If every free $R$-module has the property that independence implies extendibility, is $R$ necessarily a field?

Question

Definition. Whenever $M$ is a free $R$-module, let us call a subset $A$ of $M$ extendible iff there is a basis $B$ for $M$ such that $A \subseteq B$. (Is there a standard name for this condition?)

Then extendibility implies independence but not necessarily vice versa.

Question. Suppose a commutative ring $R$ has the property that if $M$ is a free $R$-module, then every independent subset of $M$ is extendible. Is $R$ necessarily a field?

Answer 1

The rings with the required property are called Steinitz rings.

A commutative ring $R$ is Steinitz iff $R$ is a local ring over which every flat $R$-module is free.

(For a proof see this paper, Proposition 5.4.)

Answer 2

The answer is no.

Consider $R = \mathbb{Z}/4\mathbb{Z}$ (or more generally I think, a ring where there is a finite number of noninvertible elements, all of which are zero divisors), which is of course not a field. Let $M = (\mathbb{Z}/4)^{(n)}$ be a free $R$-module for some (possibly infinite) cardinal $n$, and let $A \subset M$ a linearly independent set. Then it's possible to extend $A$ to a basis.

Reduce all the elements of $A$ modulo $2$ to get $\bar A \subset (\mathbb{Z}/2)^{(n)}$. This set $\bar A$ is linearly independent: if $\sum x_i \bar{a}_i = 0$ for $x_i \in \mathbb{Z}/2$, then $\sum 2 \phi(x_i) a_i = 0$ (where $\phi(0) = 0$ and $\phi(1) = 1$). By hypothesis $A$ is independent, so $2 \phi(x_i) = 0$ for all $i$, and thus all the $x_i$ were already zero mod 2. Since $\mathbb{Z}/2$ is a field, $\bar{A}$ can be extended to a basis $\bar{B}$; lift all the elements of $\bar{B} \setminus \bar{A}$ through $\phi$ to get a basis of $M$.

(Informally: throw out all the $2$ in the vector representation of the elements of $A$ and identify $-1 \sim 1$; extend that to a basis of $(\mathbb{Z}/2)^{(n)}$ and lift the new elements to $(\mathbb{Z}/4)^{(n)}$)

Answer 3

Let $R,S$ be two commutative rings over which finite independent subsets are extendable. I claim that this also holds for $R \times S$. (In particular, finite products of fields have the property.)

In fact, let $M$ be a free $R \times S$-module and $\{\ell_i\} \subseteq M$ be independent and finite. Then $M = A \times B$ for a free $R$-module $A$ and a free $S$-module $B$, which have the same rank. Write $\ell_i=(a_i,b_i)$. Then $\{a_i\} \subseteq A$ is independent: $\sum_i r_i a_i = 0 \Rightarrow \sum_i (r_i,0) \ell_i = 0 \Rightarrow \forall i ~ (r_i,0)=0 \Rightarrow \forall i ~ r_i=0$. Thus, $\{a_i\}$ extends to a basis $E$ of $A$. Similarly, $\{b_i\}$ extends to a basis $F$ of $B$. Since $A,B$ have the same rank, there is a bijection between $E$ and $F$, say $\tau : E \to F$. We may arrange $\tau(a_i)=b_i$. It is now easy to check that $\{(e,\tau(e)) : e \in E\}$ is a basis of $A \times B=M$, and it contains $\{\ell_i\}$.

Answer 4

There are even positive-dimensional examples. Consider the following observation: let $x$ be a nonzerodivisor in $R$, i.e. $\{x\}$ is an independent set in $R$ (viewed as a rank-1 free module), since there are no nontrivial relations $r\cdot x = 0$. (You haven't defined independent, but I'm assuming that's what you mean).

Then $\{x\}$ is a basis iff $x$ is a unit.

In particular, if $R$ has this extensibility property, then it consists only of units and zerodivisors. If $R$ is Noetherian, this is equivalent to "every maximal ideal is associated" (in particular $R$ should have only finitely-many maximal ideals).

Conversely, for local rings this property is sufficient. One proof is by row reduction (I'm sure there is a standard proof possible by considering minors and/or Plücker coordinates, but my mind is sluggish this morning): consider an independent set ${\bf v_1, \ldots, v_k}$ in $R^n$, written as a $k \times n$ matrix

$$\begin{bmatrix} v_{11} & v_{12} & \cdots & v_{1n} \\ \vdots & & & \vdots \\ v_{k1} & v_{k2} &\cdots & v_{kn} \end{bmatrix}.$$

Note that since the maximal ideal is associated, there exists $f$ such that $f \cdot \mathfrak{m} = 0$. By independence, we know that $f \cdot {\bf v_1} \ne 0$, so some entry in the first row is outside the maximal ideal, hence a unit. So WLOG we have $v_{11} = 1$, and by row operations we can clear out the rest of the first column.

Now repeat this argument on the second row ${\bf v_2} - v_{12} \cdot {\bf v_1} = \begin{bmatrix}0 & v'_{22} & v'_{23} & \cdots & v'_{2n}\end{bmatrix}$, and so on, reducing to a matrix where, WLOG, the first $k$ columns form a $k\times k$ identity matrix.

Now it is clear that we can complete to a basis, by adding in the $i$-th coordinate vectors $(i = k+1, \ldots, n)$.

(For what it's worth, this boils down to proving that independence implies that some Plücker coordinate is a unit. But I can't think of a non-inductive proof of this fact.)

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So this gives a class of "big" (non-Artinian) examples, by taking local rings in which the maximal ideal is associated, such as $k[x,y]_{(x,y)} / (x^2,xy)$. Or more generally, $k[x_1, \ldots, x_n]_{\mathfrak{m}} / (\mathfrak{m}^n \cap P)$, where $\mathfrak{m}$ is some maximal ideal, $P \subset \mathfrak{m}$ is prime, and $n > 1$.