Brian Bi
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## Section 11.9. Algebraic Geometry

Exercise 11.9.1

1. The ideal $$(x-1, y-4)$$ in $$R$$ is the image of the ideal $$(x-1, y-4, y^2 + x^3 - 17)$$ under the quotient map. Since $$x = 1, y = 4$$ is a zero of $$y^2 + x^3 - 17$$, it follows that $$y^2 + x^3 - 17 \in (x-1, y-4)$$. So $$(x-1, y-4, y^2 + x^3 - 17) = (x-1, y-4)$$, which is a maximal ideal in $$\C[x, y]$$. So the ideal $$(x-1, y-4)$$ is maximal in $$R$$ by the Correspondence Theorem.
2. The ideal $$(x+1, y+4)$$ in $$R$$ is the image of the ideal $$(x+1, y+4, y^2 + x^3 - 17)$$ under the quotient map. These three polynomials have no common zeroes, so this ideal is the unit ideal. Therefore $$(x+1, y+4) = R$$, and is not maximal in $$R$$.
3. The ideal $$J = (x^3 - 17, y^2)$$ in $$R$$ is the image of the ideal $$J' = (x^3 - 17, y^2, y^2 + x^3 - 17)$$ in $$\C[x,y]$$ under the quotient map, but $$J'$$ is just $$(x^3 - 17, y^2)$$. Now the ideal $$(x - a, y - b)$$ in $$\C[x, y]$$ has the property that its members have $$(a, b)$$ as their sole common zero, but all polynomials in $$J'$$ have three common zeroes, namely $$\{(\omega^k \sqrt[3]{17}, 0) \mid k \in \{0, 1, 2\}\}$$, where $$\omega$$ is a primitive cube root of unity. So $$J'$$ is not equal to $$(x-a, y-b)$$ for any $$a, b \in C$$, therefore $$J'$$ isn't maximal in $$\C[x, y]$$. By the Correspondence Theorem, we conclude that $$(x^3 - 17, y^2)$$ is not a maximal ideal in $$R$$.

Exercise 11.9.2 Each polynomial in $$R = \C[x_1, \ldots, x_n]$$ can be identified with a continuous function from $$\C^n$$ to $$\C$$ in the obvious way, which restricts to a function from $$V$$ to $$\C$$. Let $$\varphi$$ denote this map from polynomials to functions $$V \to \C$$. Clearly $$\varphi$$ is a homomorphism of rings. If $$f \in I$$, then the evaluation of $$f$$ yields zero for all points of $$V$$, therefore $$\varphi(f)$$ is the zero function. Therefore, $$I \subseteq \ker \varphi$$. By the fundamental homomorphism theorem, there is a unique homomorphism $$\overline{\varphi}$$ from the ring $$R/I$$ to the continuous functions from $$V \to \C$$, such that $$\overline{\varphi} \circ \pi = \varphi$$, with $$\pi$$ being the canonical projection from $$R$$ to $$R/I$$. The map $$\overline{\varphi}$$ operates in the obvious way: $$\overline{\varphi}(\overline{f}) = \varphi(f)$$ where $$f$$ is some element of $$R$$ such that $$\pi(f) = \overline{f}$$.

Exercise 11.9.3 This directly follows from the definition of the Cartesian product.

Exercise 11.9.4 Let $$S = \{f_1, \ldots, f_m\}$$ be a set of polynomials whose common zeros form the variety $$U$$. Let $$T = \{g_1, \ldots, g_n\}$$ be a set of polynomials whose common zeros form the variety $$V$$.

The common zeros of the union $$S \cup T$$ are the set $$U \cap V$$; so $$U \cap V$$ is a variety.

