被本原多项式整除的整系数多项式所得的商是整系数多项式

abababa

设$f(x)$是本原多项式，$g(x)$是整系数多项式，若$f(x)\mid g(x)$，则$\frac{g(x)}{f(x)}$是整系数多项式。

这里的本原多项式定义为：一个整系数多项式的各项系数互素，则称为本原多项式。
这个证明如下，有一步我没有看懂。先证明一个命题：设$f(x),g(x)$是整系数多项式，若素数$p$能整除$f(x)g(x)$的所有系数，则$p$或者能整除$f(x)$的所有系数，或者能整除$g(x)$的所有系数。
证明：假设$p$既不能整除$f(x)$的所有系数，也不能整除$g(x)$的所有系数，设$f(x)=a_mx^m+\cdots+a_0$中次数最低的、系数不能被$p$整除的单项式为$a_ix^i$，设$g(x)=b_nx^n+\cdots+b_0$中次数最低的、系数不能被$p$整除的单项式为$b_jx^j$。考察$f(x)g(x)$中$x^{m-i+n-j}$的系数$\sum_{s+t=m-i+n-j}a_sb_t$，由$i,j$的选法知，当$s < i$时$p \mid a_s$，当$t < j$时$p \mid b_j$。由于$p$能整除$f(x)g(x)$的所有系数，因此必能整除$\sum_{s+t=m-i+n-j}a_sb_t$，而$\sum_{s+t=m-i+n-j}a_sb_t$的和式中除$a_ib_j$外，都满足$s < i$或$t < j$，这样的$a_sb_t$都能被$p$整除，因此$p\mid a_ib_j$，但$p$是素数，因此$p\mid a_i$或$p\mid b_j$，这与$p$既不能整除$f(x)$的所有系数，也不能整除$g(x)$的所有系数矛盾。因此命题成立。

然后证明原命题：由于$f(x) \mid g(x)$，因此$g(x) = f(x)h(x)$，由于$f(x)$和$g(x)$都是有理系数多项式，因此$h(x)$也必是有理系数多项式（假设不是，可以设$h(x)$中次数最低的、非有理系数的单项式为$b_ix^i$，相乘后考察$x^i$项的系数即得矛盾），于是存在整数$c$使得$k(x)=ch(x)$是整系数多项式，此时$cg(x)=f(x)k(x)$，设$c = \prod_{i}c_i$，其中$c_i$都是素数，由于$c_i$能整除$cg(x)$的所有系数，因此$c_i$能整除$f(x)k(x)$的所有系数，由前一命题知$c_i$或者整除$f(x)$的所有系数，或者整除$k(x)$的所有系数，然而$f(x)$是本原多项式，因此$c_i$不能整除$f(x)$的所有系数，因此必有$c_i$整除$k(x)$的所有系数，从而$c = \prod_{i}c_i$能整除$k(x)$的所有系数，由于$k(x)$是整系数多项式，因此$\frac{k(x)}{c}$也是整系数多项式，即$h(x)$是整系数多项式。

就是我加红色的地方，那个$c$在分解素因数时，不是可以分解出$c=c_1^{r_1}c_2^{r_2}\cdots$这样的吗？然后第二处加红色的地方，$c_i$能整除$k(x)$的所有系数没错，但无法保证乘起来还能整除吧，比如$c=2\cdot2\cdot3$，其中$c_1=c_2=2,c_3=3$，这时根据前面那个命题可以得到$c_1=2$能整除$k(x)$的所有系数，但得不出$c_1c_2=4$也能整除$k(x)$的所有系数吧。但我又觉得这个命题是对的，具体要怎么证明呢？最好是根据前面那个命题，用相对初等的方法证明。

hbghlyj

回复 1# abababa
任取c的素因子p，k(x)的系数均为p倍数
那么在等式cg=fk两边约去p，得到c1*g=f*k1，这里c1＜c
对c递降就完事了

abababa

回复 2# hbghlyj

原来如此，谢谢，我明白了，还需要把$k(x)$约掉单个的$c_i$得到一个$k'(x)$，之后$k'(x)$还是整系数多项式，之后继续用前一个命题才行。

hbghlyj

mathreference
When Gauss' Lemma Fails
令 $q=\sqrt{-5}$, $R=ℤ[q]$，$F$ 为 $R$ 的分式域。
$R$是一个 dedekind domain，但不是 ufd。
高斯引理表示 $R[x]$ 中的多项式在 $R[x]$ 与 $F[x]$ 中的分解相同，但前提是 $R$ 是一个 ufd。
令 $p(x) = 3x^2+4x+3$。$p = [3x+(2+q)] [x+(2-q)/3]$，因此 $p$ 在 $F$ 可约。假设它也在 $R$ 可约。这个二次方程分成两个单项式，因为 3 是不可约元，所以每个单项式的系数都是 ±1 或 ±3。只有几种组合，而且它们都不会产生 $p(x)$。在 $F[x]$ 可约并不意味着在 $R[x]$ 可约。

如果你想要一个不满足高斯引理的一元多项式，我们需要一个不整闭的环。设 $q = \sqrt{-3}$，$x^2+x+1=[x+(1+q)/2][x+(1-q)/2]$，但在 环 $ℤ[q]$ 上不可约。

