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# 数学代写|数论代写Number Theory代考|Unique factorization in Euclidean and principal ideal domains

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## 数学代写|数论代写Number Theory代考|Unique factorization in Euclidean and principal ideal domains

Our proofs of the unique factorization property in both $\mathbb{Z}$ and $F[\mathrm{x}]$ hinged on the division with remainder property for these rings. This notion can be generalized, as follows.

Definition 17.28. $D$ is said to be a Euclidean domain if there is a “size function” $S$ mapping the non-zero elements of $D$ to the set of non-negative integers, such that for $a, b \in D$ with $b \neq 0$, there exist $q, r \in D$, with the property that $a=b q+r$ and either $r=0$ or $S(r)<S(b)$.

Example 17.13. Both $\mathbb{Z}$ and $F[\mathrm{x}]$ are Euclidean domains. In $\mathbb{Z}$, we can take the ordinary absolute value function $|\cdot|$ as a size function, and for $F[\mathbf{X}]$, the function $\operatorname{deg}(\cdot)$ will do.
Example 17.14. Recall again the ring
$$\mathbb{Z}[i]={a+b i: a, b \in \mathbb{Z}}$$

of Gaussian integers from Example 9.22. Let us show that this is a Euclidean domain, using the usual norm map $N$ on complex numbers (see Example 9.5) for the size function. Let $\alpha, \beta \in \mathbb{Z}[i]$, with $\beta \neq 0$. We want to show the existence of $\xi, \rho \in \mathbb{Z}[i]$ such that $\alpha=\beta \xi+\rho$, where $N(\rho)<N(\beta)$. Suppose that in the field $\mathbb{C}$, we compute $\alpha \beta^{-1}=r+s i$, where $r, s \in \mathbb{Q}$. Let $m, n$ be integers such that $|m-r| \leq 1 / 2$ and $|n-s| \leq 1 / 2$ – such integers $m$ and $n$ always exist, but may not be uniquely determined. Set $\xi:=m+n i \in \mathbb{Z}[i]$ and $\rho:=\alpha-\beta \xi$. Then we have
$$\alpha \beta^{-1}=\xi+\delta$$
where $\delta \in \mathbb{C}$ with $N(\delta) \leq 1 / 4+1 / 4=1 / 2$, and
$$\rho=\alpha-\beta \xi=\alpha-\beta\left(\alpha \beta^{-1}-\delta\right)=\delta \beta,$$
and hence
$$N(\rho)=N(\delta \beta)=N(\delta) N(\beta) \leq \frac{1}{2} N(\beta) .$$

## 数学代写|数论代写Number Theory代考|Unique factorization in D[X]

In this section, we prove the following:
Theorem 17.36. If $D$ is a UFD, then so is $D[\mathrm{X}]$.
This theorem implies, for example, that $\mathbb{Z}[\mathrm{X}]$ is a UFD. Applying the theorem inductively, one also sees that for any field $F$, the ring $F\left[\mathrm{X}_1, \ldots, \mathrm{X}_n\right]$ of multi-variate polynomials over $F$ is also a UFD.

We begin with some simple observations. First, recall that for an integral domain $D, D[\mathrm{X}]$ is an integral domain, and the units in $D[\mathrm{X}]$ are precisely the units in $D$. Second, it is easy to see that an element of $D$ is irreducible in $D$ if and only if it is irreducible in $D[\mathrm{X}]$. Third, for $c \in D$ and $f=\sum_i a_i \mathrm{X}^i \in D[\mathrm{X}]$, we have $c \mid f$ if and only if $c \mid a_i$ for all $i$.

We call a non-zero polynomial $f \in D[\mathrm{X}]$ primitive if the only elements in $D$ that divide $f$ are units. If $D$ is a UFD, then given any non-zero polynomial $f \in D[\mathrm{X}]$, we can write it as $f=c f^{\prime}$, where $c \in D$ and $f^{\prime} \in D[\mathrm{X}]$ is a primitive polynomial: just take $c$ to be a greatest common divisor of all the coefficients of $f$.
It is easy to prove the existence part of Theorem 17.36 :

Theorem 17.37. Let $D$ be a UFD. Any non-zero, non-unit element of $D[\mathrm{x}]$ can be expressed as a product of irreducibles in $D[\mathrm{x}]$.

Proof. Let $f$ be a non-zero, non-unit polynomial in $D[\mathrm{x}]$. If $f$ is a constant, then because $D$ is a UFD, it factors into irreducibles in $D$. So assume $f$ is not constant. If $f$ is not primitive, we can write $f=c f^{\prime}$, where $c$ is a non-zero, non-unit in $D$, and $f^{\prime}$ is a primitive, non-constant polynomial in $D[\mathrm{x}]$. Again, as $D$ is a UFD, $c$ factors into irreducibles in $D$.

## 数学代写|数论代写Number Theory代考|Unique factorization in Euclidean and principal ideal domains

$$\mathbb{Z}[i]=a+b i: a, b \in \mathbb{Z}$$

$N(\rho)<N(\beta)$. 假设在野外 $\mathbb{C}$, 我们计算 $\alpha \beta^{-1}=r+s i$ ，在哪里 $r, s \in \mathbb{Q}$. 让 $m, n$ 是这样的整数 $|m-r| \leq 1 / 2$ 和 $|n-s| \leq 1 / 2-$ 这样的整数 $m$ 和 $n$ 永远存在，但可能不是唯一确定的。放 $\xi:=m+n i \in \mathbb{Z}[i]$ 和 $\rho:=\alpha-\beta \xi$. 然后我们有
$$\alpha \beta^{-1}=\xi+\delta$$

$$\rho=\alpha-\beta \xi=\alpha-\beta\left(\alpha \beta^{-1}-\delta\right)=\delta \beta$$

$$N(\rho)=N(\delta \beta)=N(\delta) N(\beta) \leq \frac{1}{2} N(\beta) .$$

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