Consider the set $$ST = \{fg \mid (f, g) \in S \times T\}$$. Let $$x \in \C^n$$. If $$x \in U$$, so that $$f(x) = 0$$ for all $$f \in S$$, then $$fg = 0$$ for all $$(f, g) \in S \times T$$, so $$x$$ belongs to the variety of common zeroes of $$ST$$. The same holds if $$x \in V$$, so that $$g(x) = 0$$ for all $$g \in T$$. On the other hand, if $$x \notin U$$ and $$x \notin V$$, then there is some $$f \in S$$ and some $$g \in T$$ such that $$f(x) \ne 0$$ and $$g(x) \ne 0$$, consequently $$fg$$ is an element of $$ST$$ for which $$x$$ is not a zero, and $$x$$ does not belong to the variety of common zeros of $$ST$$. Therefore, the variety of common zeros of $$ST$$ is $$U \cup V$$.

Algebraically, when $$U \cap V = \emptyset$$ it means that there's no solution to the set of equations $$f_1(x) = \ldots = f_m(x) = g_1(x) = \ldots = g_n(x) = 0$$. When $$U \cup V = \C^n$$, it means every $$x \in \C^n$$ is a common zero of either $$S$$ or $$T$$.

Exercise 11.9.5 Let $$I$$ be the ideal generated by the polynomials $$S = \{f_1, \ldots, f_r\}$$ and let $$V$$ be the variety of their common zeros.

Suppose $$x \in \C^n$$ has the property that $$f(x) = 0$$ for all $$f \in I$$. Then, certainly $$f(x) = 0$$ for each $$f \in S$$. Therefore $$x \in V$$.

Conversely, suppose $$x \in V$$. Then $$f(x) = 0$$ for all $$f \in S$$, consequently $$f(x) = 0$$ for all $$f \in I$$.

We have established that $$V$$ consists precisely of those points $$x \in \C^n$$ such that $$f(x) = 0$$ for all $$f \in I$$. So $$V$$ depends on $$I$$ only.

Exercise 11.9.6 We will state and prove a lemma:

Lemma: The intersection of two algebraic curves in $$\C^2$$ is the union of an algebraic curve with finitely many points.

Proof: If either curve is the entire set $$\C^2$$, then this is obvious. Suppose that this is not the case. Each given algebraic curve is therefore the locus of zeros of some nonzero polynomial in $$\C[x, t]$$. Denote these two polynomials by $$f, g$$. If $$f, g$$ have a nonconstant common factor $$h$$, then write $$f = f'h, g = g'h$$. If $$f'$$ and $$g'$$ still have a nonconstant common factor, then divide it out likewise, and continue this process until there is no nonconstant common factor left. (Each division must lower either the $$x$$-degree or the $$t$$-degree of the two polynomials, so this process cannot continue indefinitely). In this way, write $$f = f_0 h_0, g = g_0 h_0$$, where $$f_0, g_0, h_0 \in \C[x, t]$$ and the two polynomials $$f_0, g_0$$ have no nonconstant common factor. Let $$x, t \in \C^2$$. Then we will have $$f(x, t) = g(x, t) = 0$$ if and only if at least one of these two conditions holds:

1. $$h_0(x, t) = 0$$
2. $$f_0(x, t) = g_0(x, t) = 0$$

Therefore the intersection of the two algebraic curves given, which is the locus of common zeros of $$f$$ and $$g$$, is the union of the algebraic curve on which $$h_0$$ vanishes, and the set of common zeros of $$f_0$$ and $$g_0$$, which, according to Theorem 11.9.10, is finite.

Suppose a variety in $$\C^2$$ is the locus of common zeros of the polynomials $$f_1, \ldots, f_r$$. Therefore, it is the intersection of $$r$$ algebraic curves. We can now prove that this set is the union of an algebraic curve and at most finitely many points by induction on $$r$$:

• Base case: $$r = 1$$. The intersection is just the single algebraic curve itself.
• Inductive case: suppose there are $$r$$ algebraic curves, $$C_1 \ldots C_r$$. The intersection $$C_1 \cap \ldots \cap C_{r-1}$$ is, by the induction hypothesis, the union $$C' \cup P'$$ where $$C'$$ is an algebraic curve and $$P'$$ is a finite set of points. Let $$S = C_1 \cap \ldots \cap C_r$$. Then $$S = (C' \cup P') \cap C_r = (C' \cap C_r) \cup (P' \cap C_r)$$. Using the Lemma, write $$C' \cap C_r = C \cup P$$ where $$C$$ is an algebraic curve and $$P$$ is finite. Then $$S = C \cup (P \cup (P' \cap C_r))$$, where $$C$$ is an algebraic curve and $$P \cup (P' \cap C_r)$$ is finite, so we are done.

