Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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 $m\ge |k|+1$.


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 $m\ge|k|+1$?

share|cite|improve this answer
Excellent answer. Thank you. – Jiangwei Xue Jul 18 '11 at 6:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.