# Finitely generated R-module is a field iff R is a field? [duplicate]

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

## marked as duplicate by rschwieb abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Aug 13 '18 at 20:10

• See Atiyah–Macdonald, Prop 5.1 and 5.7. – lhf Aug 13 '18 at 0:11
• This is one of the most duplicated ring-theory questions on the site that I know of. – rschwieb Aug 13 '18 at 20:12

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 $$s \in R$$.

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 $R$.