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Let $R$ be an integral domain and $S$ be subring of $R$ with $1_R=1_S.$ Let $T=\{f(x) \in R[X]: f(0) \in S\}.$ Suppose $R[X]$ satisfies ascending chain condition for principal ideals, $ACCP.$

Could anyone advise me on how to prove that if $T$ satisfies $ACCP$, then $U(R) \cap S =U(S)$, where $U(R), U(S)$ refers to set of units of $R$ and $S$ respectively?

(Hints will suffice.)

Thank you.

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  • $\begingroup$ I didn't use the condition that $R[X]$ satisfies ACCP, so please check in case I made some silly mistake in my answer! $\endgroup$ – user21820 Feb 15 '14 at 15:31
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$U(S) \subseteq U(R) \subseteq U(R) \cap S$

If $T$ satisfies ACCP,

  If $a \in U(R) \cap S$,

    $ab=1$ for some $b \in U(R)$

    If $b \notin S$,

      $\left<X\right> \subset \left<bX\right> \subset \left<b^2X\right> \subset \cdots$

      $T$ does not satisfy ACCP because this is a strictly ascending chain of principal ideals

      Contradiction

    Therefore $b \in S$

    $a \in U(S)$

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