Exercise 11.9.7

1. Substitute $$y = 1 - x$$ into the first equation to obtain $$1 = (1 - x)^2 - x^3 + x^2 = 1 - 2x + 2x^2 - x^3$$, or $$x^3 - 2x^2 + 2x = 0$$, which is easily factored to give the three solutions $$x = 0, x = 1 + i, x = 1 - i$$. This gives three solutions to the original system for $$(x, y)$$, namely $$(0, 1), (1 + i, -i), (1 - i, i)$$.
2. Subtract the two equations to obtain $$y^2 - xy = 0$$. This tells us that there are two possible classes of solutions: those with $$y = 0$$ and those with $$x = y$$. When $$y = 0$$, both equations reduce to $$x^2 = 1$$, so this gives us two solutions. When $$x = y$$, the two equations reduce to $$3x^2 = 1$$, so this gives us another two solutions. There are a total of four solutions for $$(x, y)$$, namely $$(1, 0), (-1, 0), (\sqrt{1/3}, \sqrt{1/3}), (-\sqrt{1/3}, -\sqrt{1/3})$$.
3. Substitute $$y = x^{-1}$$ into the first equation to obtain $$x^{-2} = x^3$$, or $$1 = x^5$$. This yields five possible solutions for $$x$$, all of which are valid for the original system because we multiplied by $$x^2$$, which is not zero. So the five solutions to the original system for $$(x, y)$$ are $$\{(\omega^k, \omega^{-k}) \mid k \in \{0, 1, 2, 3, 4\}\}$$ where $$\omega$$ is a primitive fifth root of unity.
4. Substitute $$x = -y^2$$ into the second equation to obtain $$0 = y + (-y^2)^2 + 2(-y^2)y^2 + y^4 = y$$. Conclude that there is a single solution, namely $$x = 0, y = 0$$.

Exercise 11.9.8 A solution can be found here.

Exercise 11.9.9 The constant polynomials are not considered irreducible, so $$f$$ is not constant. Therefore $$\partial f/\partial x$$ or $$\partial f/\partial y$$ is nonzero (possibly both). Applying Theorem 11.9.10 to $$f$$ and one of its nonzero partial derivatives yields the desired result.

Exercise 11.9.10 We will prove a slightly stronger statement, which will be useful for the next problem. Let $$\ell \in \C[x, y]$$ be linear and let $$L$$ be the locus of its zeros. If $$\ell$$ doesn't divide $$f$$, then we'll show that $$f$$ has at most $$d$$ zeroes on $$L$$.

Write $$\ell = ax + by + c$$, where $$a$$ and $$b$$ aren't both zero. Without loss of generality, suppose $$b$$ is nonzero. The complex line $$L$$ is then equivalent to the locus of points of the equation $$y = px + q$$ where $$p = -a/b$$ and $$q = -c/b$$. Let $$g \in \C[x]$$ be the polynomial such that $$g(x) = f(x, px + q)$$. Since $$f$$ has degree $$d$$, it follows that $$g$$ has degree at most $$d$$, or is zero. If $$g = 0$$, then this is saying that the substitution $$y = px + q$$ in the ring $$\C[x][y]$$ yields the zero element of $$\C[x]$$, which implies that the monic polynomial $$y - (px + q)$$ divides $$f$$ in $$\C[x, y]$$. We assumed this not to be the case, therefore $$g$$ is nonzero and has degree at most $$d$$, so there are at most $$d$$ values of $$x$$ with $$g(x) = 0$$, giving at most $$d$$ solutions $$(x, px + q)$$ to the set of equations $$\{ax + by + c = 0, f(x, y) = 0\}$$, that is, at most $$d$$ zeroes of $$f$$ that lie on $$L$$.

