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Let $P$ be a prime ideal in a commutative ring $R$ with unity such that an ideal $Q$ is $P$-primary and some power of $P$ is a subset of $Q$. I want to show that $\sqrt {Q[[x]]}=P[[x]]$.

If a power series $f=a_0+a_1x+a_2x^2+\cdots$ is in $\sqrt {Q[[x]]}$ then some power $f^k$ is in $Q[[x]]$ so that $a_0^k$ belongs to $Q$ whence, by the hypothesis, $a_0$ would be in $P$. But for the other coefficients $a_i$ I could not reach the same result. Thanks for any help!

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From $Q\subseteq P$ we get $Q[[X]]\subseteq P[[X]]$, and therefore $\sqrt{Q[[X]]}\subseteq\sqrt{P[[X]]}=P[[X]]$ (why?).

Now if we want to show $P[[X]]\subseteq\sqrt{Q[[X]]}$ start with a power series $f\in P[[X]]$, say $f=a_0+a_1X+\cdots$, and consider the ideal $I=(a_0,a_1,\dots)$. Since $P^k\subseteq Q$ we also have $I^k\subset Q$, and therefore $f^k\in Q[[X]]$, that is, $f\in\sqrt{Q[[X]]}$.

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  • $\begingroup$ Now, could we infer that $Q[[X]]$ is $P[[X]]$-primary? $\endgroup$ – karparvar May 29 '15 at 11:16
  • $\begingroup$ @karparvar Under the condition that some power of $P$ is contained in $Q$ it follows that $Q[[X]]$ is $P[[X]]$-primary. (For a proof see this paper, especially Corollary 4, page 432.) $\endgroup$ – user26857 May 30 '15 at 14:59
  • $\begingroup$ It really works. Thank you for introducing the paper. $\endgroup$ – karparvar May 30 '15 at 18:11

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