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Gauss' lemma
For a concrete example one can take $R = \Bbb Z[i\sqrt5], p = 1 + i\sqrt5, a = 1 − i\sqrt5, q = 2, b = 3$. In this example the polynomial $3 + 2X + 2X^2$ (obtained by dividing the right hand side by $q = 2$) provides an example of the failure of the irreducibility statement (it is irreducible over $R$, but reducible over its field of fractions $\Bbb Q[i\sqrt5]$). Another well-known example is the polynomial $X^2 − X − 1$, whose roots are the golden ratio $φ = (1 + \sqrt5)/2$ and its conjugate $(1 − \sqrt5)/2$ showing that it is reducible over the field $\Bbb Q[\sqrt5]$, although it is irreducible over the non-UFD $ℤ[\sqrt5]$ which has $\Bbb Q[\sqrt5]$ as field of fractions. In the latter example the ring can be made into an UFD by taking its integral closure $\Bbb Z[φ]$ in $\Bbb Q[\sqrt5]$ (the ring of Dirichlet integers), over which $X^2 − X − 1$ becomes reducible, but in the former example $R$ is already integrally closed.

hbghlyj

我也看到了一个证明，发在下面，看起来比1#的更短更简单?
它也证明了“$fg$为本原多项式则$f$或$g$为本原多项式”, 但是没有用到乘积的展开式
有人能解释一下两个证明之间的关系吗?

hbghlyj

rings.pdf
Theorem 3.41 (Gauss' Lemma). Suppose that $f \in \mathbb{Z}[X]$. Then $f$ is non-constant and irreducible in $\mathbb{Z}[X]$ if and only if $f$ is primitive and irreducible in $\mathbb{Q}[X]$.

Proof. Suppose that $f$ is irreducible in $\mathbb{Z}[X]$. This immediately tells us that $f$ is primitive since if $p$ were a prime dividing all the coefficients of $f$ then $p \mid f$ in $\mathbb{Z}[X]$. Since $p \nsim 1$ we conclude that $p \sim f$ (in $\mathbb{Z}[X]$ ) by irreducibility of $f$, contradicting the fact that $f$ is non-constant.

Now, suppose that $f=g h$ for $g, h \in \mathbb{Q}[X]$. Then let $\lambda \in \mathbb{N}^*$ be minimal such that there is $q \in \mathbb{Q}^*$ with $\lambda q^{-1} g$ and $q h$ both in $\mathbb{Z}[X]$. Suppose that $p \in \mathbb{Z}$ is prime with $p \mid \lambda$. Then $p$ is prime as a constant polynomial in $\mathbb{Z}[X]$ and since $p \mid \lambda f=\left(\lambda q^{-1} g\right)(q h)$, we have $p \mid \lambda q^{-1} g$ or $p \mid q h$ (both in $\mathbb{Z}[X]$ ). The former contradicts minimality of $\lambda$ directly, and the latter once we note that $(q / p) h \in \mathbb{Z}[X]$ and $(\lambda / p)(q / p)^{-1} g=\lambda q^{-1} g \in \mathbb{Z}[X]$. We conclude that $\lambda$ has no prime factors and hence (since $\mathbb{Z}$ is a UFD) is a unit. Thus $q^{-1} g \mid f$ in $\mathbb{Z}[X]$ and so by irreducibility of $f$ in $\mathbb{Z}[X]$ we conclude that either $q^{-1} g \sim 1$ or $q^{-1} g \sim f$ in $\mathbb{Z}[X]$. Hence either $g \sim 1$ in $\mathbb{Q}[X]$ or $g \sim f$ in $\mathbb{Q}[X]$ and finally, since $f$ is non-constant we have $f \nsim 1$ in $\mathbb{Q}[X]$ and so $f$ is irreducible in $\mathbb{Q}[X]$.

Conversely, suppose $f \in \mathbb{Z}[X]$ is primitive and irreducible in $\mathbb{Q}[X]$. First, $f \nsim 1$ in $\mathbb{Q}[X]$ and so $f$ is non-constant. Suppose $g \mid f$ in $\mathbb{Z}[X]$. By irreducibility of $f$ in $\mathbb{Q}[X]$, either $g \sim 1$ in $\mathbb{Q}[X]$ so $\operatorname{deg} g=0$, and since $f$ is primitive $g \sim 1$ in $\mathbb{Z}[X]$; or $g \sim f$ in $\mathbb{Q}[X]$, then $\operatorname{deg} g=\operatorname{deg} f$ and writing $f=g h$ for $h \in \mathbb{Z}[X]$ we have $\operatorname{deg} h=0$, and since $f$ is primitive $h \sim 1$ in $\mathbb{Z}[X]$, whence $g \sim f$ in $\mathbb{Z}[X]$. The result is proved.

hbghlyj

LS07.pdf
Proposition 2. Given $f ∈\Bbb Z[t]$. Then $f$ is irreducible over $\Bbb Q$ if and only if $f$ is irreducible over $\Bbb Z$.
Proof of Proposition 2. We show that if $f$ is reducible over $\mathbb{Q}$, then it is also reducible over $\mathbb{Z}$. First off, we may as well assume $f$ is primitive, because if not we can divide through by the gcd of the coefficients of $f$, creating a primitive polynomial which is reducible iff the original was.

Now suppose $f=g h$ for some $g, h \in \mathbb{Q}[t]$. There exists some $\alpha, \beta \in \mathbb{Z}$ such that $\alpha g, \beta h \in \mathbb{Z}[t]$, and $\alpha g, \beta h$ are both primitive. Now observe that
$$
\alpha \beta f=(\alpha g)(\beta h) .
$$
This implies that $\alpha \beta f$ is the product of two primitive polynomials, hence must itself be primitive. But, since we also assumed $f$ to be primitive, we deduce that $\alpha, \beta=\pm 1$. Therefore, it must have been the case that $g, h \in \mathbb{Z}[t]$, and the theorem is proved.