Question Number 174684 by mnjuly1970 last updated on 08/Aug/22
$$ \\ $$$$\boldsymbol{\mathrm{Q}}:\:\:\boldsymbol{\mathrm{I}}\:,\:\boldsymbol{\mathrm{J}}\:\:\boldsymbol{{are}}\:\boldsymbol{{two}}\:\boldsymbol{{ideals}}\:\boldsymbol{{of}}\:\:\boldsymbol{{commutative}}\: \\ $$$$\:\:\:\:\boldsymbol{{ring}}\:,\:\left(\:\boldsymbol{\mathrm{R}}\:,\oplus,\: \:\right)\:.\boldsymbol{{prove}}\:\boldsymbol{{that}}\:: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\sqrt{\:\boldsymbol{\mathrm{I}}\:\cap\:\boldsymbol{\mathrm{J}}\:}\:\:\overset{?} {=}\:\sqrt{\:\boldsymbol{\mathrm{I}}\:}\:\:\cap\:\:\sqrt{\:\boldsymbol{\mathrm{J}}\:}\:\:\:\:\:\boldsymbol{{m}}.\boldsymbol{{n}} \\ $$$$\:\:\:\:\boldsymbol{{note}}\::\:\sqrt{\boldsymbol{\mathrm{I}}\:}\:=\:\left\{\:\boldsymbol{{x}}\:\in\:\boldsymbol{\mathrm{R}}\:\mid\:\exists\:\boldsymbol{{n}}\in\:\mathbb{N}\:,\:\boldsymbol{{x}}^{\:\boldsymbol{{n}}} \:\in\:\boldsymbol{\mathrm{I}}\:\right\}\: \\ $$$$\: \\ $$
Answered by mindispower last updated on 09/Aug/22
$${I}\cap{J}\:{ideal} \\ $$$$\sqrt{{I}\cap{J}}=\left\{{x}\in{R}\:\mid\exists{n}\in{N}\:{x}^{{n}} \in{I}\cap{J}\right\} \\ $$$${x}^{{n}} \in{I}\Rightarrow{x}\in\sqrt{{I}} \\ $$$${x}^{{n}} \in{J}\Rightarrow{x}\in\sqrt{{j}} \\ $$$$\Rightarrow{x}\in\sqrt{{I}}\cap\sqrt{{J}}\Rightarrow\sqrt{{I}\cap{J}}\subset\sqrt{{I}}\cap\sqrt{{J}} \\ $$$${x}\in\sqrt{{I}}\cap\sqrt{{J}}\Rightarrow\exists{n},{m}\:\mid{x}^{{n}} \in{I}\:,{x}^{{m}} \in{J} \\ $$$$\left.{x}^{{n}+{m}} \in{I}“\:{x}^{{n}} .{x}^{{m}} \in{x}^{{m}} .{I}={I},“{x}^{{n}+{m}} \in{J},{x}^{{m}+{n}} ={x}^{{n}} .{x}^{{m}} \in{x}^{{n}} {J}={J}\right\} \\ $$$$\Rightarrow{x}^{{n}+{m}} \in{I}\cap{J}\Rightarrow{x}\in\sqrt{{I}\cap{J}}\Rightarrow\sqrt{{I}}\cap\sqrt{{J}}\subset\sqrt{{I}\cap{J}} \\ $$$$\sqrt{{I}\cap{J}}=\sqrt{{I}}\cap\sqrt{{J}} \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 14/Aug/22
$$\:\:{thanks}\:{alot}\:\:{sir}\:{power}… \\ $$