### 数学代写|代数数论代写Algebraic number theory代考|Direct Product of Rings

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## 数学代写|代数数论代写Algebraic number theory代考|Direct Product of Rings

Suppose $B_{1}, \ldots, B_{r}$ are commutative rings with 1 . We define their direct product as the Cartesian product $B=B_{1} \times \cdots \times B_{r}$, with addition and multiplication taken component-wise. Each $B_{j}$ may be regarded as a subring of $B$ via the obvious inclusion map, e.g. $B_{1} \ni b_{1} \rightarrow\left(b_{1}, 0, \ldots, 0\right) \in B$.

If $A$ is a subring of each $B_{j}$, then $A$ may be regarded as a subring of the direct product $B=B_{1} \times \cdots \times B_{r}$, via the map $A \ni a \rightarrow(a, \ldots, a) \in B$.

Theorem 4.25. Suppose $A$ with 1 is a subring of each $B_{j}$ and every $B_{j}$ is a free A-module of rank $n_{j}$. Then the direct product $B=B_{1} \times \cdots \times B_{r}$ is a free module of rank $n_{1}+\cdots+n_{r}$. Moreover
$$\mathfrak{d}{B / A}=\mathfrak{d}{B_{1} / A} \cdots \mathfrak{d}{B{r} / A}$$
Proof. We only need to prove (4.6). To simplify notation, we prove it for $r=2$. For $r>2$, the proof is similar.

Put $n_{1}=m$ and $n_{2}=n$. Let $\alpha_{1}, \ldots, \alpha_{m}$ be a basis of $B_{1}$ over $A$ and $\beta_{1}, \ldots, \beta_{n}$ be a basis of $B_{2}$ over $A$. As $A$-modules, if we identify $B_{1}$ and $B_{2}$ with the submodules $B_{1} \times{0}$ and ${0} \times B_{2}$ of $B=B_{1} \times B_{2}$, then $\left{\alpha_{1}, \ldots, \alpha_{m} ; \beta_{1}, \ldots, \beta_{n}\right}$ is a basis of $B$ over $A$. Moreover, for all $i, j$, we have $\alpha_{i} \beta_{j}=0$. Hence $\Delta\left(\alpha_{1}, \ldots, \alpha_{m} ; \beta_{1}, \ldots, \beta_{n}\right)$ is the determinant of the matrix
$$\left(\begin{array}{c|l} \operatorname{tr}{B{1} / A}\left(\alpha_{i} \alpha_{j}\right) & \ \hline & \operatorname{tr}{B{2} / A}\left(\beta_{i} \beta_{j}\right) \end{array}\right)$$
This shows that
$$\Delta\left(\alpha_{1}, \ldots, \alpha_{m} ; \beta_{1}, \ldots, \beta_{n}\right)=\Delta\left(\alpha_{1}, \ldots, \alpha_{m}\right) \Delta\left(\beta_{1}, \ldots, \beta_{n}\right)$$
Therefore, $\mathfrak{d}{B / A}=\mathfrak{d}{B_{1} / A} \mathfrak{d}{B{2} / A}$.
Suppose $A$ is a subring of $B$. Let $\mathfrak{a}$ be an ideal of $A$ and $\mathfrak{b}=\mathfrak{a} B$ be the ideal of $B$ generated by a. For $\alpha$ in $A$ and $\beta$ in $B$, let $\bar{\alpha}$ and $\bar{\beta}$ denote the residue class of $\alpha$ in $A / a$ and that of $\beta$ in $B / \mathfrak{b}$, respectively.

Definition 4.30. An element of a commutative ring $A$ with 1 is nilpotent if $a^{m}=0$ for some $m$ in $\mathbb{Z}$.

Theorem 4.31. The set nil $(A)$ of all nilpotent elements of $A$ is an ideal of A.

The ideal nil $(A)$ is called the nilradical of $A$.
Proof. Let $x, y \in \operatorname{nil}(A)$. Then for some $m, n$ in $\mathbb{N}, x^{m}=y^{n}=0$. If $l=m+n$, then it follows from the Binomial Theorem, that $(x+y)^{l}=0$. On the other hand, if $a \in A$, then $(a x)^{m}=a^{m} x^{m}=0$. This proves that nil $(A)$ is an ideal of $A$.

Theorem 4.32. The nilradical, $\operatorname{nil}(A)$, is the intersection of all prime ideals of $A$.

Proof. If $x$ in $A$ is nilpotent, then for some $m$ in $\mathbb{N}, x^{m}=0$. Hence $x \in \mathfrak{p}$, for all prime ideals $p$ of $A$.

Conversely, suppose $x$ is not nilpotent, that is $x^{m} \neq 0$ for all $m$ in $\mathbb{N}$. We show that there is at least one prime ideal $\mathfrak{p}$ such that $x \notin p$. Let $S$ be the set of ideals a of $A$, such that $x^{m} \notin \mathfrak{a}$ for all $m$ in $\mathbb{N}$. Clearly, $S$ is not empty, since the zero ideal $(0) \in S$. By Zorn’s Lemma, let $p$ be a maximal element of $S$. We shall show that $\mathfrak{p}$ is prime. If not, then there are $x, y$ in $A \backslash \mathfrak{p}$ with $x y$ in $\mathfrak{p}$. Then the ideals $\boldsymbol{a}=(\mathfrak{p}, x)$ and $\mathbf{b}=(\mathfrak{p}, y)$ both properly contain $\mathfrak{p}$. By the choice of $\mathfrak{p}$, for some $m, n$ in $\mathbb{N}, x^{m} \in \mathfrak{a}, x^{n} \in \mathfrak{b}$. This shows that $x^{m+n} \in \mathfrak{a b} \subseteq \mathfrak{p}$, implying $\mathfrak{p} \notin S$. This contradiction proves that $\mathfrak{p}$ is prime.

