Riemann surface
matlab
还在面临交换代数的学习挑战吗？别担心！我们的commutative-algebra-guide团队专业为您解决环论、模论、整体函数环等方面的问题。我们拥有深厚的专业背景和丰富的经验，能帮您完成高水平的作业和论文，让您的学习之路一帆风顺！
以下是一些我们可以帮助您解决的问题：
环和模：环的定义、性质和分类，模的定义和性质，如交换环、域、无零因子环等。
环同态和理想：环同态的定义、性质和应用，理想的定义、性质和分类，如极大理想、主理想等。
域论：域的定义、性质和扩域，域的结构和分类，如代数闭域、有限域等。
整体函数环：整体函数环的定义和性质，整体函数环上的运算和理想，整体函数环与代数几何的关系。
代数簇：代数簇的定义和性质，代数簇上的函数环和理想，代数簇的几何性质和分类。
无论您面临的交换代数问题是什么，我们都会尽力为您提供专业的帮助，确保您的学习之旅顺利无阻！
Proposition 1.14. The radical of an ideal a is the intersection of the prime ideals which contain a.
Proof. We already know $r(a)=\phi^{-1}\left(\Re_{A / a}\right)$, where $\Re$ is nilradical, and $\phi: A \rightarrow A / a$ a canonical projection. Since $\Re_{A / a}$ is the intersection of all prime ideals in $A / a$, from the one-to-one correspondence and theorem 1.8 (preimage of prime ideal is prime), $r(a)$ is intersection of prime ideals containing a.
Proposition 1.14. The radical of an ideal $a$ is the intersection of the prime ideals which contain $a$.
Proof. We already know that $r(a) = \phi^{-1}(\Re_{A/a})$, where $\Re$ is the nilradical, and $\phi: A \rightarrow A/a$ is the canonical projection. Since $\Re_{A/a}$ is the intersection of all prime ideals in $A/a$, by using the one-to-one correspondence and Theorem 1.8 (the preimage of a prime ideal is prime), we can conclude that $r(a)$ is the intersection of prime ideals containing $a$.
Proposition 1.15 (part of proof). $D=r(D)=\bigcup_{x \neq 0} \operatorname{Ann}(x)$
Proof. $D \subseteq r(D)$ is clear from above exercise. If $f \in r(D)$, then $f^n \in D$, so $f^n h=0$ for some $h \in A$, so $f\left(f^{n-1} h\right)=0$ implies $f \in D$.
Also, $f \in D$ implies $f x=0$ for some $x \in A \backslash{0}$ implies $f \in \operatorname{Ann}(x)$. Conversely, if $f \in \operatorname{Ann}(x)$, then $f \in D$. Since $x$ was chosen arbitrarily, done.
Proposition 1.15 (part of the proof): $D = r(D) = \bigcup_{x \neq 0} \operatorname{Ann}(x)$
Proof: The inclusion $D \subseteq r(D)$ is clear from the previous exercise. Now, let’s show the other inclusion.
Suppose $f \in r(D)$. This means that $f^n \in D$ for some positive integer $n$. Then, we have $f^n h = 0$ for some $h \in A$. By multiplying both sides of the equation by $f^{n-1}$, we get $f(f^{n-1} h) = 0$. This implies that $f$ is a zero-divisor in $D$, and hence $f \in D$.
Next, let’s show that $f \in D$ implies $f x = 0$ for some $x \in A \backslash {0}$, which further implies $f \in \operatorname{Ann}(x)$. This shows that $D \subseteq \bigcup_{x \neq 0} \operatorname{Ann}(x)$.
Conversely, suppose $f \in \operatorname{Ann}(x)$ for some $x \in A \backslash {0}$. This means that $f x = 0$. Since $x \neq 0$, we have found an element $x$ such that $f x = 0$, and thus $f \in D$.
Since $f$ was chosen arbitrarily, we conclude that $D = \bigcup_{x \neq 0} \operatorname{Ann}(x)$.
Therefore, combining both inclusions, we have $D = r(D) = \bigcup_{x \neq 0} \operatorname{Ann}(x)$, as desired.
E-mail: help-assignment@gmail.com 微信:shuxuejun
help-assignment™是一个服务全球中国留学生的专业代写公司
专注提供稳定可靠的北美、澳洲、英国代写服务
专注于数学，统计，金融，经济，计算机科学，物理的作业代写服务