Yuchen Liu
Reputation
1,734
Next privilege 2,000 Rep.
 8h awarded Yearling Jul2 awarded Curious Apr18 awarded Yearling Mar7 answered Does $\,f_* \mathcal{O}_{X_T} \cong \mathcal{O}_{T}$ hold in this situation? Jan13 awarded Nice Answer Nov19 answered Why is it better to have the induced map by a line bundle $L$ into projective space map into $\mathbb P |L|^*$? Nov9 answered Poles of abelian differentials Oct25 comment Is the complement of a codimension 2 subvariety of an affine variety affine Thanks for your answer! Oct25 answered Is the complement of a codimension 2 subvariety of an affine variety affine Oct23 comment Is a scheme with a single closed point affine? I think you should require $X$ to be connected, otherwise the disjoint union of a scheme without closed points and the spectrum of a local ring is a counterexample. Oct21 comment Embedding $\mathbb A^2-(0,0)$ into $k^n$, the image is not closed @AsalBeagDubh: Sure, thanks. Oct21 answered Embedding $\mathbb A^2-(0,0)$ into $k^n$, the image is not closed Oct14 comment Dimension of a meromorphic differentials space Do these degree $k$ meromorphic differential forms have only simple poles at $z_i$? Oct13 revised Let $X=\operatorname{Spec} k[w,x,y,z]/(wz−xy)$. Show the Weil divisor cut out by $w=x=0$ is not locally principal. added 122 characters in body Oct13 comment Let $X=\operatorname{Spec} k[w,x,y,z]/(wz−xy)$. Show the Weil divisor cut out by $w=x=0$ is not locally principal. Dear @Gazerun, I have edited my answer by adding a proof of the affiness of $X-Z$. Oct13 revised Let $X=\operatorname{Spec} k[w,x,y,z]/(wz−xy)$. Show the Weil divisor cut out by $w=x=0$ is not locally principal. add a proof of the affiness of $X-Z$ Oct12 answered Let $X=\operatorname{Spec} k[w,x,y,z]/(wz−xy)$. Show the Weil divisor cut out by $w=x=0$ is not locally principal. Sep17 comment A Question on the bijection between ideal sheaf and closed subscheme What does $\mathcal{O}_X/\mathcal{I}=\mathcal{O}_X/\mathcal{J}$ mean in the context? I think it is not saying that they are isomorphic (different closed embeddings from a same scheme can have same set-theoretic image), but they are isomorphic via the quotient of the identity map of $\mathcal{O}_X$, so of course $\mathcal{I}=\mathcal{J}$. Sep13 comment Computing germs of a projective curve So that is why completion is necessary to study singularities? Sep9 comment The Disk and the Punctured Disk Do you mean $\mathbb{C}[[t]]$ by the ring of formal power series? This ring is a discrete valuation ring, so $D$ contains only two elements $(0)$ and $(t)$, why do you say $D$ is a disk?