16
$\begingroup$

Is it true that the cardinality of every maximal linearly independent subset of a finitely generated free module $A^{n}$ is equal to $n$ (not just at most $n$, but in fact $n$)? Here $A$ is a nonzero commutative ring. I know that it's true if $A$ is Noetherian or integral domain. I thought it was not true in general but I came up with something that looks like a proof and I can't figure out where it went wrong.

$\endgroup$
7
  • $\begingroup$ I was in fact kicking myself for using the word "rank" and confusing everyone. $\endgroup$
    – ashpool
    Jul 25, 2010 at 17:42
  • 1
    $\begingroup$ kwan, I have taken the liberty to remove the word "rank" from the title. I hope that this will help readers discern your true intent. $\endgroup$ Jul 25, 2010 at 18:15
  • 5
    $\begingroup$ It says the result is false (without giving an example) at this link: books.google.com/… $\endgroup$
    – KConrad
    Jul 25, 2010 at 20:06
  • 1
    $\begingroup$ The link by KConrad gives a reference to: M. Lazarus. Les familles libres maximales d'un module ont-elles même cardinal? Pub. Sém. Math. Rennes 4, 1973. It is not referred by MathSciNet/ZbMath. But it's accessible at Numdam here $\endgroup$
    – YCor
    Mar 3, 2020 at 6:32
  • 1
    $\begingroup$ Having a look at Lazarus's paper: It addresses general modules (asking whether all maximal linearly independent subsets have the same cardinal). He notably proves it's true (a) for free modules of infinite rank (b) for arbitrary modules over $A$ with finite Specmin (e.g., $A$ domain, or $A$ noetherian reduced). It also solves the OP's question in Rem. 2.4, using a reduced local $A$ with compact Specmin with f.g. ideal that is faithful with no regular element (constructed by Quentel, Bull SMF 1971, also on Numdam). $\endgroup$
    – YCor
    Mar 3, 2020 at 6:46

4 Answers 4

19
$\begingroup$

I think I have a counter-example. Let $A$ be the ring of functions $f$ from $\mathbb{C}^2 \setminus (0,0) \to \mathbb{C}$ such there is a polynomial $\widetilde{f} \in \mathbb{C}[x,y]$ such that $\widetilde{f}(x,y)=f(x,y)$ for all but finitely many $(x,y)$ in $\mathbb{C}^2$.

Map $A$ into $A^2$ by $f \mapsto (fx, fy)$. We check that this is injective: If $fx=0$ then $f$ is zero off of the $x$-axis. Similarly, if $fy=0$, then $f$ is zero off of the $y$-axis. So $(fx, fy) = (0,0)$ implies that $f$ is zero everywhere on $\mathbb{C}^2 \setminus (0,0)$.

We now claim that there do not exist $(u,v)$ in $A^2$ such that $(f,g) \mapsto (fx+gu, \ fy+gv)$ is injective. Suppose such a $(u,v)$ exists. Let $\widetilde{u}$ and $\widetilde{v}$ be the polynomials in $\mathbb{C}[x,y]$ which coincide with $u$ and $v$ at all but finitely many points. Let $\Delta=\widetilde{u} y - \widetilde{v} x$. Since $\Delta$ is a polynomial which vanishes at $(0,0)$, it is not a non-zero constant. Thus, $\Delta$ vanishes on an entire infinite subset of $\mathbb{C}^2$. Let $(p,q)$ be a point in $\mathbb{C}^2 \setminus (0,0)$ such that $\Delta(p,q)=0$, $\widetilde{u}(p,q)= u(p,q)$ and $\widetilde{v}(p,q)=v(p,q)$.

So $q u(p,q) - p v(p,q) =0$. Since $(p,q) \neq (0,0)$, there is some $k \in \mathbb{C}$ such that $(u(p,q), v(p,q)) = (kp, kq)$. Take $f$ to be $-k$ at $(p,q)$ and $0$ elsewhere; let $g$ be $1$ at $(p,q)$ and $0$ elsewhere. So $(fx+gu, fy+gv)=0$, and the map $(f,g) \mapsto (fx+gu, \ fy+gv)$ is not injective.

