I'm trying to solve a problem from Jenia Tevelev's notes on GIT. (Problem 5 at the end of this pdf.)
Compute $$\operatorname{Proj}\frac{\mathbb{C}[x,y,z]}{(x^5+y^3+z^2)}$$ where $\operatorname{wt.}x,y,z=12,20,30$.
Here is what I have so far.
Let $R=\mathbb{C}[x,y,z]$ with the above weighting, so we have the natural graded ring structure given by the weighted degree: $R=\bigoplus_{n\geqslant0}R_n$.
Then $R\cong R^{(2)}\cong R^{(4)}\cong R^{(20)}\cong R^{(60)}$, where $$R^{(k)}=\bigoplus_{n\geqslant0}R_{kn}.$$
Under these transformation, the ideal given by the polynomial in the question transforms as $$(x^5+y^3+z^2)\mapsto (x^5+y^3+z^2)\mapsto (x^5+y^3+z)\mapsto (x+y^3+z)\mapsto (x+y+z).$$
So, with $S=\frac{\mathbb{C}[x,y,z]}{(x^5+y^3+z^2)}$ we have that $$S^{(60)} = \left(\frac{\mathbb{C}[x,y,z]}{(x^5+y^3+z^2)}\right)^{(60)} = \frac{\mathbb{C}[x,y,z]^{(60)}}{(x+y+z)} = \frac{\mathbb{C}[x^5,y^3,z^2]}{(x+y+z)}.$$
Questions:
- Is all of the above right? In particular, is it true that $$\left(\frac{\mathbb{C}[x,y,z]}{(x^5+y^3+z^2)}\right)^{(60)} \neq \frac{\mathbb{C}[x,y,z]^{(60)}}{(x^5+y^3+z^2)}.$$
That is, we can't just 'carry through' the truncation, we have to see how the ideal changes as well. - How can I proceed from here?
Edit:
It seems that the rather useful fact that I missed is that
$$\operatorname{Proj}\frac{k[x_0,\ldots,x_n]}{I}\cong\mathbb{V}(I)\subset\mathbb{P}(a_0,\ldots,a_n)$$
where $\operatorname{wt.}x_i=a_i$.
Then the problem boils down to understanding $$\frac{\mathbb{V}_{\mathrm{aff}}(x^5+y^3+z^2)\setminus\{0\}}{\mathbb{G}_m}$$ where $\mathbb{G}_m$ acts on $\mathbb{A}^3$ via the standard way for weighted projective space.
But I'm not sure if this is true when the generators are of different degrees, in which case we do need to look at the truncation first.
(If this is not the case then I'd be super interested to know how both methods proceed – via the truncation and just jumping straight in.)
Then it looks like it might make sense to guess that $$\operatorname{Proj}\frac{\mathbb{C}[x^5,y^3,z^2]}{(x+y+z)} = \mathbb{V}(x+y+z)\subset\mathbb{P}(60,60,60)$$ but the issue is that $(x+y+z)$ is not homogeneous, and our ideal is given in terms of elements that we can't make from our generators.
Best Answer
From Working with Weighted Complete Intersections by Iano-Fletcher, which appears to be available freely here:
First, note that this is exactly your case. Multiplying all of the weights by two amounts to making the $\mathbb{G}_m$ action non-effective (the subgroup $\{\pm 1\} \subseteq \mathbb{G}_m$ stabilizes everything).
I think what you missed above (and which confused me thinking about it as well) is that "$x$", "$y$", etc. don't mean the same thing after each isomorphism. Let's do the first one explicitly, and the others will be similar.
We know that $\mathbb{P}(6,10, 15) = \text{Proj}(k[x,y,z])$ with the grading $|x| = 6, |y| = 10, |z| = 15$ is isomorphic to $\mathbb{P}(6, 2, 3)$. However, this isomorphism is not as simple as the isomorphism of graded rings showing $\mathbb{P}(6,10,15) \simeq \mathbb{P}(12, 20, 30)$ (for example). It's somewhat non-canonical (this has something to do with the fact that the affine cones of the resulting varieties aren't isomorphic). Let $S$ be the first graded ring and $T$ the second. Now, we know that $\text{Proj}(S) \simeq \text{Proj}(S^{(5)})$. How does this isomorphism act on closed sets? (This is all spelled out in EGA II.2.4.7). The isomorphism is induced by the inclusion $S^{(5)} \hookrightarrow S$. Under this map, the homogeneous primes of $S$ containing an element $f$ of degree $d$ correspond exactly to the homogeneous primes of $S^{(5)}$ containing $f^5$ (note $V_+(f) = V_+(f^d)$!). Our element, $x^2 + y^3 + z^5$ is actually homogeneous of degree $30$, so $x^2 + y^3 + z^5 \in S^{(5)}$ and thus, in $S^{(5)}$, our homogeneous ideal is just the ideal of $S^{(5)}$ generated by the same polynomial.
