Riemann surface
matlab

Exercise 1.12.
i) $a \subseteq(a: b)$.
For each $a \in a$ we have $a b \subseteq \mathfrak{a b} \subseteq \mathfrak{a}$, so $a \in(\mathfrak{a}: \boldsymbol{b})$.
ii) $(\boldsymbol{a}: \mathbf{b}) \mathbf{b} \subseteq \mathfrak{a}$.
By definition, for $x \in(\mathfrak{a}: \mathbf{b})$ we have $x \mathfrak{b} \subseteq \mathfrak{a}$.
iii) $((a: b): c)=(a: b c)=((a: c): b)$.
\begin{aligned} & x \in((\mathfrak{a}: \mathbf{b}): \mathbf{c}) \Longleftrightarrow x \mathbf{c} \subseteq(\mathbf{a}: \mathfrak{b}) \Longleftrightarrow x \mathfrak{c b} \subseteq \mathfrak{a} \Longleftrightarrow x \in(\mathfrak{a}: \mathbf{b} c) ; \ & x \in((\mathfrak{a}: \mathrm{c}): \mathrm{b}) \Longleftrightarrow x \mathfrak{b} \subseteq(\mathrm{a}: \mathrm{c}) \Longleftrightarrow x \mathfrak{b c} \subseteq \mathfrak{a} \Longleftrightarrow x \in(\mathrm{a}: \mathrm{bc}) . \ & \end{aligned}
iv) $\left(\bigcap_i \mathfrak{a}_i: \boldsymbol{b}\right)=\bigcap_i\left(\mathfrak{a}_i: \mathbf{b}\right)$.
$$x \in\left(\bigcap_i \mathfrak{a}_i: \mathbf{b}\right) \Longleftrightarrow x \mathbf{b} \subseteq \bigcap_i \mathfrak{a}_i \Longleftrightarrow \forall i\left(x \mathbf{b} \subseteq \mathfrak{a}_i\right) \Longleftrightarrow x \in \bigcap_i\left(\mathfrak{a}_i: \mathfrak{b}\right) .$$
v) $\left(a: \sum_i b_i\right)=\bigcap_i\left(a: b_i\right)$.
$$x \in\left(\mathfrak{a}: \sum_i \mathfrak{b}_i\right) \Longleftrightarrow \mathfrak{a} \supseteq x\left(\sum_i \mathbf{b}_i\right)=\sum_i x \mathfrak{b}_i \Longleftrightarrow \forall i\left(x \mathbf{b}_i \subseteq \mathfrak{a}\right) \Longleftrightarrow x \in \bigcap\left(\mathfrak{a}: \mathfrak{b}_i\right) .$$
For an $A$-module $M$ and subsets $N \subseteq M$ and $E \subseteq A$, define $(N: E):={m \in M: E m \subseteq N}$; for subsets $N, P \subseteq M$ and $E \subseteq A$, define $(N: P):={a \in A: a P \subseteq N}$.

Note for future use that then ii) holds equally well for subsets $a, b \subseteq M$, or $b \subseteq A$ and $a \subseteq M$; iii) holds for $\mathfrak{a}, \mathfrak{b} \subseteq M$ and $\boldsymbol{c} \subseteq A$; and iv) and $\mathbf{v}$ ) hold for modules $\mathfrak{a}, \mathfrak{a}_i$ and modules or ideals $\boldsymbol{b}, \mathfrak{b}_i$.

Exercise 1.13.
$-i) \mathfrak{a} \subseteq \mathfrak{b} \Longrightarrow r(\mathfrak{a}) \subseteq r(\mathfrak{b})$
If $x \in r(\mathfrak{a})$, for some $n>0$ we have $x^n \in \mathfrak{a} \subseteq \mathfrak{b}$, so $x \in r(\mathfrak{b})$.
0) $r\left(a^n\right)=r(a)$ for all $n>0$.
$\mathfrak{a}^n \subseteq \mathfrak{a}$, so by part – i) we have $r\left(\mathfrak{a}^n\right) \subseteq r(\mathfrak{a})$. If $x \in r(\mathfrak{a})$, then for some $m>0, x^m \in \mathfrak{a}$. But then $x^{m n} \in \mathfrak{a}^n$ and $x \in r\left(\overline{a^n}\right)$.
i) $r(a) \supseteq a$.

For each $a \in$ a we have $a^1 \in \mathfrak{a}$, so $a \in r(a)$.
ii) $r(r(a))=r(a)$.
$$x \in r(r(\mathfrak{a})) \Longleftrightarrow \exists n>0\left(x^n \in r(\mathfrak{a})\right) \Longleftrightarrow \exists n, m>0\left(\left(x^n\right)^m=x^{m n} \in \mathfrak{a}\right) \Longleftrightarrow x \in r(\mathfrak{a}) .$$
iii) $r(\mathfrak{a b})=r(\mathfrak{a} \cap \mathfrak{b})=r(\mathfrak{a}) \cap r(\mathfrak{b})$.
For the first equality, note $(a \cap b)^2 \subseteq a b \subseteq a \cap b$, so by parts 0$)$ and -1$), r(a \cap b)=r\left((a \cap b)^2\right) \subseteq r(a b) \subseteq r(a \cap b)$. For the second, note that if $m, n>0$ are such that $x^m \in \mathfrak{a}$ and $x^n \in \mathbf{b}$, then $x^{\max {m, n}} \in \mathfrak{a} \cap \mathbf{b}$, and conversely.
iv) $r(a)=(1) \Longleftrightarrow a=(1)$.
If $r(a)=(1)$, then $1 \in r(a)$, so for some $n$ we have $1=1^n \in a$, and then $a=(1)$.
v) $r(\mathrm{a}+\mathrm{b})=r(r(\mathrm{a})+r(\mathrm{~b}))$.
Since $\mathbf{a}, \mathbf{b} \subseteq \mathbf{a}+\mathbf{b}$, by part $-\mathrm{i})$ we have $r(\mathfrak{a}), r(b) \subseteq r(\mathfrak{a}+\mathbf{b})$, so $r(\mathfrak{a})+r(\mathbf{b}) \subseteq r(\mathfrak{a}+\mathbf{b})$. By parts $-\mathrm{i})$, and ii), we see $r(r(a)+r(b)) \subseteq r(r(a+b))=r(a+b)$. Conversely, by part $i)$, we have $a \subseteq r(a)$ and $\mathfrak{b} \subseteq r(b)$, so adding, $\mathfrak{a}+\mathbf{b} \subseteq r(\mathfrak{a})+r(b)$. By part $-\mathrm{i}), r(\mathfrak{a}+\mathbf{b}) \subseteq r(r(a)+r(b))$
vi) If $\mathfrak{p}$ is prime, $r\left(\mathfrak{p}^n\right)=\mathfrak{p}$ for all $n>0$
By part 0$), r\left(\mathfrak{p}^n\right)=r(\mathfrak{p})$. By $(1.14), r(\mathfrak{p})=\mathfrak{p}$

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