Let $F_1(\mathbf{x}, \mathbf{y}), \ldots, F_r(\mathbf{x}, \mathbf{y})$ be bihomogeneous polynomials with rational coefficients with bidegree $(d_1, d_2)$, which means $$ F_i( s x_1, \ldots, s x_{n_1} ; t y_1, \ldots, t y_{n_2} ) = s^{d_1} t^{d_2} F_i(\mathbf{x} ; \mathbf{y} ). $$ Let $V$ be the algebraic set defined by $F_1, \ldots, F_r$ in $\mathbb{C}^{n_1+ n_2}$, and let us denote $V_1, \ldots, V_T$ to be the irreducible components of $V$. Then I am guessing that the following is true, but I wasn't really sure how to prove or where to find a reference for: Let $1 \leq i \leq T$.

  1. Suppose the point $(u_1, \ldots, u_{n_1}, v_{1}, \ldots, v_{n_2}) \in \mathbb{C}^{n_1+n_2}$ is contained in $V_i$. Then for any $s,t \in \mathbb{C}$ the point $(su_1, \ldots, su_{n_1}, tv_{1}, \ldots, tv_{n_2})$ is also contained in $V_i$.

  2. $V_i$ contains points all points of the form $(0,\ldots, 0, t_1, \ldots, t_n)$ where $t_j \in \mathbb{C}$.

I would appreciate any comments or references. Thank you very much!

  1. 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 certainly the point $(su_1,\dotsc,su_{n_1},tv_1,\dotsc,tv_{n_2}) \in V$. The map $\mathbb{C}^2 \to \mathbb{C}^{n_1+n_2}$ given by $(s,t)\mapsto (su_1,\dotsc,su_{n_1},tv_1,\dotsc,tv_{n_2})$ is regular, so the image is irreducible. This irreducible set is contained in $V$, so it must be contained in an irreducible component of $V$. Since $(u_1,\dotsc,u_{n_1},v_1,\dotsc,v_{n_2}) \in V_i$, so is $(su_1,\dotsc,su_{n_1},tv_1,\dotsc,tv_{n_2})$ for every $s,t$.

  2. This is not true. It's true that every such point is in $V$ (assuming $d_1>0$) but those points are not necessarily in each irreducible component $V_i$ of $V$. For example, with $n_1=n_2=d_1=d_2=1$, $V=V(xy)$ has two components, one of which contains the point $(0,t)$ if and only if $t=0$.

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  • $\begingroup$ I don't know what I was thinking in part (1). The irreducible set is contained in at least one irreducible component of $V$. But I don't know why it can't intersect other irreducible components. So, just containing one point of $V_i$ doesn't imply containment in $V_i$. The answer above is bogus. Sorry. $\endgroup$ – Zach Teitler Aug 21 '19 at 15:41
  • $\begingroup$ However it can be replaced. An affine variety is defined by a multihomogeneous ideal if and only if it's invariant under scaling by a certain product of tori. Then its irreducible components are each, individually, also invariant; check on general points (automatic on intersections). This is a positive answer to question (1). ... If I still feel good about this idea in a few days, and if I remember, I'll update my answer above. If you're reading this comment in the year 2099, I forgot... $\endgroup$ – Zach Teitler Aug 21 '19 at 15:47

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