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# 数学代写|代数数论代写Algebraic Number Theory代考|MAS6220 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
$$\left(\begin{array}{l|l} \operatorname{tr}{B_1 / A}\left(\alpha_i \alpha_j\right) & \ \hline & \operatorname{tr}{B_2 / A}\left(\beta_i \beta_j\right) \end{array}\right)$$
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 $A / \mathfrak{a}$ 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 $\operatorname{nil}(A)$ of all nilpotent elements of $A$ is an ideal of A.

The ideal $\operatorname{nil}(A)$ 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 $\mathfrak{p}$ 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 $x \notin \mathfrak{p}$. 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 $\mathfrak{p}$ 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 $\mathfrak{a}=(\mathfrak{p}, x)$ and $\mathfrak{b}=(\mathfrak{p}, y)$ 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 $x^{m+n} \in \mathfrak{a} \mathfrak{b} \subseteq \mathfrak{p}$, implying $\mathfrak{p} \notin S$. This contradiction proves that $\mathfrak{p}$ is prime.

## 数学代写|代数数论代写代数数论代考|环的直积

$$\mathfrak{d}{B / A}=\mathfrak{d}{B_1 / A} \cdots \mathfrak{d}_{B_r / A}$$

Put $n_1=m$ 和 $n_2=n$。让 $\alpha_1, \ldots, \alpha_m$ 成为…的基础 $B_1$ 结束 $A$ 和 $\beta_1, \ldots, \beta_n$ 成为…的基础 $B_2$ 结束 $A$。As $A$-模块，如果我们识别 $B_1$ 和 $B_2$ 通过子模块 $B_1 \times{0}$ 和 ${0} \times B_2$ 的 $B=B_1 \times B_2$，那么 $\left{\alpha_1, \ldots, \alpha_m ; \beta_1, \ldots, \beta_n\right}$ 是一个基础 $B$ 结束 $A$。此外，对于所有人来说 $i, j$，我们有 $\alpha_i \beta_j=0$。因此 $\Delta\left(\alpha_1, \ldots, \alpha_m ; \beta_1, \ldots, \beta_n\right)$ 矩阵的行列式是否
$$\left(\begin{array}{l|l} \operatorname{tr}{B_1 / A}\left(\alpha_i \alpha_j\right) & \ \hline & \operatorname{tr}{B_2 / A}\left(\beta_i \beta_j\right) \end{array}\right)$$

$$\Delta\left(\alpha_1, \ldots, \alpha_m ; \beta_1, \ldots, \beta_n\right)=\Delta\left(\alpha_1, \ldots, \alpha_m\right) \Delta\left(\beta_1, \ldots, \beta_n\right) .$$
$\mathfrak{d}{B / A}=\mathfrak{d}{B_1 / A} \mathfrak{d}_{B_2 / A}$

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