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Not sure how to do this one. If $S$ is a field, then I was considering that $\exists r_1,\ldots, r_n\in R$ s.t. $1 = r_1s_1+\cdots+r_ns_n$ so for $r = rr_1s_1+\cdots+rr_ns_n$. Maybe that is somehow useful for taking inverses of elements.

The assumption that $S$ is an integral domain is necessary because otherwise we could have $S = \mathbb{Z}_p[x]/f(x)$ where $f(x)$ is not irreducible. This is still a finitely generated $\mathbb{Z}_p$-module, but its not a field.

Any hints or solutions would be much appreciated. I feel like this isn't that hard and I'm missing something simple

Source: https://dornsife.usc.edu/assets/sites/363/docs/F17_510ab.pdf

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marked as duplicate by rschwieb abstract-algebra Aug 13 '18 at 20:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ See Atiyah–Macdonald, Prop 5.1 and 5.7. $\endgroup$ – lhf Aug 13 '18 at 0:11
  • $\begingroup$ This is one of the most duplicated ring-theory questions on the site that I know of. $\endgroup$ – rschwieb Aug 13 '18 at 20:12
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Uncover the spoilers for solutions completing the hints below:

  • Suppose $R$ is a field, and let $s \in S$ be a nonzero element. Then multiplication by $s$ is an $R$-linear endomorphism of $S$, which is injective since $s$ is nonzero and $S$ is a domain.

    Since $S$ is a finite-dimensional $R$-vector space, it follows that multiplication by $s$ is also surjective, and so $1$ is in the image of this map.

  • Suppose $S$ is a field. This solution I have in mind for this direction is a bit trickier. Let $r \in R$ be nonzero, and let $s$ be the inverse to $r$ in $S$. As before, consider the $R$-linear map $\varphi_{s} \colon S \to S$ corresponding to multiplication by $s$. Since $S$ is a finitely generated $R$-module, $\varphi_{s}$ satisfies a monic polynomial relation with coefficients in $R$ by Cayley-Hamilton. That is, there exist $r_{1}, \ldots, r_{n} \in R$ such that multiplication by $s^{n}+r_{1}s^{n-1} + \cdots +r_{n}$ is the zero element of $\mathrm{End}_{R}(S)$.

    Since $S$ is a faithful $R$-module ($R$ is a subring of $S$,and so contains $1$), this implies that $s^{n}+r_{1}s^{n-1} + \cdots +r_{n} = 0$. Now multiply both sides by $r^{n-1}$ to conclude that $s \in R$.

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Hints:

$\Rightarrow$: If $R$ is a field, let $s\in S$, and consider multiplication by $s$ in $S$. Check this is an injective $R$-linear map. What can you conclude, knowing $S$ is a finite dimensional $R$-vector space?

$\Leftarrow$: If $S$ is a field, consider $r\in R$; you know $r^{-1}\in S$, hence it is a root of a monic polynomial in $R[X]$. Deduce from this polynomial equation that $r^{-1}$ is a polynomial in $r$, hence it belongs to $R$.

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