Finite affine incidence projection and selection of curve components

Exact radical contraction. Realize $T\times {\mathbb{C}}^{m_b}$ as the closed affine set with coordinates $z_i,v_i,u_j$ and equations $z_i{v}_i=1$. Its coordinate ring is $R[u_1,\dots ,u_{m_b}]$: replace each $z_i^{-1}$ by $v_i$, and reduce any mixed positive powers using $z_i{v}_i=1$. Thus the relative assertion of [F1] gives $I(X_b)=\sqrt{J_b}$ in exactly this coordinate ring. A Laurent polynomial $f\in R$ vanishes on the projection of $X_b$ exactly when its pullback, the same expression independent of $u$, vanishes on $X_b$. Its ideal is consequently $\sqrt{J_b}\cap R$. Vanishing on a finite union means vanishing on every member, so $I(S)$ is the stated intersection. Moreover $I(S)=I(\bar{S})$: each polynomial zero set containing $S$ contains its closure, and $S\subseteq \bar{S}$ gives the reverse inclusion. This proves the first formula without asserting that a projection is closed. For no charts, $S$ is empty and every polynomial vanishes on it.

Select actual curve components. The set $Z$ is a closed subset of the displayed affine model for $T$, so [F2] gives its finite component list $Z_i$. Select precisely the members of dimension one, with dimension as in [F3]. This is a finite union of closed sets, hence closed. A polynomial vanishes on that union exactly when it vanishes on each selected member, giving the second formula directly. Each component ideal is radical, since $f^k$ vanishing at all its points forces $f$ to vanish; their intersection is radical as well. If no component is selected its vanishing ideal is the whole ring. The need for selection is literal: in $({\mathbb{C}}^{\ast }{)}^2$, the closed set $V(L-1)\cup \{(2,2)\}$ has ideal $(L-1)\cap (M-2,L-2)$, whereas its selected curve ideal is $(L-1)$. These differ since $L-1$ has value $1$ at $(2,2)$. The line ideal formula follows by substitution $L=1$ in a Laurent polynomial (multiply by a monomial and divide by $L-1$); the point ideal follows by subtracting its value successively in $M$ and $L$. Thus even radical elimination does not itself discard isolated points.

Rational descent commutes with selection. A field automorphism $\sigma$ fixing $\mathbb{Q}$ induces a bijection of $T$ taking $V(f_1,\dots ,f_r)$ to $V(\sigma f_1,\dots ,\sigma f_r)$; the same assertion for arbitrary collections of equations makes it a Zariski homeomorphism. If $S$ is stable, its closure $Z$ is stable as well. This homeomorphism permutes the maximal irreducible closed subsets by maximality and preserves their dimensions by bijecting all strict chains of irreducible closed subsets, the convention in [F3]. Therefore it preserves $C$. Apply [F4] separately to the stable point sets $Z$ and $C$ to obtain both asserted radical rational contractions. Finally, rational generators for each $J_b$ make $X_b$ stable by applying $\sigma$ to the defining equations and their inverse implications; projection commutes with $\sigma$, so their union $S$ is stable. All arguments include $n=0$, empty charts, empty zero sets and the whole torus.