Riemann surface
matlab

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.

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