# 换元代数代写|COMMUTATIVE ALGEBRA MATH247 University of Liverpool Assignment

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Commutative algebra is a branch of abstract algebra that studies commutative rings, which are algebraic structures that have two binary operations, usually denoted as addition and multiplication. Commutative algebra plays an important role in several areas of mathematics, including number theory, algebraic geometry, and linear algebra.

In commutative algebra, one studies commutative rings, which are rings in which the multiplication operation is commutative. Examples of commutative rings include the ring of integers, polynomial rings, and the field of real numbers. Commutative algebra also studies modules over commutative rings, which are generalizations of vector spaces over a field.

The theory of commutative algebra has many applications to other areas of mathematics. For example, it is used in algebraic geometry to study algebraic varieties, which are geometric objects defined by systems of polynomial equations. Commutative algebra is also used in number theory to study algebraic number fields, which are extensions of the rational numbers that are closed under addition, subtraction, multiplication, and division.

Finally, commutative algebra has applications to linear algebra, especially in the study of linear transformations of vector spaces over fields. For example, if one has a linear transformation of a vector space that is represented by a matrix, then commutative algebra can be used to study the properties of that matrix, such as its eigenvalues and eigenvectors.

Show that $k\left[\mathbb{A}^2 \backslash{0}\right]=k\left[\mathbb{A}^2\right]$. Conclude that $\mathbb{A}^2 \backslash{0}$ is not affine.
[Hint: Use the covering by two affine open subsets given by $x \neq 0$ and $y \neq 0$, where $x, y$ are coordinates on $\left.\mathbb{A}^2\right]$.

Let $U_1$ and $U_2$ be the affine open subsets of $\mathbb{A}^2$ defined by $x\neq0$ and $y\neq0$, respectively. We want to show that $k[U_1\cup U_2] = k[\mathbb{A}^2]$. Since $U_1$ and $U_2$ cover $\mathbb{A}^2\backslash{0}$, it is sufficient to show that $k[U_1\cup U_2] = k[\mathbb{A}^2\backslash{0}]$.

Note that $\mathbb{A}^2\backslash{0} = U_1\cup U_2 \cup (U_1\cap U_2)$. Therefore, we have the following chain of inclusions: \begin{align*} k[\mathbb{A}^2\backslash{0}] &\subseteq k[U_1\cup U_2 \cup (U_1\cap U_2)]\ &\subseteq k[U_1] \cap k[U_2] \cap k[U_1\cap U_2]\ &= k[\mathbb{A}^2]. \end{align*} The first inclusion follows from the fact that $k[\mathbb{A}^2\backslash{0}]$ is the subring of $k(\mathbb{A}^2)$ consisting of rational functions that are regular on $\mathbb{A}^2\backslash{0}$, and every such function is clearly regular on $U_1\cup U_2 \cup (U_1\cap U_2)$. The second inclusion follows from the fact that a rational function that is regular on $U_1\cup U_2 \cup (U_1\cap U_2)$ must be regular on each of these sets individually.

Since we have $k[U_1\cup U_2] = k[\mathbb{A}^2]$, it follows that $\mathbb{A}^2\backslash{0}$ cannot be affine, because an affine variety cannot be covered by two non-empty affine open subsets.

Let $C$ be a curve in $\mathbb{P}^2, x$ be a point in $C$ and $L$ a line passing through $x$. Let $m$ be the multiplicity of $C$ at $x$ and $M$ the multiplicity of intersection of $C$ and $L$ at $x$. Show that $m \leq M$ and that for given $C, x$ the equality $m=M$ holds for all but finitely many lines $L$ as above.

Let $f$ be a homogeneous polynomial of degree $d$ defining $C$. We may assume that $x = [1:0:0]$. Up to projective transformation we may also assume that $L$ is given by $y = 0$. Then $C \cap L$ is defined by the ideal $(f, y)$ in $k[x,y,z]$.

Since $f$ vanishes at $x$, we may write $$f(x,y,z) = z^m g(x,y,z)$$ where $g(x,y,z)$ is a homogeneous polynomial of degree $d-m$ and $g(1,0,0) \neq 0$. Then $$(f, y) = (z^m g(x,y,z), y) = (z^m, y) + (g(x,y,z), y).$$ Since $y$ does not divide $z^m$, we have $(z^m, y) = (y)$, and so $$(f, y) = (y) + (g(x,y,z), y).$$ Thus the intersection multiplicity $M$ is equal to the order of vanishing of $g(x,y,z)$ at $[1:0:0]$, which is at least $m$.

To show that $m = M$ for all but finitely many lines $L$, it suffices to show that the intersection multiplicity $M$ is upper semicontinuous in $L$. In other words, we need to show that if $L_t$ is a family of lines converging to $L$ (in the Zariski topology) then $M_t \geq M$ for $t$ sufficiently close to $0$.

We may assume that the family $L_t$ is given by $y = t$. Let $g_t(x,z)$ be the polynomial of degree $d-m$ defining $C \cap L_t$ at $[1:0:0]$. Then $g_t(x,z) = f(x,t,z)$, and so the coefficient of $z^{d-m}$ in $g_t(x,z)$ is given by $$a_t = f_{0,d-m}(x,t,1).$$ Since $f_{0,d-m}(x,t,1)$ is a polynomial in $x$ and $t$, it is upper semicontinuous in $t$. Therefore, there exists a neighborhood $U$ of $0$ such that $a_t > 0$ for all $t \in U$.

Let $Z$ be an irreducible closed subset in an algebraic variety $X$. Show that if $\operatorname{dim}(Z)=\operatorname{dim}(X)$ then $Z$ is a component of $X$.

Suppose that $Z$ is not a component of $X$. Then there exists a proper closed subset $Y$ of $X$ such that $Z$ is contained in $Y$. Since $Y$ is a proper subset of $X$, we have $\operatorname{dim}(Y) < \operatorname{dim}(X)$. By the irreducibility of $Z$, we have $Z \not\subseteq Y$. Therefore, there exists a point $p\in Z$ such that $p\notin Y$.

Since $p$ is a point of $Z$, we have $\operatorname{dim}(\mathcal{O}{X,p}) \geq \operatorname{dim}(Z)$, where $\mathcal{O}{X,p}$ is the local ring of $X$ at $p$. On the other hand, since $p\notin Y$, we have $\mathcal{O}{X,p} \subseteq \mathcal{O}{Y,p}$. This implies $\operatorname{dim}(\mathcal{O}{X,p}) \leq \operatorname{dim}(\mathcal{O}{Y,p})$. By the definition of dimension of a local ring, we have $\operatorname{dim}(Y) = \operatorname{dim}(\mathcal{O}_{Y,p})$. Combining these inequalities, we obtain $\operatorname{dim}(Z) \leq \operatorname{dim}(Y)$. This contradicts the assumption that $\operatorname{dim}(Z) = \operatorname{dim}(X)$.

Therefore, our assumption that $Z$ is not a component of $X$ must be false. Thus, $Z$ is a component of $X$.