User:J D Bowen/Math899 HW2

1.3.1) Let $$F:V\to W \ $$ be a morphism of affine algebraic varieties.  Prove $$F \ $$ is continuous in the Zariski topology.

To show: Pre-images of closed sets are closed. Same as: pre-images of AAVs are AAVs.

$$W \ $$ is an AAV. The preimage, $$V \ $$, is also an AAV. Now let $$X\subset W \ $$ be an AAV, and let $$U=F^{-1}(X) \ $$. Since $$X \ $$ is a subvariety of $$W \ $$, there are ideals $$I, J \ $$ such that $$J=\mathbb{I}(W)\subsetneq \mathbb{I}(X) = I \ $$. The morphism $$F \ $$ induces a morphism $$f:\mathbb{I}(V)\to J \ $$; this extends to a unique homomorphism of a larger ideal $$H\to I \ $$; since this map is a homomorphism it can only carry ideals to ideals, so $$f^{-1}(I)=H \ $$ is an ideal. Since $$\mathbb{I}(V)\subset H \ $$ is an ideal, it follows that $$\mathbb{V}(H)\subset V \ $$ is an affine algebraic variety. Hence the preimage of every closed set under $$F \ $$ is a closed set, so $$F \ $$ is continuous.

1.3.2) Show the twisted cubic $$V=\left\{{(t,t^2,t^3)\in\mathbb{A}^3:t\in\mathbb{C} }\right\} \ $$ is isomorphic to the affine line $$\mathbb{A}^1 \ $$.

Let $$\phi(x\in V )=\phi(t,t^2,t^3)=t\in\mathbb{A}^1 \ $$. This is an isomorphism, because it is polynomial and invertible.

2.1.6) Let $$R \ $$ be a $$\mathbb{C}- \ $$algebra, and let $$I$$ be an ideal of $$R$$. Prove that the natural surjection $$R\to R/I \ $$ is a $$\mathbb{C}- \ $$algebra map.

Let $$R \ $$ be a $$\mathbb{C} \ $$-algebra, $$I$$ any ideal of $$R$$. Let $$\phi:R\to R/I \ $$ be the natural surjection, ie, taking $$r+I\mapsto r \ $$.

Then we have for vectors $$x,y,z\in R \ $$, we have

$$(x+y)=(\overline{x}+I+\overline{y}+I)=(\overline{x}+\overline{y}+I)\mapsto (\overline{x}+\overline{y})=\overline{x+y} \ $$,

$$(x+y)z=(\overline{x}+I+\overline{y}+I)(\overline{z}+I)=(\overline{x}\overline{z}+\overline{y}\overline{z}+I)\mapsto\overline{x}\overline{z}+\overline{y}\overline{z} \ $$.

Hence, since $$R, \ R/I \ $$ are both $$\mathbb{C}$$-algebras, and $$\phi \ $$ preserves operations, it is a $$\mathbb{C}$$-algebra map.

2.5.1) Show that the pullback $$F^\sharp:\mathbb{C}[W]\to\mathbb{C}[V] \ $$ is injective iff $$F \ $$ is dominant; that is, the image set $$F(V) \ $$ is dense in $$W \ $$.

2.5.2) Show that the pullback $$F^\sharp:\mathbb{C}[W]\to\mathbb{C}[V] \ $$ is surjective iff $$F \ $$ defines an isomorphism between $$V \ $$ and some algebraic subvariety of $$W \ $$.

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2.5.3) If $$F=(F_1, \dots, F_n):\mathbb{A}^n\to\mathbb{A}^n \ $$ is an isomorphism, then the Jacobian

$$\text{det} \begin{pmatrix}

\frac{\partial F_1}{\partial x_1} & \dots & \frac{\partial F_1}{\partial x_n} \\ & \dots & \\ \frac{\partial F_n}{\partial x_1} & \dots & \frac{\partial F_n}{\partial x_n} \end{pmatrix} \ $$

is a nonzero constant polynomial.

If we take $$\mathbb{A}=\mathbb{C} \ $$, then any polynomial $$F_1:\mathbb{C}^n\to\mathbb{C} \ $$ must be linear if $$F \ $$ is an isomorphism, since otherwise there will be $$x\neq y, \ F(x)=F(y) \ $$. Since all the $$F_i \ $$ are linear in all variables, their partial derivatives in any one variable are constant.

A determinant of constants is a constant. If $$J=0 \ $$, then we would have every tiny region around a point being taken to a region of zero volume in the image. This would not be an isomorphism.

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(4) Give an example of a algebraic set V who's coordinate ring C[V] is not a UFD (Unique Factorization Domain).


 * Bonus ****

Here's a fun exercise extending ideas from the course into other areas of mathematics. The exercise is taken from Atiyah-MacDonald, Introduction to Commutative Algebra (A sadly out of print standard).

The point of this exercise is to show that studying a space X should be equivalent to the studying the ring of functions on X with the appropriate notion of functions. Here we leave the algebraic category all together (meaning our functions are not polynomial etc.) but algebra still provides the link.

Let X be a compact, Hausdorf topological space and C(X) the ring of continuous real valued functions on X. We show that we can recover X algebraically from C(X) in the sense that we identify a subset Y \in C(X) and a surjective homeomorphism \mu:X \rightarrow Y using only algebraic means. (recall that in the category of topological spaces an homeomorphism X \rightarrow Y is an isomorphism).

Take R = C(X) and define a map \mu:X \rightarrow maxSpec(R) by x \mapsto m_x, the maximal ideal of all continuous functions vanishing at x. We know that m_x is maximal because it's the kernel of a surjective map from R to \mathbb{R} given by f \mapsto f(x), evaluation at x. This m subset R is maximal iff R/m is a field).

Now we take Y = maxSpec(R). We want to show that \mu: X \rightarrow Y is a homeomorphism.

(a) Show that \mu is surjective. Let m be any maximal ideal of C(X) and we want to show that m = m_x some x. Write V = V(m) for the set of common zeros of elements of m. If V is empty then around every x \in X there is function f_x \in m with f_x(x) \neq 0. By continuity there is an open set U_x on which f_x does not vanish. By compactness X is covered by finitely many of these open sets X = U(x_1) \union ...\union U(x_n).

Use these ideas to construct an f \in m that is non-vanishing everywhere on X. Derive a contradiction. It will then follow that m \subset m_x, some x, and by maximality they are equal.

(b) Show that \mu is injective. Use Urysohn's Lemma to show that if x \neq y then there is a continuous function vanishing at x and not at y.

(c) \mu is a homeomorphism. A subasis for the topology on X is given by the distinguished sets

U_f = {x \in X | f(x) \neq 0}, f \in C(X).

A subasis for the topology on maxSpec(R) is

V_f = {m \in maxSpec(R) | f \notin m}, f \in C(X)

Show that \mu(U_f) = V_f (and finish by using that a bijective map is a homeomorphism iff it is an open map).