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Let R be a commutative ring with identity 1 not equal to 0. Let I subgroup of R be an ideal.

show that if p a subgroup of R is a prime ideal containing I, then √I is a subgroup of p.(√I is defined here Showing $\sqrt{N}$ is an ideal of $R$)

what are some properties of prime ideal that will be helpful in approaching this? I am still new to the concept of ideals, so any suggestions would be a lot of help.

Thanks

Community
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Monica
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1 Answers1

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Let $\mathfrak{p}\subseteq R$ be a prime ideal with $I\subseteq\mathfrak{p}$, and let $a\in\sqrt{I}$. Then by definition, $a^n\in I$ for some $n\in\mathbb{N}$. So, $a^n\in\mathfrak{p}$. But since $\mathfrak{p}$ is prime, $a\in\mathfrak{p}$ (either $a$ or $a^{n-1}$ is in $\mathfrak{p}$, and induction will show that in fact $a\in\mathfrak{p}$). Thus, $\sqrt{I}\subseteq\mathfrak{p}$. This sort of argument is typical with prime ideals, considering their defining property is $ab\in\mathfrak{p}$ implies that $a\in\mathfrak{p}$ or $b\in\mathfrak{p}$.

Stahl
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