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Algebraic geometry is a branch of mathematics that studies the properties of algebraic varieties, which are geometric objects defined by polynomial equations. The subject combines techniques from algebra and geometry to study the solutions of polynomial equations and their geometric properties.
In algebraic geometry, the main objects of study are algebraic varieties, which are geometric objects defined by polynomial equations. For example, the equation x^2 + y^2 = 1 defines a circle in the plane, and the equation x^3 + y^3 = 1 defines a curve known as the twisted cubic. Algebraic varieties can be defined over any field, but in classical algebraic geometry the focus is on varieties defined over the complex numbers.
One of the key tools in algebraic geometry is the notion of a scheme, which generalizes the notion of an algebraic variety to include more general types of objects. Schemes are studied using sheaf theory, which provides a way to associate to each point of a scheme a local ring, which encodes information about the geometry of the scheme near that point.
Algebraic geometry has a wide range of applications, including in number theory, cryptography, physics, and computer science. It has also been influential in other areas of mathematics, such as topology and representation theory. Some of the most famous results in algebraic geometry include the proof of Fermat’s Last Theorem by Andrew Wiles, and the proof of the Hodge Conjecture by Pierre Deligne.
(a) Prove that if $X=\operatorname{Spec}(A)$ is affine and locally factorial, then $P i c(X)$ is trivial iff $A$ is a UFD.
First, suppose that $Pic(X)$ is trivial. Since $X$ is affine and locally factorial, the divisor class group $Cl(X)$ is also trivial. In other words, every Weil divisor on $X$ is linearly equivalent to a principal divisor. Let $f$ be a nonzero element of $A$, and consider the principal divisor $(f)$. Since $Pic(X)$ is trivial, $(f)$ is the divisor of a regular function $g$ on $X$. This means that $g$ has a pole of order $1$ at every point of $V(f)$ (the vanishing locus of $f$) and is regular elsewhere. In particular, $g$ is regular on the complement of $V(f)$, which is the open set $D(f)$. Therefore, $g$ is a unit in $A_f$, the localization of $A$ at $f$. This implies that $(f)$ is the trivial divisor, since it is linearly equivalent to the zero divisor $(1)$.
Now, suppose that $A$ is a UFD. Let $D$ be a Weil divisor on $X$, given by a nonzero element $f$ of $A$ and a nonzero function $g$ on $D$. We can write $g$ as $g=au/f^n$ for some unit $u$ of $A$, some integer $n\geq 0$, and some element $a$ of $A$ not divisible by $f$. We define a principal divisor $(g)$ as the formal sum $\sum_{x\in X} v_x(g) \cdot [x]$, where $v_x(g)$ denotes the order of vanishing of $g$ at $x$ and $[x]$ denotes the corresponding point of $X$. Note that $v_x(g)$ is zero for all but finitely many points of $X$, since $g$ is nonzero. Also note that $v_x(g)$ is equal to $n$ if $x\in D$ and $v_x(f)$ if $x\not\in D$. Therefore, $(g)$ is linearly equivalent to the divisor $(f^{n-v_x(g)})$.
Let $h$ be another nonzero function on $D$, represented by a nonzero element $b$ of $A$ and a unit $v$ of $A$. We have $h=bv/f^m$ for some integer $m\geq 0$. The ratio $g/h$ is then equal to $u(v/b)(f^{m-n})$, which is a unit of $A$ (since $A$ is a UFD). Therefore, $(g)$ and $(h)$ are linearly equivalent as divisors on $X$. This shows that every Weil divisor on $X$ is linearly equivalent to a principal divisor, so $Cl(X)$ is trivial. Since $Pic(X)$ is isomorphic to $Cl(X)$ (as explained, for example, in Hartshorne’s Algebraic Geometry, Chapter II, Section 6), we conclude that $Pic(X)$ is also trivial.
(b) Let $X \subset \mathbb{P}^n$ be a projective variety. Suppose that the homogeneous coordinate ring of $X$ is a UFD. Show that $\operatorname{Pic}(X) \cong \mathbb{Z}$.
