Reorganization of Chapter 4.3

William Fulton Algebraic Curves, chapter 4, section 3

Throughout, let $φ=φ_{n+1}$, and let $τ : k[x_1,...,x_{n+1}] → k[x_1,...,x_n]$ be the surjective ring morphism given by evaluation $x_{n+1}↦1$. Given $V⊆\mathbb{A}^n$, the projective closure $V^*$ of $V$ is defined to be the smallest projective algebraic set containing $φ(V)⊆\mathbb{P}^n$ (that is, $V(I(φ(V)))$). Given $V⊆\mathbb{P}^n$, the dropdown $V_*$ of $V$ is defined to be $φ^{-1}(V)⊆\mathbb{A}^n$. Note that the dropdown of a projective algebraic set is an affine algebraic set, with defining ideal $τ(I(V))$.

Lemma: For finite unions, $(∪V_i)_* = ∪{V_i}_*$ and $(∪V_i)^* = ∪V_i^*$. Proof: The first equality is clear, as unions commute with function pullbacks. As for the second, note that $∪V_i^*$ is the smallest algebraic set containing each $V_i^*$ (as it is itself an algebraic set). Therefore, since any algebraic set containing $φ(V_i)$ contains $V_i^*$, it is the smallest algebraic set containing each $φ(V_i)$. That is, it is the smallest algebraic set containing $∪φ(V_i)=φ(∪V_i)$—and this is exactly the definition of $(∪V_i)^*$.

Proposition 3: Let $V, W$ denote algebraic sets throughout.

1. If $V⊆\mathbb{A}^n$, then $φ(V) = V^* ∩ U_{n+1}$ and $(V^*)_* = V$.
2. Projective closures and dropdowns are inclusion-preserving operations.
3. If $V⊆\mathbb{A}^n$ is a variety, then $V^*⊆\mathbb{P}^n$ is a variety.
4. If $V=∪V_i$ is the irreducible decomposition of $V$ in $\mathbb{A}^n$, then $V^* = ∪V_i^*$ is the irreducible decomposition of $V^*$ in $\mathbb{P}^n$.
5. If $V⊂\mathbb{A}^n$ is nonempty, then $V^*$ does not lie in or contain $H_∞$.
6. If $V⊆\mathbb{P}^n$ and no component of $V$ lies in or contains $H_∞$, then $V_*⊂\mathbb{A}^n$ and $(V_*)^* = V$.

Proof: (1) Suppose $[a,1]∈V^* \setminus φ(V)$. Then $a∉V$, so $f(a) = f^*([a,1]) ≠ 0$ for some $f∈I(V)$. Thus, since $f^*$ vanishes on $φ(V)$, we would have $V⊆V(f^*)∩V^*⊂V^*$, a contradiction. Now we compute $$(V^*)_* = φ^{-1}(V^*) = φ^{-1}(V^*∩U_{n+1}) = φ^{-1}(φ(V)) = V$$ (2) This is obvious.

(3) Suppose $V^* = W_1∪W_2$. Then since $$(W_1∪W_2)_* = {W_1}_*∪{W_2}_* = (V^*)_* = V$$ is a variety, we may suppose ${W_1}_* = V$ and $W_1$ contains $φ(V)$, hence contains $V^*$.

(4) By the Lemma and (3), it follows $∪V_i^*$ is an irreducible decomposition of $V^*$. To see that it is not redundant, suppose $V_i^* ⊆ V_j^*$: then by (1) and (2), we would see $V_i ⊆ V_j$.

(5) Note $V^* ⊈ H_∞$ by (1). Now, choose some nonzero $f∈I(V)$. Then we note $f^*∈I(V^*)$, the leading form $f'$ of $f$ is not zero at some point $a∈\mathbb{A}^n$, and $f^*([a,0]) = f'(a) ≠ 0$, so that $[a,0]∉V^*$ and $V^*$ does not contain $H_∞$.

(6) We may assume $V$ is nonempty. Then $V$ itself does not lie in or contain $H_∞$, for this latter is irreducible (with defining ideal $(x_{n+1})$). If we show $U_{n+1}^* = \mathbb{P}^n$, then it will follow $V_*$ must be properly contained in $\mathbb{A}^n$. As such, suppose $f$ is a form vanishing on $U_{n+1}$. Then $τ(f)$ is necessarily $0$, so write $f = (x_{n+1}-1)·h$. But the expression on the right is not a form unless $h=f=0$. Thus $I(U_{n+1}^*) = (0)$ and $U_{n+1}^* = \mathbb{P}^n$.

Note that $V_i = ({V_i}_*)^* ∪ (H_∞∩V_i)$ for each component $V_i$ of $V$, so since $V_i$ does not lie in $H_∞$, we must have $V_i = ({V_i}_*)^*$. Now we compute $$(V_*)^* = ∪({V_i}_*)^* = ∪V_i = V~~\square$$