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 $$(V_*)^* = ∪({V_i}_*)^* = ∪V_i = V~~\square$$

Plane Curve Multiplicities Under Polynomial Mappings (3.1.8)

William Fulton Algebraic Curves, chapter 3, section 1, exercise 8:

Let $k$ be an algebraically closed field, $F∈k[x,y]$, and $T : \mathbb{A}^2 → \mathbb{A}^2$ be a polynomial map with $T(P)=Q$. Show that $m_P F^T ≥ m_Q F$, with equality achieved if the Jacobian of $T$, defined as the matrix $(dT_i / dx_j (Q))$ where $T = (T_1,T_2)$, is invertible. Show that the converse is false in considering $T = (x^2,y)$ and $F=y-x^2$, $P=Q=(0,0)$.

Proof: Suppose $P = Q = (0,0)$. Then necessarily $T_1$ and $T_2$ are of the form $$T_1 = ... + ax+by$$ $$T_2 = ... + cx+dy$$ with no constant terms. Write $F = F_m + ... + F_n$ in terms of its form decomposition; then we see $F^T = F_m^T + ... + F_n^T$, and for each $i$, $F_i^T$ contributes to the sum forms of degree $≥i$. This shows the nonzero forms in $F^T$ are all of degree $≥m$, so that $m_{(0,0)} F^T ≥ m_{(0,0)} F$. Now consider the general case for points $P,Q$. Let $S_1, S_2$ be the translations mapping the origin to $P,Q$ respectively. Then $$m_Q F^T = m_{(0,0)} F^{T∘S_2} = m_{(0,0)} F^{S_1∘S_1^{-1}∘T∘S_2}$$$$= m_{(0,0)} (F^{S_1})^{S_1^{-1}∘T∘S_2} ≥ m_{(0,0)} F^{S_1} = m_P F$$ Now to consider the properties of the Jacobian: Again suppose $P = Q = (0,0)$, and factor $F_m$ into a product of linear factors. Since $F_i^T$ for $i > m$ only contribute forms of degree $> m$, it will suffice to show $F_m^T$ contributes a nonzero form of degree $m$ when the Jacobian is invertible. When the Jacobian is invertible, that is to say its rows are linearly independent, so that if $ex+fy$ is a nonzero linear term, then $T(ex+fy) = eT_1(x) + fT_2(y)$ will have a nonzero form of degree $1$ in its decomposition. Since the smallest nonzero form of a product is the product of the smallest nonzero forms, we see that $m_{(0,0)} F^T = m_{(0,0)} F$. The argument for the general case of $P, Q$ is exactly as before, together with the observations that (1) the Jacobian of a composition of polynomial maps is the same as the matrix product of the Jacobian of polynomial maps at the appropriate points, and (2) that the Jacobians of translations are the identity. As for the demonstration that the converse is not in general true, we note that in the example given, $m_{(0,0)} F = 1 = m_{(0,0)} y - x^4 = m_{(0,0)} F^T$ despite the fact that the Jacobian of $T$ at $(0,0)$ is the noninvertible $$J_{(0,0)}T = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}~~\square$$

Finite Dimensional Rings Containing Algebraically Closed Fields (2.9.47)

William Fulton Algebraic Curves, chapter 2, section 9, exercise 47:

Let $R$ be a commutative ring sharing an identity with $k⊆R$, an algebraically closed field. Show that $R$ is isomorphic to a finite direct product of local rings.

Proof: Let $r_1,...,r_n$ be a basis for $R$ over $k$. Then $k[x_1,...,x_n] / I ≅ R$ for some ideal $I$ via the natural identification $x_i ↦ r_i$. We see that distinct (surjective) morphisms $R → k$ preserving $k$ thus correspond to maximal ideals in $k[x_1,...,x_n]$ containing $I$, which themselves correspond to points in the locus of $I$. It follows that if we can prove there are only finitely many such morphisms, then Proposition 6 of this same section will imply the result. To this end, we show that the image of $r_i$ in $k$ under a morphism $φ$ may be among only finitely many elements; indeed, this follows by considering that $1,r_i,r_i^2,...$ are linearly dependent over $k$, and thus $φ(r_i)$ must be a root of the polynomial of $r_i$ over $k$ induced by this dependence.$~\square$

Multiple Representations are Required to Identify the Pole Set of a Rational Function (2.4.20)

William Fulton Algebraic Curves, chapter 2, section 4, exercise 20:

Let $k$ be algebraically closed, and consider the rational function $f = x/y = z/w$ over the algebraic variety $V = \mathcal{Z}(xw-yz)$ in $\mathbb{A}^4(k)$. Show that the pole set of $f$ is precisely $\{(t_1,0,t_2,0)~|~t_1,t_2∈k\}$, and that it is impossible to write $f = a/b$ for $b(p)≠0$ for all $p∈V$ outside the pole set of $f$.

Proof: First of all, $y$ and $w$ are nonzero on points of $V$, so they are nonzero in $\Gamma(V)$. As well, $\mathcal{I}(\mathcal{Z}(xw-yz)) = (xw-yz)$ since $xw-yz$ is irreducible. Let $J_f = \{G ∈ \Gamma(V)~|~Gf ∈ \Gamma(V)\}$. It is clear that $J_f$ is an ideal in $\Gamma(V)$, and that $y,w∈ J_f$. Suppose $gf = u ∈ \Gamma(V)$ for some $g∈k[x,z]$; then we may write $$gx - uy = 0 \mod (xw-yz)$$ $$gx = 0 \mod (xw,y)$$ from which we conclude $g = 0$, so that $J_f = (y,w)$. This allows us to conclude that the denominator of $f$ in any way that we write it will always have $\{(t_1,0,t_2,0)\}$ as zeros, so that this is precisely the pole set of $f$. Suppose we could write $f = a / b$ with $b$ having zeros precisely at the pole set of $f$; this would be to say $$\mathcal{Z}(b,xw-yz) = \mathcal{Z}(y,w,xw-yz) = \mathcal(Z)(y,w)$$ $$\mathcal{Z}(b,xw-yz) ∩ \mathcal{Z}(x,z) = \mathcal{Z}(x,y,w,z)$$ $$\mathcal{Z}(b,x,z) = \{(0,0,0,0)\}$$ Since $b∈(y,w)⊆\Gamma(V)$, it will thus suffice to show that given $b ∈ (y,w) ⊆ k[y,w]$, we may find a zero besides $(0,0)$. Indeed, this is equivalent to showing $$\mathcal{Z}(b) = \mathcal{Z}(y,w)$$ is impossible. Indeed, every nonconstant polynomial in multiple variables over an algebraically closed field has infinitely many zeros (1.2.14), while $\mathcal{Z}(y,w)$ is just a single point.$~\square$