$I_1\cap\cdots\cap I_r=I_1I_2\cdots I_r$

Matsumura, Commutative Algebra, page 3

Let $A$ be a ring, and $I_1,\cdots,I_r$ be ideals of $A$ such that $I_i +I_j=A(i\neq j)$. Then $I_1\cap\cdots\cap I_r=I_1I_2\cdots I_r$.

显然,$I_1\cap\cdots\cap I_r\supset I_1I_2\cdots I_r$.

$\because I_1 +I_2=A$, $\therefore$ 存在$x\in I_1,y\in I_2$使得$x+ y=1$,

$\forall a\in I_1\cap I_2$, $a=a(x +y)=ax+ ay\in I_1I_2$.
$\therefore I_1\cap I_2\subset I_1I_2$.
$\therefore I_1\cap I_2= I_1I_2$.
$\because I_3+ I_1=A, I_3 +I_2=A$, $\therefore$存在$c\in I_3,d\in I_1,e\in I_3,f\in I_2$使得$c+ d=1,e+ f=1$,
$\therefore (c+d)(e +f)=ce+ cf+ de +df=1$, 并且$ce+ cf+ de\in I_3, df\in I_1\cap I_2$, $\therefore (I_1\cap I_2) +I_3=A$.
$\therefore I_1\cap I_2\cap I_3=(I_1\cap I_2)\cap I_3=(I_1\cap I_2) I_3=I_1I_2I_3$.

由归纳知,$I_1\cap\cdots\cap I_r=I_1I_2\cdots I_r$.

交换代数很好学,代数几何不易学。




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