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 \ $.
workspace
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).