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Consider the Chow variety of $2$ points in $\mathbb P^1$. When the points collide, the support of that cycle is $1$ point. I.e. the map you attempted to define from the Chow variety to "the" Hilbert s …

There's a slight stupidity here which is that if the dimension of the fiber jumps up, the degree can go down. E.g. let $X = (T \times C) \cup (p \times \mathbb P^2) \subseteq T \times \mathbb P^2$, wh …

Yes, it's obvious (and due to Gel$'$fand-MacPherson). Start with $k\times n$ matrices and act with $GL(k) \times T^n$, then reduce in stages in either order. Half my thesis was about the fact that th …

No. Try $\{0\} \hookrightarrow {\mathbb P}^1$.`

This is local, of course. So $R \leftarrow S : \phi$ is an injection of domains, and $I \leq S$ is a nonradical ideal; is $R \phi(I)$ a nonradical ideal of $R$? Say $s$ descends to a nonzero nilpoten …

Let's start with the case of a torus. Since you don't require $X$ to be smooth, we can reduce to this case by replacing $X$ by $X//N$. (Though actually inferring results about the nonabelian case from …

I will answer in the contrapositive. Let $X \subseteq \prod_i {\mathbb P}^{n_i}$ be irreducible of codimension $k$. If whenever $\sum k_i = k$, you can find subspaces $\prod_i {\mathbb P}^{n_i}$ that …

Yes. Otherwise there would be an isolated ramification point in $Y$. A link of that point is $S^3$, which does not have a nontrivial double cover. So upstairs, $X$ looks near that point like two ${\ma …

As the coauthor of the relevant paper (and as Mariano already said), I consider irreducibility an important enough concept that it's worth reserving the word "variety" for only reduced, irreducible sc …

I disagree with "this works only for curves". Say we've already defined degree for schemes up to dimension $n$. Then use your rule to define it for schemes of dimension $n+1$. Dr. "This works only for …

As Felipe says, these questions are very different, and I will only address #1. As Jack commented, for the specific example you gave, the answer is well-known. There's a general reason why the equatio …

Here's a place to see the normal cone side-by-side with other familiar constructions, that I learned from Fulton's "Intersection Theory". Here $X \subset Y$. Start with the space $Y \times {\mathbb P …

First let's revisit the usual case. The $\mathbb N$-grading on $R$ says that $Spec\ R$ is a cone, and there is a map $R \to R_0$. We can rip out $Spec\ R_0$ from $Spec\ R$, take the quotient, and get …

It's easier to compute its $T$-equivariant $r$th Chern class $c_r(E) \in H^*_T(G/P)$, and use the maps $H^*_T(G/P) \twoheadrightarrow H^\ast(G/P)$ and $H_T^*(G/P) \hookrightarrow \oplus_{W/W_P} H_T^*( …

I like the way you asked to avoid. Forgive me if I describe it in polytope rather than fan language. Step 1: ${\mathbb P}^1 \times {\mathbb P}^1$'s polytope is a square (or any rectangle). The four e …