# 数学代写|数论作业代写number theory代考|Noetherian Domains

#### Doug I. Jones

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## 数学代写|数论作业代写number theory代考|Noetherian Domains

Let $I_1$ be a nonzero ideal in the domain $\mathbb{Z}$. We consider ideals $I$ such that $I_1 \subseteq I$. As $\mathbb{Z}$ is a principal ideal domain (Theorem 1.4.1), there are nonzero integers $m$ and $n$ such that $I_1=\langle m\rangle, I=\langle n\rangle$, and $n \mid m$. Now $m$ has only finitely many divisors $n$ so there exist only finitely many ideals $I$ with $I_1 \subseteq I$. Thus there cannot exist infinitely many ideals $I_k(k=2,3, \ldots)$ such that
$$I_1 \subset I_2 \subset I_3 \subset I_4 \subset \cdots$$
The importance of domains such as $\mathbb{Z}$ that do not contain infinite ascending chains of ideals of the type (3.1.1) was first recognized by the German mathematician Emmy Noether (1882-1935). Such domains are now called Noetherian domains in her honor. We note that some domains do contain infinite chains of ideals of the type (3.1.1). For example, if $F$ is a field, the domain $F\left[X_1, X_2, \ldots\right]$ contains the infinite chain of ideals
$$\left\langle X_1\right\rangle \subset\left\langle X_1, X_2\right\rangle \subset\left\langle X_1, X_2, X_3\right\rangle \subset \ldots$$

## 数学代写|数论作业代写number theory代考|Factorization Domains

Let $D$ be an integral domain that is not a field so that $D$ contains nonzero, nonunit elements. It may be the case that all of these elements are reducible so that $D$ contains no irreducibles. The next example illustrates this.

Example 3.2.1 Let $D$ be the domain of polynomials in positive rational powers of $x$ over $\mathbb{C}$, that is,
\begin{aligned} & D=\left{a_1 x^{r_1}+\cdots+a_n x^{r_n} \mid n \in \mathbb{N}, a_1, \ldots, a_n \in \mathbb{C}, r_1, \ldots, r_n \in \mathbb{Q},\right. \ & \left.0 \leq r_1<\cdots<r_n\right} \text {. } \ & \end{aligned}
Clearly $U(D)=\mathbb{C}^*$. We show that $D$ does not possesss any irreducible elements.
Suppose that
$$f(x)=a_1 x^{r_1}+\cdots+a_n x^{r_n}$$
is an irreducible element of $D$. As $f(x)$ is a nonzero element of $D$, we may suppose that $a_n \neq 0$. If $n=1$ and $r_1=0$ then $f(x)=a_1 \neq 0$ is a unit of $D$, a contradiction. If $n=1$ and $r_1>0$ then $f(x)=a_1 x^{r_1}=a_1\left(x^{r_1 / 2}\right)^2$ is reducible in $D, a$ contradiction. Hence $n \geq 2$ and $r_n>0$. Let
\begin{aligned} t= & \text { least common multiple of the (positive) denominators of the } \ & \text { rationals } r_1, \ldots, r_n \end{aligned}
so that $t$ is a positive integer such that $r_1 t, \ldots, r_n$ t are integers with
$$0 \leq r_1 t<r_2 t<\cdots<r_n t$$
Then
$$f\left(x^t\right)=a_1 x^{r_1 t}+\cdots+a_n x^{r_n t} \in \mathbb{C}[x]$$
Hence there exist $b_1, \ldots, b_{r_n t} \in \mathbb{C}$ such that
$$a_1 x^{r_1 t}+\cdots+a_n x^{r_n t}=a_n\left(x-b_1\right) \cdots\left(x-b_{r_n t}\right)$$
Thus
$$f(x)=a_n\left(x^{1 / t}-b_1\right) \cdots\left(x^{1 / t}-b_{r_n t}\right)$$
Since
$$x^{1 / t}-b_1=\left(x^{1 / 2 t}-b_1^{1 / 2}\right)\left(x^{1 / 2 t}+b_1^{1 / 2}\right),$$
where $x^{1 / 2 t} \pm b_1^{1 / 2}$ are nonzero, nonunit elements of $D, f(x)$ is reducible, a contradiction.

# 数论作业代写

## 数学代写|数论作业代写number theory代考|Noetherian Domains

$$I_1 \subset I_2 \subset I_3 \subset I_4 \subset \cdots$$

$$\left\langle X_1\right\rangle \subset\left\langle X_1, X_2\right\rangle \subset\left\langle X_1, X_2, X_3\right\rangle \subset \ldots$$

## 数学代写|数论作业代写number theory代考|Factorization Domains

\begin{aligned} & D=\left{a_1 x^{r_1}+\cdots+a_n x^{r_n} \mid n \in \mathbb{N}, a_1, \ldots, a_n \in \mathbb{C}, r_1, \ldots, r_n \in \mathbb{Q},\right. \ & \left.0 \leq r_1<\cdots0，那么f(x)=a_1 x^{r_1}=a_1\left(x^{r_1 / 2}\right)^2在D, a的矛盾中是可约的。因此有n \geq 2和r_n>0。让
\begin{aligned}
t= & \text { least common multiple of the (positive) denominators of the } \
& \text { rationals } r_1, \ldots, r_n
\end{aligned}
$$所以t是一个正整数使得r_1 t, \ldots, r_n t是整数$$
0 \leq r_1 t<r_2 t<\cdots<r_n t
$$然后$$
f\left(x^t\right)=a_1 x^{r_1 t}+\cdots+a_n x^{r_n t} \in \mathbb{C}[x]
$$因此存在b_1, \ldots, b_{r_n t} \in \mathbb{C}这样的$$
a_1 x^{r_1 t}+\cdots+a_n x^{r_n t}=a_n\left(x-b_1\right) \cdots\left(x-b_{r_n t}\right)
$$因此$$
f(x)=a_n\left(x^{1 / t}-b_1\right) \cdots\left(x^{1 / t}-b_{r_n t}\right)
$$自从$$
x^{1 / t}-b_1=\left(x^{1 / 2 t}-b_1^{1 / 2}\right)\left(x^{1 / 2 t}+b_1^{1 / 2}\right),


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