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[数论] 最大公约数乘以最小公倍数

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hbghlyj posted 2025-8-11 17:03 |Read mode
当逐元素写证明时(使用 gcd⁡、lcm 和 Bézout 等式)等式都需要乘某个unit
适宜用ideal来写在一个PID中$\langle\gcd(a,b)\rangle=\langle a\rangle+\langle b\rangle,\langle\operatorname{lcm}(a,b)\rangle=\langle a\rangle\cap\langle b\rangle$.
设$I,J$为一个PID中任意的ideal,则Ideal quotient
$$
(IJ:I∩J)=I+J
$$$$
(IJ:I+J)=I∩J
$$
$$
IJ= (I+J)(I∩J)
$$

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original poster hbghlyj posted 2025-8-11 07:54
一个包含:(⟨a⟩+⟨b⟩)(⟨a⟩∩⟨b⟩)⊆⟨a⟩⟨b⟩

设 $I=\langle a\rangle,J=\langle b\rangle,⟨d⟩ = ⟨a⟩ + ⟨b⟩$,则 $∃u, v$ 使 $d = ua + vb$.
取任意 $y\in \langle a\rangle\cap\langle b\rangle$ 则 $∃\alpha,\beta$ 使 $y=a\alpha=b\beta$.
$$
dy=(ua+vb)y
=uay+vby
=ua(b\beta)+vb(a\alpha)
=ab(u\beta+v\alpha)\in\langle ab\rangle
$$
对于每个 $y\in\langle a\rangle\cap\langle b\rangle$ 都成立,我们得到
$$
(\langle a\rangle+\langle b\rangle)(\langle a\rangle\cap\langle b\rangle)
=\langle d\rangle(\langle a\rangle\cap\langle b\rangle)\subseteq \langle ab\rangle
$$
另一个包含:⟨a⟩⟨b⟩⊆(⟨a⟩+⟨b⟩)(⟨a⟩∩⟨b⟩)

因为 $d\mid a$ 且 $d\mid b$,则 $∃a_1,b_1$ 使 $a=da_1,b=db_1$,于是
$$
\frac{ab}{d}=a b_1 = b a_1 \in \langle a\rangle\cap\langle b\rangle.
$$
因此 $ab = d\cdot \frac{ab}{d}\in \langle d\rangle(\langle a\rangle\cap\langle b\rangle) =(\langle a\rangle+\langle b\rangle)(\langle a\rangle\cap\langle b\rangle)$,这表明
$$\langle ab\rangle\subseteq
(\langle a\rangle+\langle b\rangle)(\langle a\rangle\cap\langle b\rangle).
$$

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original poster hbghlyj posted 2025-8-11 17:02

一般地,交换环中$(I+J)(I\cap J)\subseteq IJ$成立,而另一个包含不成立

上面的 PID 是 Dedekind domain 的特例。在那个世界之外等式不成立:在 $R=k[x,y]$ 中取 $I=⟨x⟩,J=⟨x^2,y⟩$,则
$$
I+J=⟨x,y⟩,\quad I\cap J=⟨x⟩\cap⟨x^2,y⟩=⟨x^2,xy⟩=x⟨x,y⟩.
$$
因此$$
(I+J)(I\cap J)=⟨x,y⟩\cdot x⟨x,y⟩=⟨x^3,x^2y,xy^2⟩,
$$对比$$
IJ=⟨x⟩\cdot⟨x^2,y⟩=⟨x^3,xy⟩.
$$
注意到 $xy\notin⟨x^3,x^2y,xy^2⟩$,所以 $(I+J)(I\cap J)\subsetneq IJ$.

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original poster hbghlyj posted 2025-8-11 07:59

何时等式成立

  • Comaximal case (any ring). If $I+J=R$, then $I\cap J=IJ$, so the identity is trivially true.
  • Prüfer domains (in particular Dedekind domains and PIDs).
    In a Prüfer domain every finitely generated ideal is invertible. For invertible ideals one has
    $$
    (I+J)(I\cap J)=IJ \quad\text{(for finitely generated } I,J).
    $$
    A slick proof: localize at any maximal ideal $\mathfrak m$. The localization of a Prüfer domain is a valuation domain, where finitely generated ideals are linearly ordered. Thus, in $R_{\mathfrak m}$ either $I_{\mathfrak m}\subseteq J_{\mathfrak m}$ or $J_{\mathfrak m}\subseteq I_{\mathfrak m}$. Say $I_{\mathfrak m}\subseteq J_{\mathfrak m}$. Then
    $$
    (I+J)_{\mathfrak m}=J_{\mathfrak m},\qquad (I\cap J)_{\mathfrak m}=I_{\mathfrak m},
    $$
    so
    $$
    \big((I+J)(I\cap J)\big)_{\mathfrak m}
    =J_{\mathfrak m}I_{\mathfrak m}
    =I_{\mathfrak m}J_{\mathfrak m}
    =(IJ)_{\mathfrak m}.
    $$
    Since this holds for every $\mathfrak m$, the ideals are equal globally:
    $(I+J)(I\cap J)=IJ$.
  • Dedekind domains (special case of 2).
    All nonzero ideals are invertible, so the identity holds for all nonzero $I,J$.

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original poster hbghlyj posted 2025-8-11 17:06
In a PID, two ideal-quotient equalities $(K:I)=J$ and $(K:J)=I$ force $K=IJ$.

Proof: write $I=⟨a⟩, J=⟨b⟩, K=⟨c⟩$. In a PID,
$$
\big(⟨x⟩:⟨y⟩\big)=\{r: ry\in⟨x⟩\}=⟨x/\gcd(x,y)⟩.
$$
Our hypotheses
$$
(K:I)=J,\qquad (K:J)=I
$$
become
\begin{align*}
(c:a)&=⟨b⟩\Rightarrow⟨c/\gcd(c,a)⟩=⟨b⟩,\\
(c:b)&=⟨a⟩\Rightarrow⟨c/\gcd(c,b)⟩=⟨a⟩.
\end{align*}
Thus, up to a unit,
$$
c=b\gcd(c,a)=a\gcd(c,b).
$$
From $c=b\gcd(c,a)$ we get $\gcd(c,b)=\gcd(b\gcd(c,a),b)=b$. Plugging into the second display gives $c\sim ab$. Hence $⟨c⟩=⟨ab⟩$, i.e.
$$
K=IJ.
$$
(Aside: the same implication holds in any ring where $I$ and $J$ are invertible ideals, e.g. Dedekind domains: from $(K:I)=J$ multiply by $I$ to get $K=IJ$. In a PID every nonzero ideal is invertible.)

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original poster hbghlyj posted 2025-8-11 17:01

One equality is not enough in general

Assume $(K:I)=J$ with $I=⟨a⟩,K=⟨c⟩,J=⟨b⟩$. Using what we just proved,
\[
(c:a)=⟨c/g⟩=⟨b⟩\quad\Rightarrow\quad b \sim c/g,\ \text{so }K=⟨c⟩=⟨bg⟩.
\]
Question: does this force $K=IJ=⟨ab⟩$? In $R=\mathbb Z$,$$(⟨6⟩:⟨4⟩)=⟨3⟩$$but $⟨6⟩\ne⟨4⟩⟨3⟩$.

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