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Let $F$ be a domain and let $R\le F$ be a subring such that $F$ is a free $R$-module of finite rank $n$.

Question: Is there an $R$-basis $\lbrace e_1,...,e_n\rbrace$ of $F$ such that at least one of the basis elements is a unit in $F$ ?

As an example consider $R = \mathbb{Z}$ and $F=\mathbb{Z}[\text{i}]$ where we can take the units $1,\text{i}$ as basis.

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  • $\begingroup$ tranquoctan -- the question needs some work as it does not seem to make sense as stated (e.g., what does "R be a subring of domain F" mean?). $\endgroup$
    – algori
    Commented Nov 27, 2012 at 1:52
  • $\begingroup$ Why doesn't the question make sense ? $F$ is a domain, $R\le F$ a subring such that $F$ is a free $R$-module with basis $e_1,...,e_n$. Can we conclude that some $e_i\;(1 \le i \le n)$ is a unit of $F$ ? As an example you may take $F=\mathbb{Z}[\text{i}], R=\mathbb{Z}$. $\endgroup$
    – Ralph
    Commented Nov 27, 2012 at 2:09
  • $\begingroup$ Ralph -- well, that is much clearer. If you edit the question then (assuming this is what the OP meant) I'll vote to reopen. $\endgroup$
    – algori
    Commented Nov 27, 2012 at 2:19
  • $\begingroup$ algori, good idea. If my interpretation of the original question should be wrong, the OP still has the opportunity to edit it again. $\endgroup$
    – Ralph
    Commented Nov 27, 2012 at 2:26
  • $\begingroup$ I mean, F is a domain, R≤F a subring such that F is a free R-module , Can we conclude that some ei(1≤i≤n) is a unit of F with basis e1,...,en.? (can We chose a base of F?) $\endgroup$
    – tranquoctan
    Commented Nov 27, 2012 at 2:30

2 Answers 2

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If $R$ is a dedekind domain or if each finitely generated projective $R$-module is free, one can choose $e_1=1$. This is explained in the answers of Angelo and Florian of this question:

Does $S$ being a free rank-$n$ $R$-algebra imply that $S/R$ is free rank $n-1$?.

However, the counter-examples given there don't apply since they aren't domains.

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First some general remarks. You're asking whether $S/R$ is free as a $R$-module. If $S$ is a $R$-algebra which is free of finite rank, then the map $R \to S$ splits as a map of $R$-modules. This fact was already noticed by Florian Eisele in his answer to the other MO question.

Now for an explicit counter-example to your question. Consider the ring $R={\bf Z}[x,y,z]/(x^2+y^2+z^2-1)$. It is an integral domain. It is known that there exists a $R$-module $M$ which is not free such that $R \oplus M \cong R^3$. For a nice construction, see e.g. Keith Conrad's notes. Explicitly we can take $M=\{(f,g,h) \in R^3 : xf+yg+zh=0\}$. Note that we can embed $M$ in $R^2$ by $(f,g,h) \mapsto (f,g)$, and the cokernel $R^2/M$ is a torsion module, so there exists $F \in R \backslash \{0\}$ such that $F \cdot R^2 \subset M$.

Now, we would like to construct a $R$-algebra structure on $R \oplus M$. We can do this by considering the $R$-algbera $S_0 = R \otimes_{\mathbf{Z}} \mathcal{O}$ where $\mathcal{O}$ is an order of a cubic field $K$. It is an integral domain, since the polynomial $x^2+y^2+z^2-1$ is irreducible over any field of characteristic not $2$. Let $(1,\alpha,\beta)$ be a $\mathbf{Z}$-basis of $\mathcal{O}$. Embed $R \oplus M$ in $S_0$ by $(f,(g,h)) \mapsto f+g\alpha+h\beta$. This won't be a subring of $S_0$ in general, but $S=R \oplus FM$ is a subring of $S_0$ since $(FM) \cdot (FM) \subset F^2 S_0 \subset R \oplus FM$. So we have constructed an integral domain $S$ over $R$ such that $S/R \cong M$ is not free over $R$.

I don't know whether it's possible to find a counterexample where $R \to S$ splits as a map of rings, in other words where $S=R \oplus I$ where $I$ is an ideal of $S$.

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  • $\begingroup$ Sorry, I didn't realize Florian had given the same proof for the splitting. I will edit my answer. $\endgroup$
    – François Brunault
    Commented Nov 28, 2012 at 10:41
  • $\begingroup$ Actually, his proof is simpler (doesn't need integrality), so I will just delete my proof. $\endgroup$
    – François Brunault
    Commented Nov 28, 2012 at 10:47

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