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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