Fredrik Meyer
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 2d comment Torus action and multigrading. See this mathoverflow answer. It answers positively in the case of a one-dimensional torus. It seems that the proof can be modified to deal with a higher-dimensional torus as well. 2d comment Torus action and multigrading. What is $\mathbb C[G]$? Is it the coordinate ring or the group ring? May 2 comment Is this toric variety the blowup of $\mathbb C^2$ at some point? No. The blowup of a point in $\mathbb C^2$ is a smooth variety, but the cone $\sigma_1 = \langle u_0,u_1 \rangle$ is not a smooth cone, so your variety has a singular point. May 1 answered Smooth curve of genus $1$ in $\mathbb{P}_{\mathbb{C}}^1\times \mathbb{P}_{\mathbb{C}}^1$. Apr 17 comment How are non-homogenous elliptic curves projective varieties? @CaptianLama Proj is not a functor ;) Apr 15 comment Listing all the ideals of a quotient ring Another hint: $1+x+x^2+x^3=(x^4-1)/(x-1)$. Apr 14 comment Betti numbers of complex “sphere” Good question! Do you have a source on the statement of the vanishing of the Betti numbers for $i \geq n$? I'd very much like to know. Apr 14 comment Maximal ideal in $\mathbb{R}[x,y]/ (x^2 + y^2 -1)$ A ideal $I$ is maximal in a ring $A$ if and only if $A/I$ is a field. So try computing the quotient. Apr 14 answered math software - permutation group elements operation Apr 14 answered Integral of Differential 1-form Apr 14 comment How to show $\text{Sym}^n(\mathbb{P}^1)=\mathbb{P}^n$ At least set-wise you can see this as follows: $n$ unordered points in $\mathbb P^1$ determine up to scalar a unique polynomial (the polynomial with those points as zeroes) in $\Gamma(\mathbb P^1, \mathscr O_{\mathbb P^1}(n))$. This vector space has dimension $n+1$, so its projectivisation is $\mathbb P^n$. Apr 11 comment Canonical scheme structure on the singular locus of a variety One high-tech way to see this is to note that $S_X$ is the same as the locus of points where $\Omega_X^1$ has rank less than $\dim X$. Since the sheaf $\Omega_X^1$ is an invariant of $X$, we're done. Apr 6 comment Vanishing Jacobian determinant @Hans Thank you, that was indeed a typo. Apr 6 revised Vanishing Jacobian determinant edited body Apr 6 comment Proving that every $n$-submanifold of $\mathbb{R}^{n}$ has a natural orientation Since $M$ is a n-dimensional submanifold of $\mathbb R^n$, it is an open set in $\mathbb R^n$. Hence you can use $M$ itself as a coordinate chart, so that the only Jacobian you get is the identity. Apr 6 answered Vanishing Jacobian determinant Apr 3 comment Induced map on cohomology Thank you. I've tried to look at the details of this: intersecting with $D$ gives a union of three disjoint $\mathbb P^1$s. How can I think of these as a class in $H^4(D)$? Apr 1 revised Induced map on cohomology added 9 characters in body Apr 1 asked Induced map on cohomology Mar 29 awarded commutative-algebra