6
votes
How weird can the level set of a polynomial be?
Let me explain an upper bound on the number of components, which is asymptotically consistent with your conjecture.
I will be using your notation where the dimension is $d$ and the degree is $n$ even ...
- 105k
6
votes
Accepted
Birational K3 surfaces
Not necessarily. Take your favorite K3 surface $X'$ of degree $2d$ in $\mathbb{P}^N$ and let $X \subset \mathbb{P}^3$ be its general linear projection. Then $X'$ is birational to $X$ and $X$ is a (...
- 19.2k
3
votes
Accepted
Are the connected components of the product of two connected schemes over a field necessarily finite?
To sum up: when $X$ and $Y$ are affine over a common base $k$, then $X \times_k Y$ is also affine over $k$, hence quasicompact; however, even if $X$ and $Y$ have finitely many connected components, $X ...
- 93.3k
3
votes
Accepted
Torsionfree quotient of vector bundles is also a vector bundle?
Consider on $\mathbb{A}^2$ the exact sequence
$$
0 \to
\mathcal{O} \stackrel{(x,y)}\to
\mathcal{O} \oplus \mathcal{O} \to
I_{(0,0)} \to
0.
$$
Here $I_{(0,0)}$, the ideal of the point $(0,0)$, is ...
- 19.2k
3
votes
Accepted
Intersection of all divisors in a linear system
I am not too familiar with Griffiths-Harris, so take this answer with a grain of salt (and please correct me if I am wrong).
Your interpretation of the statement seems correct (with the addition that ...
- 457
3
votes
Accepted
Questions / Understanding Check on the Curvature of the Chern Connection.
For a metric-compatible connection, and a local orthonormal frame, the matrix representation of the curvature is skew-hermitian. So, if they say hermitian, it's just a typo.
Effects of levels of ...
- 61.8k
2
votes
Integrating classes in Čech cohomology
This is answered in my book Introduction to Complex Manifolds, Theorem 7.14. Using the conventions I use there, the first Chern class corresponds to $-\rho$.
(See the remark following the proof of ...
- 49.1k
1
vote
Accepted
Hodge-Riemann form is non-degenerate
Let $\xi \in P^{n - k - 2i}$ and let $\eta \in P^{n - k - 2j}$, so that $L^i \xi \in H^{n - k}$ and $L^j \eta \in H^{n - k}$. Then we have
\begin{align*}Q(L^i\xi, L^j \eta) &= \int_M \omega^i \...
- 2,733
1
vote
How can I find a family of hypersurfaces in $\Bbb A^n$ with normal singular points for all $n\geq 3$?
Your proposed solution at the end of the post will work, but I think the point of the exercise here is to reduce the problem of finding a hypersurface in $\Bbb A^n$ for each $n\geq 3$ to finding a ...
- 71.1k
1
vote
Can support of finitely generated module be complement of single point?
You misunderstood something, Noetherian is not relevant here.
You can consider the $R \times S$-module $M \times N$ for an $R$-module $M$ and an $S$-module $N$. Let $R,S$ be Artinian local, and take $...
- 176k
1
vote
Accepted
Positive divisors on a projective surface
First write $L$ as a difference of effective divisors: $L = L_+ - L_-$. Now use that for any effective divisor $E$ and any ample divisor $A$, $D = E + mA$ is ample for $m \gg 0$ (you should be able to ...
- 3,798
1
vote
Accepted
Finite dominant morphism is surjective
Set for simplicity $A = A(Y), B = A(X)$. I will assume that varieties are irreducible, so that $A$ and $B$ are integral domains.
You are right that this basically follows from going-up. However, going-...
- 10.4k
1
vote
Clarifying the definition of a geometrically connected variety
The thing that's going on here is that the following are all equivalent and all equally good to take as a definition for "$\operatorname{Spec} A$ is geometrically connected":
$\operatorname{...
- 71.1k
1
vote
Space of plane cubics though (a certain arrangement of) 8 points is a subspace of the space of all plane cubics paramaterisable by $\mathbb{P}^1.$
The condition that each point $P_i$ is on the cubic gives a linear condition on the space of coefficients, so the locus of cubics through $P_1,\cdots,P_8$ is the intersection of 8 hyperplanes in the $\...
- 71.1k
1
vote
Computations of vanishing ideals
If the characteristic of $k$ is $2$, then $\newcommand{\V}{\mathbb{V}} x^2 + y^2 - 1 = (x+y+1)^2$, so
\begin{align*}
\V(x^2 + y^2 - 1) = \V((x+y+1)^2) = \V(x+y+1) \, .
\end{align*}
Then
$$
V = \V(x+y+...
- 20.1k
Only top scored, non community-wiki answers of a minimum length are eligible