I have the following problem:

Prove that a integral domain $R$ is a field iff there exist a finitely generated divisible $R$-module.

One side is easy, because if $R$ is a field then $R$ itself is a finitely generated divisible $R$-module. But I get stuck in the other side.

I want to prove that the finitely generated divisible $R$-module is isomorfic as modules to $R$ and then $R$ is a field.

Thanks for the hints!

  • $\begingroup$ Not just any finitely generated divisible $R$-module is isomorphic to $R$; i.e. a $2$-dimensional vector space over a field is f.g. and divisible, but not isomorphic to $R$. You need to construct a cyclic divisible module, and that would be your isomorphic copy of $R$. $\endgroup$ – Sean Clark Jul 27 '14 at 16:52
  • $\begingroup$ How do I construct that cyclic $R$ module? $\endgroup$ – YTS Jul 27 '14 at 17:02
  • 2
    $\begingroup$ See the reasoning in this answer (and note that only divisibility was used). This also makes clear that the statement should be "an integral domain is a field iff there is a finitely generated nonzero divisible $R$-module" $\endgroup$ – zcn Jul 27 '14 at 18:21

Proof in short: If we can construct a divisible cyclic submodule, it must be isomorphic to $R$, and therefore $R$ is a field. We will do this by showing that any element of a minimal set of generators of a divisible $R$-module generates a cyclic divisible submodule by showing that otherwise we obtain a smaller set of generators.

Proof: Suppose $x_1,\ldots, x_n$ is a minimal set of generators for a divisible $R$-module $M$.

First suppose $n=1$, so $M=Rx$ for some $x\in M$. Then for any $r\in R$, $x=ry$ for some $y\in M$; since $M$ is generated by $x$, we have $y=r'x$ for some $r'\in R$ and so $x=rr'x$. In particular, the map $r\mapsto rx$ is a $R$-module isomorphism, since $rx=0$ implies $0=r'rx=x$. In particular, we see that $r'r=1$ in $R$, so $r'=r^{-1}$, and thus $R$ is a field.

Now suppose $n>1$. Let $a\in R^\times$ and let $ay=x_1$. Then $y=r_1x_1+\ldots +r_nx_n$ for some $r_1,\ldots, r_n$; if such an expression exists where $r_2,\ldots, r_n$ are all zero for every $r$, then the submodule generated by $x_1$ is divisible and we reduce to the previous case.

Therefore, let us assume that for some $r\in R$, in any such expression for $y$ we have some $r_m\neq 0$ for $m>1$. We will show that we can pick a smaller generating set, contradicting our assumption on the generators.

We have $x_1=ar_1x_1+ar_2x_2+\ldots+ar_nx_n$. Note we cannot have $r_1=a^{-1}$, since then we could write $x_1=ar_1x_1$, contradicting our assumption that some $r_m\neq 0$ for $m>1$. Since $ar_1\neq 1$, $\lambda=1-ar_1\neq 0$ in $R$, so $x_1=\lambda (s_1x_1+\ldots s_nx_n)$ for some $s_1,\ldots, s_n\in R$. But then $$x_1=s_1\lambda x_1+s_2\lambda x_2+\ldots +s_n \lambda x_n=(s_2\lambda+as_1r_2)x_2+\ldots+(s_n\lambda+as_1r_n)x_n,$$ hence $M$ is generated by $x_2,\ldots, x_n$ contradicting our assumption.

  • $\begingroup$ How did you get the last line? I don't see it. $\endgroup$ – YTS Jul 27 '14 at 21:10
  • $\begingroup$ $\lambda x_1 =ar_2x_2+\ldots+ar_n x_n$ by the expression for $y$, and the last line is just plugging this in for $s_1\lambda x_1$ Combined with the divisibility of $x_1$ by $\lambda$. $\endgroup$ – Sean Clark Jul 28 '14 at 3:16

Let $M$ be a non-zero divisible $R$ module, finitely generated by $m_1,\ldots,m_n$. Say $r\in R$ is non-zero. We need to show $r$ has an inverse in $R$. Since $M$ is divisible, there exist elements $m’_1,\ldots,m’_n\in M$ such that $rm’_i=m_i$ for $i=1,2,\ldots,n$. We can write each $m’_i$ as an $R$-linear combination of the $m_j$, say $$ m’_i=\sum_{j=1}^n a_{i,j}m_j. $$ Let $A=(a_{i,j})\in M_n(R)$. It can be checked that $A$ has $r^{-1}$ as an eigenvalue, so $r^{-1}$ is integral over $R$. This implies $r^{-1}\in R$: if $r^{-k}+b_{k-1}r^{-(k-1)}+\ldots+b_0=0$, with $b_i\in R$, then $r^{-1}=b_{k-1}-b_{k-2}r-\ldots-b_0 r^{k-1}\in R$.


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