Q: I , J are two ideals of commutative ring , ( R ,⊕, ) .prove that : (√( I ∩ J )) =^? (√( I )) ∩ (√( J )) m.n note : (√(I )) = { x ∈ R ∣ ∃ n∈ N , x^( n) ∈ I }


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Question Number 174684 by mnjuly1970 last updated on 08/Aug/22
$Q: I , J are two ideals of commutative ring , ( R ,⊕, ) .prove that : (√( I ∩ J )) =^? (√( I )) ∩ (√( J )) m.n note : (√(I )) = { x ∈ R ∣ ∃ n∈ N , x^( n) ∈ I }$
$Q : I, J a r e t w o i d e a l s o f c o m m u t a t i v e$ $r i n g, (R, \oplus,) . p r o v e t h a t :$ $\sqrt{I \cap J} \overset{?}{=} \sqrt{I} \cap \sqrt{J} m . n$ $n o t e : \sqrt{I} = {x \in R ∣ \exists n \in N, x^{n} \in I}$
	Answered by mindispower last updated on 09/Aug/22
	$I∩J ideal (√(I∩J))={x∈R ∣∃n∈N x^n ∈I∩J} x^n ∈I⇒x∈(√I) x^n ∈J⇒x∈(√j) ⇒x∈(√I)∩(√J)⇒(√(I∩J))⊂(√I)∩(√J) x∈(√I)∩(√J)⇒∃n,m ∣x^n ∈I ,x^m ∈J x^(n+m) ∈I“ x^n .x^m ∈x^m .I=I,“x^(n+m) ∈J,x^(m+n) =x^n .x^m ∈x^n J=J} ⇒x^(n+m) ∈I∩J⇒x∈(√(I∩J))⇒(√I)∩(√J)⊂(√(I∩J)) (√(I∩J))=(√I)∩(√J)$
	$I \cap J i d e a l$ $\sqrt{I \cap J} = {x \in R ∣ \exists n \in N x^{n} \in I \cap J}$ $x^{n} \in I \Rightarrow x \in \sqrt{I}$ $x^{n} \in J \Rightarrow x \in \sqrt{j}$ $\Rightarrow x \in \sqrt{I} \cap \sqrt{J} \Rightarrow \sqrt{I \cap J} \subset \sqrt{I} \cap \sqrt{J}$ $x \in \sqrt{I} \cap \sqrt{J} \Rightarrow \exists n, m ∣ x^{n} \in I, x^{m} \in J$ $x^{n + m} \in I ‘ ‘ x^{n} . x^{m} \in x^{m} . I = I, ‘ ‘ x^{n + m} \in J, x^{m + n} = x^{n} . x^{m} \in x^{n} J = J}$ $\Rightarrow x^{n + m} \in I \cap J \Rightarrow x \in \sqrt{I \cap J} \Rightarrow \sqrt{I} \cap \sqrt{J} \subset \sqrt{I \cap J}$ $\sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$
	Commented by mnjuly1970 last updated on 14/Aug/22

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