I'm working through an introductory problem in scheme theory and just want to make sure I'm understanding the concepts correctly. The problem states to consider the following two rings, specify their Zariski open sets, as well as the restriction maps:
- $A = \mathbb{C}[t]/\left (t^2-t\right) = \{a+bt \; | \; a,b\in\mathbb{C}\}$
- $B = \mathbb{C}[t]/\left (t^3-t^2\right) = \{a+bt+ct^2 \; | \; a,b,c\in\mathbb{C}\}$
For ring $A$ we can compute $X_A = \text{Spec} A = \left \{\widetilde{(t)}, \widetilde{(t-1)}\right \}$ where the $\sim$ is to signify that we are really dealing with the quotients $\widetilde{(t)} = (t)/\left (t^2-t\right )$. I arrived at this since prime ideals in a ring $R/I$ are projections of prime ideals of $R$ containing $I$. Furthermore, I concluded that the Zariski topology on $X_A$ is discrete since both prime ideals can be written as the zero set of an ideal (namely itself) in $A$. At the risk of overdoing the notation I'll call the open sets $U_t$ and $U_{t-1}$, then the structure sheaf elements should be
\begin{eqnarray*} \mathcal{O}(U_t) & = & A_{\widetilde{(t)}} \;\; =\;\; \left \{\left .\frac{a+bt}{c+dt} \;\right | \; c+dt \not \in \widetilde{(t)}\right \} \\ \mathcal{O}(U_{t-1}) & = & A_{\widetilde{(t-1)}} \;\; =\;\; \left \{\left .\frac{a+bt}{c+dt} \;\right | \; c+dt \not \in \widetilde{(t-1)}\right \}. \end{eqnarray*}
Is my logic more or less correct? We as well have that $U_t \cap U_{t-1} = \emptyset$ so we need not worry about the sheaf element in that case.
For ring $B$ much of my logic remains the same in that $X_B = \text{Spec} B = \left \{\widetilde{(t)}, \widetilde{(t-1)}\right \}$, this is discrete as well for similar reasons, and the only difference is in the forms of the structure sheaf elements:
\begin{eqnarray*} \mathcal{O}(U_t) & = & B_{\widetilde{(t)}} \;\; =\;\; \left \{\left .\frac{a+bt+ct^2}{d+ft+gt^2} \;\right | \; d+ft+gt^2 \not \in \widetilde{(t)}\right \} \\ \mathcal{O}(U_{t-1}) & = & B_{\widetilde{(t-1)}} \;\; =\;\; \left \{\left .\frac{a+bt+ct^2}{d+ft+gt^2} \;\right | \; d+ft+gt^2 \not \in \widetilde{(t-1)}\right \}. \end{eqnarray*}
I'm studying algebraic geometry for the first time out of Bosch's Algebraic Geometry and Commutative Algebra, and I also don't have much of an algebra background aside from groups, Lie groups, and basic ring theory.
Response to Comments
I think I caught KReiser's comment about the denominators in ring $B$ being incorrect. In response to the comments about rationalization, for ring $A$ we can take
$$ \frac{a+bt}{c+dt} \;\; =\;\; \frac{at+bt^2}{ct+dt^2} \;\; =\;\; \frac{at+bt}{ct+dt} \;\; =\;\; \frac{a+b}{c+d} $$
which means that both sheaf elements in ring $A$ are just copies of $\mathbb{C}$. So can we conclude that $\mathcal{O}(U_t) \cong \mathbb{C}$ as well as $\mathcal{O}(U_{t-1}) \cong \mathbb{C}$?
Moving on to ring $B$ we can similarly compute
$$ \frac{a+bt+ct^2}{d+ft+gt^2} \;\; =\;\; \frac{at^2 + bt^3 +gt^4}{dt^2+ft^3+gt^4} \;\; =\;\; \frac{a+b+c}{d+f+g} $$
so, do we get the same result for ring $B$?