Since $X$ is a projective variety, it is a closed subset of $\mathbb{P}^n$ for some $n$, and hence we may assume that $X$ is a closed subset of $\mathbb{P}^n$ for some $n$. Let $S(X)$ be the homogeneous coordinate ring of $X$, which is a graded ring.
Since $S(X)$ is a UFD, every irreducible homogeneous element of $S(X)$ is a prime element. In particular, every irreducible hypersurface in $\mathbb{P}^n$ that intersects $X$ is a prime divisor on $X$.
Conversely, every prime divisor $D$ on $X$ corresponds to an irreducible hypersurface in $\mathbb{P}^n$ that intersects $X$. By the projective Nullstellensatz, the homogeneous ideal of this hypersurface is generated by homogeneous polynomials of the same degree, and we may assume that these polynomials are irreducible. Since $S(X)$ is a UFD, the irreducible polynomials generate a prime ideal, and hence $D$ is an irreducible element of $\operatorname{Pic}(X)$.
Thus, we have established a bijection between the prime divisors on $X$ and the irreducible elements of $\operatorname{Pic}(X)$. Moreover, every non-zero element of $\operatorname{Pic}(X)$ can be written as a product of irreducible elements, since $S(X)$ is a UFD.
Therefore, $\operatorname{Pic}(X)$ is isomorphic to the group of divisors modulo principal divisors, which is generated by the class of any prime divisor on $X$. Since $X$ is irreducible, any non-empty open subset of $X$ is dense in $X$, and hence any non-zero regular function on $X$ is a unit in $S(X)$. It follows that the group of principal divisors on $X$ is isomorphic to $\mathbb{Z}$, and hence $\operatorname{Pic}(X) \cong \mathbb{Z}$.
Let $X$ be the line with a double point at zero, thus we have a map $X \rightarrow \mathbb{A}^1$ which is an isomorphism over $\mathbb{A}^1 \backslash{0}$ and the preimage of 0 consists of two points.
(a) Let $Y=\mathbb{A}^2 \backslash{0}$. Show that the map $m: Y \rightarrow \mathbb{A}^1, m(x, y)=x y$ can be lifted to an onto map $Y \rightarrow X$; moreover, there are two distinct such liftings.
To construct a lifting, we need to find a map $f: Y \rightarrow X$ such that $m = \pi \circ f$, where $\pi: X \rightarrow \mathbb{A}^1$ is the given projection map. Since $m(x,y) = xy$ and $\pi$ has a double point at zero, it’s natural to define $f$ as follows:
- For $(x,y) \in Y$ with $xy \neq 0$, let $f(x,y) = (x,y)$, i.e., we map $(x,y)$ to itself.
- For $(x,y) \in Y$ with $xy = 0$, let $f(x,y) = (x,0)$ if $y=0$, and $f(x,y) = (0,y)$ if $x=0$.
It’s clear that $f$ is well-defined and continuous on $Y$. Moreover, since $f(x,y) = (x,y)$ for $(x,y) \in Y \cap (\mathbb{A}^1 \times \mathbb{A}^1)$, where $\mathbb{A}^1 \times \mathbb{A}^1$ is the open subset of $Y$ where both coordinates are nonzero, we have $\pi \circ f = m$ on this open subset. Therefore, $\pi \circ f = m$ on all of $Y$, since both $m$ and $\pi \circ f$ are continuous.
To show that there are two distinct liftings, note that we can also define $f’: Y \rightarrow X$ by interchanging the two coordinates in the second case above:
- For $(x,y) \in Y$ with $xy \neq 0$, let $f'(x,y) = (x,y)$, i.e., we map $(x,y)$ to itself.
- For $(x,y) \in Y$ with $xy = 0$, let $f'(x,y) = (0,y)$ if $x=0$, and $f'(x,y) = (x,0)$ if $y=0$.
Then $f’$ is also well-defined and continuous on $Y$, and satisfies $\pi \circ f’ = m$. However, $f$ and $f’$ are not equal, since $f(1,0) = (1,0)$ and $f'(1,0) = (0,0)$. Therefore, we have found two distinct liftings of $m$ to $X$.
