MY-ASSIGNMENTEXPERT™可以为您提供studier MAT4200 Commutative Algebra交换代数课程的代写代考和辅导服务！
这是奥斯陆大学交换代数课程的代写成功案例。
MAT4200课程简介
Learning outcome
After completing the course you
know the definition of commutative rings, local rings, prime and maximal ideals, and modules over commutative rings
are familiar with the notions of noetherian and artinian rings and modules
know how to localize rings and modules, and are familiar with important applications of localization
know the Hilbert basis theorem and the Hilbert Nullstellensatz
are familiar with the concepts of support and associated primes
know the definition of an exact sequence of modules, and you also know important properties and applications of exact sequences
know the concept of direct limit and you can compute this limit in some non-trivial examples
know how to define tensor products of modules and are familiar with the concept of flatness
know Krull-Cohen-Seidenberg theory
know the basic results in the dimension theory for local rings
know how to complete a ring in an ideal.
Prerequisites
Admission to the course
Students admitted at UiO must apply for courses in Studentweb. Students enrolled in other Master’s Degree Programmes can, on application, be admitted to the course if this is cleared by their own study programme.
Nordic citizens and applicants residing in the Nordic countries may apply to take this course as a single course student.
If you are not already enrolled as a student at UiO, please see our information about admission requirements and procedures for international applicants.
MAT4200 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 $K \otimes \downarrow \mathbf{Q}[x] \rightarrow K[x]$ by the specification $a \otimes \emptyset f \mapsto a f$, which is well defined because it is balanced. We define a map $K[x] \rightarrow K \otimes \mathbb{Q} \mathbb{Q}[x]$ 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) $\mathbb{Q}(\sqrt{2}) \otimes \mathbb{Q} \mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}[x] /\left(x^2-2\right) \otimes \mathbb{Q} \mathbb{Q}(\sqrt{2}) \cong \mathbb{Q}(\sqrt{2})[x] /\left(x^2-2\right)$ 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, $B=B_1$.
Let $A, B$ be left $R$-modules and let $r \in Z(R)={s \in R \mid$ st $=t s$ for all $t \in R}$, the center of $R$. Let $\mu_r: B \rightarrow B$ be multiplication by $r$. Prove that the induced map $\left(\mu_r\right)_*: \operatorname{Hom}_R(A, B) \rightarrow \operatorname{Hom}_R(A, B)$ is also multiplication by $r$.
Solution. Let $f: A \rightarrow B$ be an $R$-module homomorphism. The effect of $\left(\mu_r\right)* f$ on an element $a \in A$ is $\left(\left(\mu_r\right)* f\right)(a)=\mu_r(f(a))=r f(a)$. This equals the effect of $r f$ on $a$ by the definition of the $R$-module structure on $\operatorname{Hom}_R(A, B)$.
MY-ASSIGNMENTEXPERT™可以为您提供STUDIER MAT4200 COMMUTATIVE ALGEBRA交换代数课程的代写代考和辅导服务！