$S^{(5)}$ is generated by monomials $x^a y^b z^c$ with $15a + 10b + 6c = 5n$ for some $m$. Since $5$ doesn't divide $6$, this means that $5|c$, and thus any such monomial is of the form $x^{a} y^b z^{5c'}$. Since any monomial of this form has degree a multiple of $5$, we conclude that these span $S^{(5)}$. This shows that $S^{(5)}$ is $k[x, y, z^5]$ where $x$ has weight $15$, $y$ has weight $10$, and $z^5$ has weight $30$. This is isomorphic to the graded ring $k[s, t, r]$, up to an overall multiplicative factor in the grading, where $s$ has weight $3$, $t$ has weight $2$, and $r$ has weight $6$, by the map $s \mapsto x, t \mapsto y, r \mapsto z^5$. This map induces a canonical isomorphism on their $\text{Proj}$ (which, if you recall, is constructed by considering degree $0$ fractions, so multiplying the degree of everything by a constant doesn't change this). In particular, the isomorphism $\text{Proj}(S^{(5)}) \simeq \text{Proj}(T)$ takes $V_+(x^2 + y^3 + z^5)$ to $V_+(s^2 + t^3 + r)$. Note that this new polynomial, $s^2 + t^3 + r$ is homogeneous of degree $6$ in $T$. Continuing this way, we finally get an isomorphism of $\text{Proj}(S)$ with $\text{Proj}(k[x,y,z])$ where the latter has the usual $(1,1,1)$ grading. This isomorphism takes $x^2 + y^3 + z^5$ to $x + y + z$, which is of course a homogenous linear function in three variables. Thus, $V_+(x^2 + y^3 + z^5) \subseteq \mathbb{P}(15,10,6)$ is isomorphic to $V_+(x+y+z) \simeq \mathbb{P^1} \subseteq \mathbb{P}(1,1,1) = \mathbb{P}^2$.
Notice that what I did here is essentially the same as the algebra you did above. However, there are a few important differences to point out:
First, it appears that some of the confusion stems from the fact that you used the letters $R$, $x,y,z$ to represent different but isomorphic things.
Second, it is not the case that $R \simeq R^{(d)}$ for any $d$. It is only the case that $\text{Proj}(R) \simeq \text{Proj}(R^{(d)})$. Note that the affine cones over our initial and final varieties are very different! The former is a singular hypersurface of degree $5$, while the latter is just a plane.
Finally, I'd like to go back to my earlier comment, which is interesting in light of the understanding that we're really just dealing with $\mathbb{P}^1$. First of all, you're completely correct in the interpretation that $\text{Proj}(k[x,y,z])/f(x,y,z)$ is just the quotient of $V(f(x,y,z)) \subset \mathbb{A}^3$ by the action of the circle $\mathbb{G}_m$. In this case, the action is $\zeta: (x,y,z) \mapsto (\zeta^{15} x, \zeta^{10}y, \zeta^{5}z)$. Indeed, it's true (and I don't know a good reference for this fact; I'd appreciate if someone finds one) that the Proj of any graded algebra is the quotient of its spectrum by a circle action.
This interpretation is very interesting topologically. Knowing that the quasi-homogeneous variety is just $\mathbb{P}^1$ tells us that the link of the singularity $L:= \{(x,y,z) \in S^5 \subseteq \mathbb{C}^3 \mid x^2 + y^3 + z^5 = 0\}$ admits a fixed point-free circle action such that the quotient is just $S^2$! This circle action is "almost" a fiber bundle; the reason it isn't is that at points when, for example, $z = 0$, the fiber $\{(\zeta^{15}x, \zeta^{10}y, 0)\}$ has multiplicity $5$. Topologists refer to such $3$-manifolds as Seifert fiber spaces, and they form interesting examples of $3$-manifolds whose algebraic topology is generally manageable. Indeed, knowing the genus of the quotient surface in addition to some information about orientability (which is trivial in this case since everything is complex) and the local description of the fibers is almost enough to determine the fundamental group and homology entirely. In fact, this particular $3$-manifold is very famous! It's the Poincaré Homology Sphere, the earliest historical example of a manifold with the homology groups of $S^3$ which is not homeomorphic to it.