When we are told that $$f$$ is irreducible, it is clear that either $$f$$ is a nonzero constant multiple of $$\ell$$, giving $$C = L$$, or else $$f$$ is not divisible by $$\ell$$, in which case, as proven above, $$C \cap L$$ can contain at most $$d$$ points.

Exercise 11.9.11

1. Let $$r$$ be a third point on $$L$$. We can choose constants $$c_1, c_2$$, not both zero, such that $$c_1 f_1(r) + c_2 f_2(r) = 0$$, as follows. If $$f_1(r) = 0$$, then choose $$c_1 = 1, c_2 = 0$$. If $$f_2(r) = 0$$, then choose $$c_1 = 0, c_2 = 1$$. If $$f_1(r)$$ and $$f_2(r)$$ are both nonzero, then choose $$c_1 = f_2(r), c_2 = -f_1(r)$$. No matter how we choose $$c_1, c_2$$, it is clear that $$c_1 f_1 + c_2 f_2$$ already vanishes on $$p$$ and $$q$$. So the polynomial $$g = c_1 f_1 + c_2 f_2$$ vanishes at three distinct points on $$L$$. Without loss of generality, suppose the equation of $$L$$ is $$y = ax + b$$. By Exercise 11.9.10, since $$g$$ has total degree at most 2, and yet has three distinct zeroes on $$L$$, it follows that $$y - ax - b$$ divides $$g$$; so, in fact, $$g$$ vanishes identically on $$L$$.

If $$g$$ is quadratic in $$y$$, then performing division of $$g$$ by $$y - ax - b$$ in the variable $$y$$ yields a polynomial $$h$$ which is linear in $$y$$ with constant coefficient. Write $$h = h_1 y + h_0(x)$$ where $$h_1$$ is a constant and $$h_0$$ is a polynomial. So $$g = (h_1 y + h_0(x)) (y - ax - b)$$. If $$h_0$$ has degree greater than 1, the highest-degree term in $$h_0(x)y$$ can't be cancelled by any other terms in the product, resulting in a contradiction. Therefore $$h$$ has total degree 1. If on the other hand $$g$$ is only linear in $$y$$, $$g = g_1(x)y + g_0(x)$$ where $$g_1$$ has degree at most 1 and $$g_0$$ has degree at most 2, then division by $$y - ax - b$$ must yield $$g_1$$. In either case, $$g$$ is a product of linear factors.

2. It is obvious that any common zeros of $$f_1$$ and $$f_2$$ are also zeros of $$g$$. If $$g$$ is linear, then, by the conditions of the problem, it doesn't divide both of $$f_1$$ and $$f_2$$. Without loss of generality, suppose $$g$$ doesn't divide $$f_1$$; in that case, by Exercise 11.9.10, $$g$$ and $$f_1$$ have at most two common zeroes, therefore $$f_1$$ and $$f_2$$ have at most two common zeroes.

If on the other hand $$g$$ is quadratic, then write $$g = g_1 g_2$$, the product of linear factors, as we argued we could do in part (a). The set of points $$(x, y)$$ on which $$f_1(x, y) = f_2(x, y) = g_1(x, y) g_2(x, y) = 0$$ is the union of the sets of points $$S_1$$, on which $$f_1(x, y) = f_2(x, y) = g_1(x, y) = 0$$, and $$S_2$$, on which $$f_1(x, y) = f_2(x, y) = g_2(x, y) = 0$$. By the argument given in the previous paragraph, each of $$S_1$$ and $$S_2$$ has size at most 2, so there are a total of at most 4 common zeros of $$f_1, f_2, g$$, and hence a total of at most 4 common zeros of $$f_1$$ and $$f_2$$.

