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I come across this question as I consider a problem dealing with semilocal rings.

Suppose that $k$ is a field, $R=\oplus_{i=1}^m k$ is a finite $k$-algebra via the diagonal embedding $k\to R$. Let $V$ be a free $R$-module of rank $n$. So we have $V=\oplus_{i=1}^m V_i$ where each $V_i$ is a $k$-vector space of $n$-dimension. Let $W$ be a $k$-subspace of $V$ that generates $V$ over $R$. Is it true that we may always find an $R$-basis of $V$ in $W$?

I found the answer is yes in each of the following situations (1) $m=2$, (2)$k$ has $\infty$-many elements, e.g, $k$ is algebraically closed. But I don't have any good picture of what's going on in the case $k$ is a finite field and $m\geq 3$. Could anyone give a proof or counterexample in this case?

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up vote 2 down vote accepted

Would the following be the kind of counterexample you have in mind?

Let $k=\mathbf{Z}/2\mathbf{Z}$ be the field of two elements, and let $m=3$. Let further $V$ be a free $R$-module of rank $n=1$, so as a vector space over $k$ we have $V\cong k^3$. Let $W$ be the zero sum $k$-subspace of $V$: $$W=\{(a_1,a_2,a_3)\in k^3\mid a_1+a_2+a_3=0\}=\{000,110,101,011\}.$$ We easily see that $W$ generates $V$ as an $R$-module, because for each coordinate position there is a vector in $W$ such that its component in that position is non-zero. Yet no element of $W$ generates $V$ alone as an $R$-module. Any element has at least a single zero component, so the cyclic $R$-module generated by that element cannot be all of $V$.

Edit1: If $k=\{x_1,x_2,\ldots,x_q\}$ then the two-dimensional subspace $W\subseteq k^{q+1}$ spanned by the vectors $\vec{a}=(0,1,1,1,1,\ldots,1)$ and $\vec{b}=(1,x_1,x_2,\ldots,x_q)$ has the property that any vector of $W$ has at least a single component equal to zero. This is because a non-zero scalar multiple of $\vec{b}$ has all the $q$ elements permuted in the last $q$ positions, so one of them will get cancelled, when we add a non-zero multiple of $\vec{a}$. Also, obviously all the positions have something non-zero in either $\vec{a}$ or $\vec{b}$. The argument works the same as in the earlier case $q=2$.

I don't know yet how to prove that $m=q+1$ is the shortest length, where such a subspace $W$ can be found. I think that is the case, though. Two remarks:

1) We can make vectors of $W$ longer by replicating one of the coordinates as many times as we need.

2) Does this show that counterexamples with a 2-dimensional $k$-space $W$ exist, whenever $mn\ge |k|+1$? This is wrong, but in the case $n=1$ we do get counterexamples like this, if $m\ge |k|+1$.

Edit2:

Ok, here's the missing part.

Lemma. Assume $|k|=q$, and $W$ is an $\ell$-dimensional subspace of $k^m$ such that for all the $m$ coordinate positions there is a vector $w\in W$ with a non-zero component in that position, but also every vector of $W$ has at least one coordinate equal to zero. Then $\ell\ge2$ and $m\ge q+1$.

Proof. Obviously $\ell=1$ doesn't work, so $\ell\ge2$. Let $S_i$ be the subspace of $W$ consisting of those vectors that have a zero in position $i$. Clearly $\dim_k S_i=\ell-1$, so $|S_i|=q^{\ell-1}$. From our assumptions it follows that the union of all the subsets $S_i$ covers all of $W$. OTOH the zero vector is contained in all of the sets $S_i$, so there is some overlap. Therefore $$ \sum_i|S_i|= mq^{\ell-1}>|W|=q^\ell. $$ For this to hold we must have $m\ge q+1.$ Q.E.D.

2') Doesn't it follow that counterexamples exist, iff $m\ge|k|+1$?

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Excellent answer. Thank you. –  Jiangwei Xue Jul 18 '11 at 6:07
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