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MAT4200 commutative Algebra课程简介

Commutative rings and their modules are important structures in algebraic geometry and number theory, so understanding their properties and structures is crucial for further studies in these areas.

The concept of localization is particularly important in commutative ring theory. Given a commutative ring $R$ and a multiplicatively closed subset $S$ of $R$, the localization of $R$ at $S$, denoted $S^{-1}R$, is a new ring obtained by formally inverting the elements of $S$. Localization allows us to study properties of a ring by restricting attention to a smaller subset of elements. For example, we can use localization to study the behavior of a ring at a prime ideal, which is a fundamental concept in algebraic geometry.

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 commutative Algebra HELP（EXAM HELP， ONLINE TUTOR）

(1) First, we prove $\sqrt{I+J}\supseteq\sqrt{\sqrt{I}+\sqrt{J}}$. Let $x\in \sqrt{I}+\sqrt{J}$, i.e., there exist $a\in I, b\in J$ such that $x=a+b$. Since $\sqrt{I}$ and $\sqrt{J}$ are both radical ideals, we have $a^n\in I, b^m\in J$ for some $n,m>0$. Thus, $(a+b)^{n+m-1}=\sum_{i+j=n+m-1}\binom{n+m-1}{i}a^ib^j\in I+J$, which implies that $x^{n+m-1}\in I+J$. Therefore, $x\in\sqrt{I+J}$. This shows that $\sqrt{\sqrt{I}+\sqrt{J}}\subseteq\sqrt{I+J}$.

Now we prove $\sqrt{I+J}\subseteq\sqrt{\sqrt{I}+\sqrt{J}}$. Let $x\in\sqrt{I+J}$, i.e., $x^n=\sum_{i=1}^m a_ib_i$ for some $a_i\in I, b_i\in J$ and $n,m>0$. We may assume that $m$ is minimal among all such representations of $x^n$. Let $I’=\langle a_1,\dots,a_m\rangle$ and $J’=\langle b_1,\dots,b_m\rangle$. Then $I’+J’=A$ since $1\in I’+J’$. Since $x^n\in I’+J’$, we have $x^n=\sum_{i=1}^k c_id_i$ for some $c_i\in I’, d_i\in J’$ and $k\leq m$. Note that $k$ must be greater than $1$, otherwise we would have $x\in I$ or $x\in J$ contradicting $x\in\sqrt{I+J}$.

Let $S={i\in{1,\dots,k}\mid c_i\in I}$ and $T={i\in{1,\dots,k}\mid d_i\in J}$. Since $I’$ and $J’$ are ideals, $S$ and $T$ are nonempty. Let $s=\sum_{i\in S}c_ia_i$ and $t=\sum_{i\in T}d_ib_i$. Then $x^n=s+t$ with $s\in I$ and $t\in J$. Since $x^n$ is a minimal sum of elements in $I$ and $J$ that add up to $x^n$, we must have $s\neq0$ and $t\neq 0$.

Now we have $s^n=\left(\sum_{i\in S}c_ia_i\right)^n\in I^n$ and $t^n=\left(\sum_{i\in T}d_ib_i\right)^n\in J^n$, since $I$ is a prime ideal. Thus, $x^n=s^n+t^n\in I^n+J^n$, and $x\in\sqrt{I^n+J^n}=\sqrt{I}+\sqrt{J}$, where the last equality follows from the fact that $I$ is a prime ideal. Therefore, we have $\sqrt{I+J}\subseteq\sqrt{\sqrt{I}+\sqrt{J}}$.

(1) Suppose $I \subseteq J$, and let $\mathfrak{p} \in Z(J)$. This means $J \subseteq \mathfrak{p}$. But $I \subseteq J$, so $I \subseteq \mathfrak{p}$, which shows that $\mathfrak{p} \in Z(I)$. Therefore, $Z(J) \subseteq Z(I)$.

(2) Let $\mathfrak{p} \in Z(\sqrt{I})$. Then $\sqrt{I} \subseteq \mathfrak{p}$, which means that $\mathfrak{p}$ contains every element of $I$ and all of its roots. In particular, $\mathfrak{p}$ contains $I$, so $\mathfrak{p} \in Z(I)$. This shows that $Z(\sqrt{I}) \subseteq Z(I)$.

Conversely, let $\mathfrak{p} \in Z(I)$. Then $I \subseteq \mathfrak{p}$. Since $\sqrt{I}$ is the intersection of all prime ideals containing $I$, we have $\sqrt{I} \subseteq \mathfrak{p}$, which shows that $\mathfrak{p} \in Z(\sqrt{I})$. This shows that $Z(I) \subseteq Z(\sqrt{I})$.

Combining these inclusions, we have $Z(I) = Z(\sqrt{I})$.

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

On the other hand, if $A \subseteq \mathfrak{p}$ for some prime ideal $\mathfrak{p}$, then $1 \in A$ is not in $\mathfrak{p}$, which contradicts the definition of a prime ideal. Therefore, there is no prime ideal containing $A$, so $Z(A) = \varnothing$.

(4) Let $I = \sum_{\alpha \in \mathcal{T}} I_\alpha$. Suppose $\mathfrak{p} \in \bigcap_{\alpha \in \mathcal{T}} Z(I_\alpha)$, so $I_\alpha \subseteq \mathfrak{p}$ for all $\alpha \in \mathcal{T}$. Then $I \subseteq \mathfrak{p}$, since $I$ is the sum of the $I_\alpha$. Therefore, $\mathfrak{p} \in Z(I)$.

Conversely, suppose $\mathfrak{p} \in Z(I)$. Then $I \subseteq \mathfrak{p}$, which means that $I_\alpha \subseteq \mathfrak{p}$ for all $\alpha \in \mathcal{T}$. Therefore, $\mathfrak{p} \in Z(I_\alpha)$ for all $\alpha \in \mathcal{T}$, which shows that $\mathfrak{p} \in \bigcap_{\alpha \in \mathcal{T}} Z(I_\alpha)$.

Combining these inclusions, we have $\bigcap_{\alpha \in \mathcal{T}} Z(I_\alpha) = Z(I)$.

(5) Suppose $\mathfrak{p} \in Z(I \cap J)$. Then $(I \cap J) \subseteq \mathfrak{p}$. Since $I \cap J \

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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