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Suppose $S$ is an $R$-algebra (associative, commutative, with unit...) such that $S$ is free of finite rank $n$ over $R$. Is it necessarily the case that we can find an $R$-basis $y_1, ..., y_n$ for $S$ with $y_1 = 1$?

I can prove this when $n = 2$:

Suppose $x_1, ..., x_n$ is an $R$-basis for $S$, and write $1 = \sum_{i=1}^n a_ix_i$, with $a_i \in R$. We have $S$ faithfully flat over $R$, so we have $(a_1, ..., a_n) = (a_1, ..., a_n)S \cap R = (1)$, so we can find $b_1, ..., b_n$ in $R$ such that $1 = \sum_{i=1}^n a_ib_i$. Now I use the assumption $n = 2$: the matrix $\left( \begin{array}{cc} a_1 & a_2 \\\ -b_2 & b_1 \end{array} \right)$ is unimodular, so $y_1 = a_1x_1+a_2x_2,$ $y_2 = -b_2x_1+b_1x_2$ is an $R$-basis of $S$ with $y_1 = 1.$

More generally, as long as the unimodular row $(a_1, ..., a_n)$ can be completed to a unimodular matrix over $R$, we can use the unimodular matrix to produce an $R$-basis of $S$ with $y_1 = 1$.

Since not every unimodular row can be completed to a unimodular matrix when $n \ge 3$ and $R$ is arbitrary, it seems like it may be possible to construct a counterexample, but I haven't been able to do it (specifying an algebra structure on a rank $3$ $R$-module is surprisingly tricky).

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2 Answers 2

up vote 6 down vote accepted

As Will Sawin says in his answer, if $R$ has a module $M$ such that the module $R\oplus M$ is free of rank 3 but $M$ is not free then there is a counterexample: $R\oplus M$ with zero multiplication in $M$. For an example, let $R$ be the continuous (or just polynomial) functions on the $2$-sphere and let $M$ be the continuous (or polynomial) tangent vector fields.

Conversely, suppose $R$ has no such module. Whenever $S$ is an $R$-algebra such that $S$ is free of rank 3, then the ring map $R\to S$, considered as an $R$-module map, has a left inverse because the composed map $R\to S\to End_R(S)$ has a left inverse. (There is an $R$-basis for the $3\times 3$ matrices over $R$ starting with the identity matrix.) Therefore as an $R$-module $S$ is the direct sum of $R$ and a free module; $S$ has a basis starting with $1$.

Note: We didn't need commutativity in $S$. We didn't need commutativity of $R$, either; we could just assume $R\to S$ is a homomorphism of associative unital rings for the last argument. But you did need commutativity of $R$ to say that if the module $M$ is such that $R\oplus M$ is free of rank $2$ then $M$ is free of rank one.

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If $R$ has a finite-dimensional vector bundle that is nontrivial but stably trivial, then obviously not. If $M$ is the corresponding module, then $R[M]/M^2$ is a counterexample.

If $R$ has no such bundle, and the rank of $S$ as an $R$-module is a unit in $R$, then we can use the trace map to split the exact sequence $0 => R => S => S/R => 0$, then find a basis for $S/R$.

Thus the only unclear case is when $R$ has no such bundle but is not over the rationals.

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By non-trivial stably-trivial fin. dim. vector bundle over $R$ you mean non-free stably-free fin. gen. projective module, I guess? –  Mariano Suárez-Alvarez Nov 17 '12 at 4:25
    
That is correct. –  Will Sawin Nov 17 '12 at 6:07

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