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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）
Let $(x)=\left(x_0, \ldots, x_n\right),(y)=\left(y_0, \ldots, y_m\right)$ be $n+m+2$ elements of A. By considering the iterated boundary monoid of $(y)$ in $\mathbf{A} /\langle a\rangle$, we obtain that $\mathcal{S}{\mathbf{A}}^{\mathrm{K}}(y)$ contains a multiple of $a$, say $b a$. By considering the iterated boundary ideal of $(x)$ in $\mathbf{A}[1 / a]$, we obtain that $\mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x)$ contains a power of $a$, say $a^e$.
Then $(b a)^e \in \mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x) \cap \mathcal{S}{\mathrm{A}}^{\mathrm{K}}(y)$, so $1 \in \mathcal{I}_{\mathbf{A}}^{\mathrm{K}}(x, y)$ by Fact XIII-2.9, item
The proof is by induction on $n+m$. If $n+m=0$, then there is nothing to prove. Now suppose $n+m>0$. Without loss of generality, assume $a$ does not divide any $y_i$. Let $\mathbf{A}=\mathbf{k}[y_0,\ldots,y_m]$, and let $\mathbf{B}=\mathbf{A}/\langle y_0-a\rangle$. Then $y’=(y_1,\ldots,y_m)\in\mathbf{B}$, and $\mathcal{S}{\mathbf{A}}^{\mathrm{K}}(y)\subseteq\mathcal{S}{\mathbf{B}}^{\mathrm{K}}(y’)$ and $\mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x,y)\supseteq\mathcal{I}{\mathbf{B}}^{\mathrm{K}}(x,y’)$. Since $\mathbf{B}$ is a polynomial ring in $m$ variables, the induction hypothesis applies to $(x,y’)$, yielding $\mathcal{I}{\mathbf{B}}^{\mathrm{K}}(x,y’)=\mathcal{S}{\mathbf{B}}^{\mathrm{K}}(y’)$. Thus, it suffices to show that $\langle y_0-a\rangle$ is in the radical of $\mathcal{I}_{\mathbf{A}}^{\mathrm{K}}(x,y)$.
By considering the iterated boundary monoid of $(y)$ in $\mathbf{A}/\langle y_0-a\rangle$, we obtain that $\mathcal{S}{\mathbf{A}}^{\mathrm{K}}(y)$ contains a multiple of $y_0-a$, say $(y_0-a)b$. By considering the iterated boundary ideal of $(x)$ in $\mathbf{A}[1/(y_0-a)]$, we obtain that $\mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x)$ contains a power of $y_0-a$, say $(y_0-a)^e$. Then $(y_0-a)^{e+1}b\in\mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x,y)\cap\mathcal{S}{\mathbf{A}}^{\mathrm{K}}(y)$, so $y_0-a\in\mathrm{rad}(\mathcal{I}_{\mathbf{A}}^{\mathrm{K}}(x,y))$, as desired.
By using the previous question, we show by induction on $i \in \llbracket 0 . . k+1 \rrbracket$ that we have $\operatorname{Kdim} \mathbf{A}\left[1 / a_i\right] \leqslant i-1$; for $i=k+1$, we obtain $\operatorname{Kdim} \mathbf{A} \leqslant k$.
We proceed by induction on $i$. The base case $i=0$ is clear since $\mathbf{A}[1/a_0] = \mathbf{A}$ and $\operatorname{Kdim} \mathbf{A} \leqslant -1$ is impossible.
For the inductive step, assume that $\operatorname{Kdim} \mathbf{A}[1/a_{i-1}] \leqslant i-2$ for some $i \in \llbracket 1, k+1 \rrbracket$. By the previous question, we have $\mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x, a{i-1}) \subseteq \mathcal{I}{\mathbf{A}}^{\mathrm{K}}(x, a)$ and $\mathcal{S}{\mathrm{A}}^{\mathrm{K}}(y, a_{i-1}) \subseteq \mathcal{S}_{\mathrm{A}}^{\mathrm{K}}(y, a)$. Therefore, we have
\begin{aligned} \operatorname{Kdim} \mathbf{A}[1/a_i] &= \operatorname{Kdim} \left(\mathbf{A}[1/a_{i-1}]\right)[1/a_i] \\ &\leqslant \operatorname{Kdim} \mathbf{A}[1/a_{i-1}] + 1 \\ &\leqslant i-2 + 1 \\ &= i-1. \end{aligned}KdimA[1/ai]=Kdim(A[1/ai−1])[1/ai]⩽KdimA[1/ai−1]+1⩽i−2+1=i−1.
This completes the induction. Setting $i=k+1$, we obtain $\operatorname{Kdim} \mathbf{A} \leqslant k$.
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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