Remark: In the first edition of the text, the exercise simply asked to show that two quadratic polynomials in two variables with no nonconstant factor in common have at most four common zeros. The exercise is much easier in the second edition, since part (a) is a hint to part (b).

Exercise 11.9.12 Suppose $$f_2(x, t) = 0$$. Then we can write $$t = 1/x$$ and substitute this into the equation $$f_1(x, t) = 0$$ to obtain $$x^{-2} + x^2 - 2 = 0$$, which implies $$x^4 - 2x^2 + 1 = 0$$. This factorizes as $$(x+1)^2(x-1)^2 = 0$$, so the only possible common zeros of $$f_1$$ and $$f_2$$ are $$x = 1, t = 1$$ and $$x = -1, t = -1$$. By inspection, neither is a zero of $$f_3$$, so the three polynomials together have no common zeros.

The problem also asks us to explicitly write 1 as a linear combination of $$f_1, f_2, f_3$$ with polynomial coefficients. Solutions can be found here and here, although it's not clear to me how they are derived.

Exercise 11.9.13 We are given that $$x(t)$$ and $$y(t)$$ are not both constant. Without loss of generality, suppose $$x(t)$$ is not constant. If $$\ker\varphi$$ is the zero ideal, it is principal. We proceed assuming that this is not the case.

Let $$d_0$$ be the minimum $$y$$-degree of nonzero elements of $$\ker\varphi$$, and let $$S \subseteq \ker\varphi$$ consist of the elements with $$y$$-degree equal to $$d_0$$. The coefficient of the $$y^{d_0}$$ term of each polynomial in $$S$$ is a polynomial in $$x$$. Choose $$P_0 \in S$$ such that the degree of the coefficient of the $$y^{d_0}$$ term is minimal.

Suppose there is some $$\alpha \in \C$$ such that $$P_0 = (x-\alpha)Q$$ with $$Q \in \C[x, y]$$. Then $$0 = \varphi(P_0) = \varphi(x-\alpha)\varphi(Q) = (x(t) - \alpha)\varphi(Q)$$. Since we assumed that $$x(t)$$ is not constant, and since $$\C[t]$$ is an integral domain, it follows that $$\varphi(Q) = 0$$. But $$Q$$ has the same $$y$$-degree as $$P_0$$ and the $$x$$-degree of its $$y^{d_0}$$ term is smaller by 1, which gives a contradiction. So there is in fact no such $$\alpha$$. (This will allow us to apply Proposition 11.9.9 shortly.)

Suppose $$f \in \C(x)[y]$$ so that $$f(x(t), y(t)) = 0$$. We can always multiply $$f$$ by some polynomial in $$x$$ to clear denominators, giving an element of $$\C[x][y]$$ with the same $$y$$-degree as $$f$$. It follows that if $$f$$ is nonzero, then the $$y$$-degree of $$f$$ is at least $$d_0$$.

Let $$P \in \ker\varphi$$ be given. We may divide $$P$$ by $$P_0$$ in the variable $$y$$ in $$\C(x)[y]$$, giving $$P = P_0 Q + R$$ where $$Q, R \in \C(x)[y]$$, and either $$R$$ is zero or the $$y$$-degree of $$R$$ is less than $$d_0$$. Now $$R(x(t), y(t)) = P(x(t), y(t)) - P_0(x(t), y(t)) Q(x(t), y(t)) = 0$$, so if $$R$$ is nonzero, then by the argument in the previous paragraph, the $$y$$-degree of $$R$$ is at least $$d_0$$. This would be a contradiction. Therefore $$R = 0$$, and $$P_0$$ divides $$P$$ in $$\C(x)[y]$$. By Proposition 11.9.9, $$P_0$$ divides $$P$$ in $$\C[x, y]$$. As this holds for all choices of $$P$$, we conclude that $$\ker\varphi = (P_0)$$.