal is a prime ideal, but need not be maximal.

Recall that the coordinates on $U_{\sigma_0}$ are $y,x/y$ and on $U_{\sigma_1}$ are $x,y/x.$ On their overlap $U_{\tau}$ we have the correspondence $y\leftrightarrow x\cdot y/x$, $y\cdot x/y\leftrightarrow x$, and $x/y\leftrightarrow (y/x)^{-1}.$ Thus, the gluing isomorphism mapping say $U_{\sigma_1}\cap U_{\tau}\to U_{\sigma_0}\cap U_{\tau}$ takes $(x,y/x)\mapsto (y\cdot x/y,(x/y)^{-1})$, i.e., $\mathbb C\times\mathbb C^\times\to\mathbb C\times\mathbb C^\times,(u,v)\to(s\cdot t,t^{-1})$ choosing some different names for the coordinates. This is basically what you have written, and it depends only on the ring homomorphism. However, note that this expression explicitly describes what happens only to closed points of these schemes.

In particular, we are gluing two copies of $\mathbb C^2$ (the spectrum version) along $\mathbb C\times\mathbb C^\times$ (which implies the charts share $\mathbb C^\times\times\mathbb C^\times$ as well, of course).

$U_{\tau}$ is embedded in $U_{\sigma_0}$ in the obvious way, which would take a closed point $(a,b)$ to $(a,b)$. Similarly, $U_\tau$ is embedded in $U_{\sigma_1}$ in the obvious way, and this takes $(c,d)$ to $(c,d)$. The trick is that this uses the local coordinates. So $(a,b)$ (with say $a$ and $b$ nonzero) describes two different points of $U_{\tau}\subseteq \mathbb C^2$ depending on which chart $U_{\sigma_i}$ we are in. We have to tread a little carefully here, but it is no serious difficulty.

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  • $\begingroup$ It is known that the above variety is 'isomorphic' (how can a scheme and affine/projective variety be $\cong$???) to $Z=V(xt_1\!-\!yt_0)\subseteq\mathbb{C}^2\!\times\!\mathbb{P}^1$. In this description, for $U_0\!=\!\{(x,y,t_0:t_1)\!\in\!V;\, t_0\!\neq\!0\}$ and $U_1\!=\!\{(x,y,t_0:t_1)\!\in\!V;\, t_1\!\neq\!0\}$, we have $Z=U_0\!\cup\!U_1$ and $U_0\!\cong\!\mathbb{C}^2\!\cong\!U_1$ and $U_0\!\cap\!U_1\!\cong\!\mathbb{C}^\ast\!\times\!\mathbb{C}^\ast$. How can this be: once we glue along $\mathbb{C}^\ast\!\times\!\mathbb{C}^\ast$ and once along $\mathbb{C}^\ast\!\times\!\mathbb{C}$? $\endgroup$ – Leo Jun 7 '13 at 3:32
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    $\begingroup$ Dear @LeonLampret, in scheme theory an 'abstract variety' is a type of scheme (pp.104-5 in Hartshorne). So we are saying that the scheme is isomorphic to the image of the variety under the functor $t$ (from Hartshorne Prop. II.4.10). Intuitively, we pretend the geometric points of the scheme are the same thing as the original variety. But it seems to me that $U_0\cap U_1=\mathbb C\times\mathbb C^\times,$ for if $t_0,t_1\neq0$, we can set $t_0=1,$ and then $t_1\in\mathbb C^\times$ can be anything. Then $x$ determines $y$ and vice-versa by $xt_1=yt_0=y,$ but we can take any $x\in\mathbb C.$ $\endgroup$ – Andrew Jun 7 '13 at 19:42
  • $\begingroup$ One last question: how is $U_\tau\!=\!\mathbb{C}^\ast\!\times\!\mathbb{C}$ embedded in $U_{\sigma_0}\!=\!\mathbb{C}^2$ and in $U_{\sigma_1}\!=\!\mathbb{C}^2$, i.e. what is its presentation $\mathbb{C}^2\!\setminus\!\mathcal{Z}(\ldots)$ as a complement of an affine algebraic set? More precisely, how do morphisms $\mathbb{C}^2\overset{\iota_0^\ast}{\leftarrow} \mathbb{C}^\ast\!\times\!\mathbb{C} \!\equiv\! \mathbb{C}\!\times\!\mathbb{C}^\ast\overset{\iota_1^\ast}{\rightarrow}\mathbb{C}^2$ map points $(u,v)$? $\endgroup$ – Leo Jun 8 '13 at 0:56
  • $\begingroup$ Dear @LeonLampret, this may sound silly, but $\iota_0^*$ and $\iota_1^*$ are both inclusions $\mathbb C\times\mathbb C^\times\hookrightarrow\mathbb C^2,$ so whatever the coordinates are for $\mathbb C^2,$ say $(u,v),$ the maps will take $(u,v)$ to $(u,v).$ However, the two copies of $\mathbb C^2$ have different coordinates, so if we want to change coordinates, we must use the map $(u,v)\mapsto(s\cdot t,t^{-1})$ that I mentioned, where $(u,v)$ and $(s,t)$ are the two sets of coordinates (or in our case $x,y/x$ and $x/y,y$). $\endgroup$ – Andrew Jun 12 '13 at 13:51

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