## 数学代写|代数数论代写Algebraic number theory代考|Reduced Rings

Definition 4.33. A commutative ring $A$ with 1 is reduced if $\operatorname{nil}(A)=(0)$.
Example 4.34.

1. An integral domain is reduced.
2. The product $A_{1} \times \ldots \times A_{r}$ is reduced if all $A_{j}$ are reduced.
Theorem 4.35. Suppose $K$ is a number field and $\mathfrak{P}$ a prime ideal of $\mathcal{O}=\mathcal{O}_{K}$. The quotient ring $\mathcal{O} / \mathfrak{P}^{e}$ is reduced if and only if $e=1$.

Proof. If $e=1$, then $\mathcal{O} / \mathfrak{F}$ is a field, hence reduced. On the other hand, if $e>0$, choose $\pi$ in $\mathfrak{P}-\mathfrak{P}^{2}$. Then $\pi \neq 0$ in $\mathcal{O} / \mathfrak{P}^{e}$, but $\pi^{e}=0$ in $\mathcal{O} / \mathfrak{P}^{e}$. Therefore, $\mathcal{O} / \mathfrak{P}^{e}$ is not reduced.

Now let $A$ be a subring of a commutative ring $B$, both with 1 . Suppose $B$ is a free $A$-module of rank $n$.

Theorem 4.36. If $A$ is a finite field, then $B$ is reduced if and only if $\mathfrak{D}_{B / A} \neq$ $(0)$.

Proof. First suppose that $B$ is not reduced, that is, it has a nilpotent element $\alpha \neq 0$. $A$ being a field, $\alpha$ can be completed into a basis $\alpha_{1}=\alpha, \alpha_{2}, \ldots, \alpha_{n}$ of the vector space $B$ over $A$. Now all the elements $\alpha_{1} \alpha_{j}, j=1, \ldots, n$ are also nilpotent and since the matrix for a nilpotent element is also nilpotent, its trace is zero (why?). Hence the first row of the matrix $\left(\operatorname{tr}\left(\alpha_{1} \alpha_{j}\right)\right)$ consists of zeros only, which shows that
$$\Delta\left(\alpha_{1}, \ldots, \alpha_{n}\right)=\operatorname{det}\left(\operatorname{tr}\left(\alpha_{i} \alpha_{j}\right)\right)=0$$
Therefore, $\mathfrak{o}{B / A}=(0)$. Conversely, suppose $B$ is reduced, i.e. $\operatorname{nil}(B)={0}$. Since $\operatorname{nil}(B)$ is the intersection of all prime ideals and $B$ is finite, $$(0)=\mathfrak{F}{1} \cap \ldots \cap \mathfrak{P}{r}\left(\mathfrak{P}{i} \neq \mathfrak{P}{j} \text { for } i \neq j\right) .$$ Every $B / \mathfrak{P}{j}$, being a finite integral domain, is a field, hence all $\mathfrak{P}{j}$ are maximal and therefore coprime in pairs, and $$\mathfrak{F}{1} \cap \ldots \cap \mathfrak{P}{r}=\mathfrak{P}{1} \ldots \mathfrak{P}{r}$$ By the Corollary $4.28$, $$B=B /(0)=B / \mathfrak{P}{1} \ldots \mathfrak{P}{r} \cong B / \mathfrak{P}{1} \times \ldots \times B / \mathfrak{P}{r} .$$ By Theorem 4.25, $$\mathfrak{o}{B / A}=\mathfrak{o}{\left(B / \mathfrak{F}{1}\right) / A} \cdots \mathfrak{o}{\left(B / \mathfrak{F}{\mathrm{r}}\right) / A} .$$
Since $A$ is a field, each $\mathfrak{o}{\left(B / \mathfrak{P}{j}\right) / A} \neq(0)$. Hence $\mathfrak{d}_{B / A} \neq(0)$.

## 数学代写|代数数论代写Algebraic number theory代考|Direct Product of Rings

d乙/一个=d乙1/一个⋯d乙r/一个

\left(\begin{array}{c|l} \operatorname{tr}{B{1} / A}\left(\alpha_{i} \alpha_{j}\right) & \ \hline & \operatorname{ tr}{B{2} / A}\left(\beta_{i} \beta_{j}\right) \end{array}\right)\left(\begin{array}{c|l} \operatorname{tr}{B{1} / A}\left(\alpha_{i} \alpha_{j}\right) & \ \hline & \operatorname{ tr}{B{2} / A}\left(\beta_{i} \beta_{j}\right) \end{array}\right)

Δ(一个1,…,一个米;b1,…,bn)=Δ(一个1,…,一个米)Δ(b1,…,bn)

## 数学代写|代数数论代写Algebraic number theory代考|Reduced Rings

1. 减少了一个积分域。
2. 产品一个1×…×一个r如果全部减少一个j被减少。
定理 4.35。认为ķ是一个数字字段并且磷的首要理想○=○ķ. 商圈○/磷和当且仅当和=1.

Δ(一个1,…,一个n)=这⁡(tr⁡(一个一世一个j))=0

(0)=F1∩…∩磷r(磷一世≠磷j 为了 一世≠j).每一个乙/磷j，作为有限积分域，是一个域，因此所有磷j是最大的，因此成对互质，并且

F1∩…∩磷r=磷1…磷r由推论4.28,

○乙/一个=○(乙/F1)/一个⋯○(乙/Fr)/一个.

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