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MAT4200 Representation theory课程简介

This course gives an introduction to commutative rings and their modules. We study concepts such as localization, decomposition of modules, chain conditions for rings and modules, and dimension theory. The course gives a relevant background for studies in algebraic geometry, but also relates the theory to problems in number theory.

PREREQUISITES 

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.

MAT4200 Representation theory HELP（EXAM HELP， ONLINE TUTOR）

(1) Firstly, we show that $\sqrt{I+J}\subseteq \sqrt{\sqrt{I}+\sqrt{J}}$. Suppose $x\in \sqrt{I+J}$, then there exists a positive integer $n$ such that $x^n\in I+J$. Write $x^n=\sum_i a_i+b_i$, where $a_i\in I$ and $b_i\in J$. Since $\sqrt{I}$ is an ideal, we have $(\sum_i a_i)^m\in I$ for some positive integer $m$, and similarly $(\sum_i b_i)^k\in J$ for some positive integer $k$. Then we have \begin{align*} x^n&=(\sum_i a_i+b_i)^n\ &=\sum_{i_1,i_2,\dots,i_n}\binom{n}{i_1,i_2,\dots,i_n}a_{i_1}a_{i_2}\dots a_{i_n}(\sum_i b_i)^{n-i_1-i_2-\dots-i_n}\ &\in I^n+J^n+\sum_{i=1}^{n-1}I^{i}J^{n-i}+\sum_{i=1}^{n-1}J^{i}I^{n-i}\ &\subseteq (I+\sqrt{I}J)^n+(J+\sqrt{J}I)^n\ &\subseteq (\sqrt{I}+\sqrt{J})^n \end{align*} where the first inclusion is due to the binomial expansion of $(\sum_i a_i+b_i)^n$ and the second inclusion is due to the fact that $I\subseteq \sqrt{I}$ and $J\subseteq \sqrt{J}$. Therefore, we have $x\in \sqrt{\sqrt{I}+\sqrt{J}}$.

Secondly, we show that $\sqrt{I+J}\supseteq \sqrt{\sqrt{I}+\sqrt{J}}$. Suppose $x\in \sqrt{\sqrt{I}+\sqrt{J}}$, then there exists a positive integer $n$ such that $x^n\in \sqrt{I}+\sqrt{J}$. Write $x^n=a+b$ where $a\in \sqrt{I}$ and $b\in \sqrt{J}$. Since $\sqrt{I}$ and $\sqrt{J}$ are ideals, we have $a^m\in I$ and $b^k\in J$ for some positive integers $m$ and $k$. Then we have \begin{align*} x^n&=(a+b)^n\ &=\sum_{i_1,i_2,\dots,i_n}\binom{n}{i_1,i_2,\dots,i_n}a^{i_1}b^{i_2}\dots a^{i_{n-1}}b^{i_n}\ &\in I^n+J^n+\sum_{i=1}^{n-1}I^{i-1}J^{n-i}a^i+\sum_{i=1}^{n-1}J^{i-1}I^{n-i}b^i\ &\subseteq (I+J)^n+\sum_{i=1}^{n-1}I^{i-1}J^{n-i}(I^m+J^k)+\sum_{i=1}^{n-1}J^{i-1}I^{n-i}(I^m+J^k)

(1) Suppose $I \subset J$ and let $\mathfrak{p} \in Z(J)$, i.e., $J \subset \mathfrak{p}$. Since $I \subset J$, we have $I \subset \mathfrak{p}$, so $\mathfrak{p} \in Z(I)$. Thus, $Z(J) \subset Z(I)$.

(2) Let $\mathfrak{p} \in Z(\sqrt{I})$, i.e., $\sqrt{I} \subset \mathfrak{p}$. We want to show that $I \subset \mathfrak{p}$. Suppose not, then there exists $x \in I$ such that $x \notin \mathfrak{p}$. Since $\sqrt{I}$ is the intersection of all prime ideals containing $I$, it follows that $x^n \in \mathfrak{p}$ for some $n > 0$. But then $(x^n)^m \in \mathfrak{p}$ for all $m > 0$, so $x \in \sqrt{I} \subset \mathfrak{p}$, a contradiction. Thus, $I \subset \mathfrak{p}$ and hence $\mathfrak{p} \in Z(I)$. This shows that $Z(\sqrt{I}) \subset Z(I)$.

Conversely, suppose $\mathfrak{p} \in Z(I)$, i.e., $I \subset \mathfrak{p}$. Then $\sqrt{I} \subset \mathfrak{p}$ since $\mathfrak{p}$ is prime. Thus, $\mathfrak{p} \in Z(\sqrt{I})$. This shows that $Z(I) \subset Z(\sqrt{I})$. We have shown that $Z(\sqrt{I}) = Z(I)$.

(3) For any prime ideal $\mathfrak{p}$, we have $0 \subset \mathfrak{p}$, so $\mathfrak{p} \in Z(0)$. This shows that $Z(0) = \operatorname{Spec} A$.

On the other hand, if $A \subset \mathfrak{p}$ for some prime ideal $\mathfrak{p}$, then $1 \in A \subset \mathfrak{p}$, which implies that $\mathfrak{p} = A$. Thus, there are no prime ideals containing $A$, so $Z(A) = \varnothing$.

(4) Let $\mathfrak{p} \in \bigcap_{\alpha \in \mathcal{T}} Z(I_\alpha)$. Then $I_\alpha \subset \mathfrak{p}$ for all $\alpha$, so $\sum_{\alpha \in \mathcal{T}} I_\alpha \subset \mathfrak{p}$. This shows that $\mathfrak{p} \in Z\left(\sum_{\alpha \in \mathcal{T}} I_\alpha\right)$, so $\bigcap_{\alpha \in \mathcal{T}} Z(I_\alpha) \subset Z\left(\sum_{\alpha \in \mathcal{T}} I_\alpha\right)$.

Conversely, suppose $\mathfrak{p} \in Z\left(\sum_{\alpha \in \mathcal{T}} I_\alpha\right)$, i.e., $\sum_{\alpha \in \mathcal{T}} I_\alpha \subset \mathfrak

Textbooks

• An Introduction to Stochastic Modeling, Fourth Edition by Pinsky and Karlin (freely
available through the university library here)
• Essentials of Stochastic Processes, Third Edition by Durrett (freely available through
the university library here)
To reiterate, the textbooks are freely available through the university library. Note that
you must be connected to the university Wi-Fi or VPN to access the ebooks from the library
links. Furthermore, the library links take some time to populate, so do not be alarmed if
the webpage looks bare for a few seconds.



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