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The Grassmannian in its Plücker embedding spans the space. The space $\mathbb{P}^{\binom{n}{k}-1}$ of alternating tensors is spanned by the simple wedges (also called decomposable) $v_1 \wedge \dotsb …

I don't know if this is really research-level math, but here is a quick answer. For each $i=1,\dotsc,6$, let $P_i$ be the center of circle (i) and write $P_i=(x_i,y_i)$. We have $P_1=(-1,0)$, $P_3=(1, …

In characteristic zero, a point $a$ is in $\mathcal{Q}$ if and only if $p$ vanishes identically on each line through $a$ parallel to one of the coordinate axes, i.e., all lines given by fixing all but …

Yes, it is possible for the degree to increase. Say $V \subset \mathbb{C}^3$ is reducible: a union of a curve of degree $d$ plus one more point that doesn't lie on the curve. Then $V$ has degree $d$. …

No. The pre-closure of the join of two irreducible projective varieties is NOT necessarily quasi-projective. Let $X$ be a smooth plane conic and let $Y$ be a single point of $X$. The pre-closure of th …

Theorem 2.2.26 of Lazarsfeld's Positivity in Algebraic Geometry, 2004, is the following: Theorem: The big cone is the interior of the pseudoeffective cone and the pseudoeffective cone is the closure …

If $Z$ is a pure one-dimensional closed subscheme of degree $d$ in projective space, with $d$-dimensional tangent space at every closed point, then $Z$ is supported on a line. To see this, let $H$ be …

The curve upstairs (hyperplane section) is smooth I think so its projection (the curve downstairs) is evidently resolved by a single blowup along $Z$. So I believe that makes it nodal --- non-nodal si …

Since each $F_i(u_1,\dotsc,u_{n_1},v_1,\dotsc,v_{n_2})=0$, then $F_i(su_1,\dotsc,su_{n_1},tv_1,\dotsc,tv_{n_2}) = s^{d_1}t^{d_2} F_i(u_1,\dotsc,u_{n_1},v_1,\dotsc,v_{n_2})=s^{d_1}t^{d_2}0 = 0$. So ce …

See: Mordechai Katzman, Counting monomials This paper presents two enumeration techniques based on Hilbert functions. The paper illustrates these techniques by solving two chessboard problems.

One way to think of the Segre map $\mathbb{P}^{n_1} \times \dotsm \times \mathbb{P}^{n_s} \to \mathbb{P}^N$ is as the map corresponding to multiplication of linear forms. By this I mean, for each $i=1 …

The basepoint freeness part has been proved in dimension $5$ by Fei Ye and Zhixian Zhu, On Fujita's freeness conjecture in dimension $5$, Adv. Math., 2020 DOI:10.1016/j.aim.2020.107210, arXiv:1511.091 …

Yes, the flattening rank is a lower bound for border rank. First note that flattening rank is a lower bound for rank. If $T$ is a decomposable tensor (simple tensor, rank one tensor) then every flatt …

You asked: Is $W(P \otimes Q) \leq W(P) W(Q)$ for two homogeneous polynomials $P$ and $Q$? First, I think you have to be a little bit careful about the difference between tensor product and mult …

Some authors write $\operatorname{in}_<(f)$ to mean the largest term of $f$, and other authors write $\operatorname{in}_<(f)$ to mean the smallest term of $f$. (Some authors say "leading term".) In yo …