MY-ASSIGNMENTEXPERT™可以为您提供users.cse.umn.edu Math8211 Commutative Algebra交换代数课程的代写代考和辅导服务！
Math8211课程简介
TEXT: Commutative algebra. With a view toward algebraic geometry by David Eisenbud, 1995.
Prerequisite: Math 8201-02 General Algebra or some previous experience with groups, rings, and fields.
DESCRIPTION: Commutative algebra stands at the crossroads of algebra, number theory, and algebraic geometry. It is subsumed by algebraic geometry as the local study of algebraic varieties, somewhat similar to analysis in R^n succumbing to the theory of manifolds. Homological algebra is a powerful algebraic tool used in many fields of mathematics, including commutative and noncommutative algebra, group theory, Lie theory, several complex variables, geometry and topology, PDE, combinatorics, functional analysis, numerical analysis, and mathematical physics, to name a few.
Prerequisites
CONTENT: In the Fall Semester, we will study commutative algebra. This will include commutative rings and modules over them, Noetherian rings, Krull dimension theory, Noether normalization, the so-called Nullstellensatz, the spectrum of a ring, rings of fractions and localization, primary decomposition, discrete valuation rings, normal integral domains, and regular local rings. The geometric view of a commutative ring as the ring of functions on a space will be emphasized.
The homological algebra part of the course will be taught in the Spring Term by Bernard Badzioch. The topics will include complexes, homology, resolutions and derived functors. These notions will be put into the context of two different axiomatic approaches to homological algebra: via triangulated categories and via closed model categories. Additional topics will include Koszul complex, Hochschild homology and cyclic homology. Applications to commutative algebra (such as the notion of depth and Cohen-Macaulay rings), algebraic geometry and topology will be discussed.
Math8211 Commutative Algebra HELP（EXAM HELP， ONLINE TUTOR）
(a) Let $K \supseteq \mathbb{Q}$ be a field containing $\mathbb{Q}$. Show that $K \otimes \mathcal{Q} \mathbb{Q}[x] \cong K[x]$ as rings
(b) Show that, as a ring, $\mathbb{Q}(\sqrt{2}) \otimes d \mathbb{Q}(\sqrt{2})$ is the direct sum of two fields.
[The ring multiplication is $(a \otimes b)(c \otimes d):=a c \otimes b d$ on basic tensors. Use the isomorphism $\left.\mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}[x] /\left(x^2-2\right).\right]$
Solution. (a) We define a map $K \otimes \downarrow \quad \mathbb{Q}[x] \rightarrow K[x]$ by the specification $a \otimes \emptyset f \mapsto a f$, which is well defined because it is balanced. We define a map $K[x] \rightarrow K \otimes \mathcal{Q} \mathbb{Q}[x]$ on monomials $b x^n$ by $b x^n \mapsto b \otimes x^n$ extended by linearity because these monomials span $K[x]$. These two maps are ring homomorphisms and are mutually inverse.
(b) $\mathbb{Q}(\sqrt{2}) \otimes \mathcal{Q} \mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}[x] /\left(x^2-2\right) \otimes \mathbb{Q} \mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{2})[x] /\left(x^2-2\right)$ by an extension of part (a). In $\mathbb{Q}(\sqrt{2})[x]$ the polynomial $x^2-2$ factors as $(x-\sqrt{2})(x+\sqrt{2})$, so by the Chinese Remainder Theorem $\mathbb{Q}(\sqrt{2})[x] /\left(x^2-2\right) \cong \mathbb{Q}(\sqrt{2})[x] /(x-\sqrt{2}) \oplus \mathbb{Q}(\sqrt{2})[x] /(x+\sqrt{2}) \cong$ $\mathbb{Q}(\sqrt{2}) \oplus \mathbb{Q}(\sqrt{2})$
Let $0 \rightarrow A \stackrel{\alpha}{\longrightarrow} B \stackrel{\beta}{\longrightarrow} C \rightarrow 0$ be a short exact sequence of $R$-modules, for some ring $R$. Suppose that $A$ can be generated as an $R$-module by a subset $X \subseteq A$ and that $C$ can be generated as an $R$-module by a subset $Y \subseteq C$. For each $y \in Y$, choose $y^{\prime} \in B$ with $\beta\left(y^{\prime}\right)=y$. Prove that $B$ is generated by the set $\alpha(X) \cup\left{y^{\prime} \mid y \in Y\right}$.
Solution. Let $B_1$ be the submodule of $B$ generated by $\alpha(X) \cup\left{y^{\prime} \mid y \in Y\right}$. Now $B_1$ contains $\alpha(A)$ and by the correspondence theorem corresponds to a submodule of $B / \alpha(A)$, which is isomorphic to $C$ via a map induced by $\beta$. Because $C$ is generated by $Y$ that submodule is $B / \alpha(A)$. Because $B$ also corresponds to this submodule, $B=B_1$.
MY-ASSIGNMENTEXPERT™可以为您提供UNIVERSITY OF ILLINOIS URBANA-CHAMPAIGN MATH2940 linear algebra线性代数课程的代写代考和辅导服务！ 请认准MY-ASSIGNMENTEXPERT™. MY-ASSIGNMENTEXPERT™为您的留学生涯保驾护航。