$\endgroup$
6
  • 1
    $\begingroup$ Looks good. How did you come up with this? $\endgroup$
    – KConrad
    Jul 25, 2010 at 20:36
  • 1
    $\begingroup$ After a lot of dead ends, I decided to concentrate on $(m,n)=(1,2)$. So I wanted a ring with two elements $(x,y)$ such that nothing nonzero is annihilated by the ideal $(x,y)$, but such that every element of the ideal $(x,y)$ annihilates something nonzero. I decided to build $A$ as a subring of $\mathbb{C}^Z$ for some set $Z$. I needed there to be no nonzero function supported on $\{ x=y=0 \}$, but a nonzero function supported on $\{ ay-bx =0 \}$ for any $(a,b) \in A^2$. I tried to find clever ways to do this for a while, before realizing I could just impose it by fiat. $\endgroup$ Jul 25, 2010 at 20:48
  • 6
    $\begingroup$ This extends to other examples. Let A be 2-dim. Noeth. domain with inf. many max. ideals. Pick a max. ideal M in A generated by two elements, and let R be the sequences (a_m) indexed by max. ideals m other than M, where a_m is in A/m and there is an a in A such that a_m = a mod m for all but finitely many m (i.e., R is seq. looking like reduction mod m of an elt. of A for all but fin. many m). The ring A embeds in R, so A^2 embeds in R^2. Let M have generators x and y. In R^2, the vector (x,y) is lin. indep. over R but is not part of a 2-elt. lin. indep. subset of R^2. Use C[x,y], Z[x],... $\endgroup$
    – KConrad
    Jul 25, 2010 at 23:49
  • $\begingroup$ Very nice examples! $\endgroup$ Jul 26, 2010 at 7:47
  • 2
    $\begingroup$ Bravo, David. This is a perfect illustration of our site working at its best: ask a rather difficult question and get a clever answer within a few hours. Long live MathOverflow! $\endgroup$ Jul 26, 2010 at 19:29
4
$\begingroup$

We have to prove that $m \leq n$ if there is a monomorphism $A^m \to A^n$. Since this is given by a $n \times m$ matrix with entries in $A$ and every finitely generated ring is noetherian, it is enough to consider the case that $A$ is noetherian.

Now you already know the proof for this case, but I just add it. Pick a minimal prime ideal $\mathfrak{p} \subseteq A$. This exists since $A \neq 0$. Now localize at $\mathfrak{p}$. Then we may replace $A$ by $A_{\mathfrak{p}}$, and thereby assume that $A$ is a $0$-dimensional noetherian ring, thus artinian. For such a ring it is known that the length of finitely generated modules is finite, and additive on short exact sequences. In particular $m * l(A) \leq n * l(A)$. Since $l(A) \neq 0$ is finite, we get $m \leq n$.

By the way, the assertion can be generalized to the infinite case:

Let $M$ be a free module with basis $B$ and $L \subseteq M$ a linearly independent subset. Then $|L| \leq |B|$.

Proof: Let $B$ be infinite. Representing elements of $L$ as linear combinations of elements in $B$ yields a map $f : L \to E(B)$, where $E(B)$ denotes the set of finite subsets of $B$. Now let $F$ be such a finite subset with $n$ elements. The finite case yields that there are at most $n$ linearly independent elements in $\langle F \rangle$, thus also in $f^{-1}(F)$. Now we use cardinal arithmetics:

$|L| = \sum_{n > 0} \sum_{F \in E(B), |F|=n} |f^{-1}(F)| \leq \sum_{n > 0} |B^n| = \sum_{n > 0} |B| = |B|.$

EDIT: See the comments; this does not answer kwan's question yet.

$\endgroup$
5
  • $\begingroup$ Martin, how does the inequality imply every maximal linearly independent subset have the same cardinality $n$? $\endgroup$
    – ashpool
    Jul 25, 2010 at 17:54
  • $\begingroup$ Ok I realize this is more complicated than I thought ... cf. your recent question mathoverflow.net/questions/30066/…. With the proof above we know that every maximal linearly independent subset of $A^n$ has at most $n$ elements. If it had $m<n$ elements, we should enlarge it somehow ... $\endgroup$ Jul 25, 2010 at 18:02
  • $\begingroup$ One has to prove more than that an injection $A^m\to A^n$ entails that $m\le n$. One has to show that any linearly independent set of $m<n$ elements in $A^n$ is a subset of a linearly independent set of $n$ elements. $\endgroup$ Jul 25, 2010 at 18:06
  • $\begingroup$ Yes, see also my previous comment. Perhaps we can use the Noetherian case ... $\endgroup$ Jul 25, 2010 at 18:20
  • $\begingroup$ Incidentally the infinite case you mentioned can be also inferred backward from the fact that every maximal linearly independent subset of a free module of infinite rank has the same cardinality. $\endgroup$
    – ashpool
    Jul 25, 2010 at 22:19
-1
$\begingroup$

Assuming that $A$ has a maximal ideal $\mathfrak{m}$ (for example, by using Zorn's Lemma), one can proceed as follows: if $M$ is a free $A$-module with basis $(v_i)_{i\in I}$, then $M \cong A^I$, whence $M / \mathfrak{m} M \cong A^I / \mathfrak{m} A^I \cong (A / \mathfrak{m} A)^I$. This is a vector space over $k := A / \mathfrak{m} A$ of dimension $|I|$. Since over fields, all vector space bases of the same vector space have the same length, and since the $k$-vector space structure of $M / \mathfrak{m} M$ is independent of the choice of the basis, this shows that all $A$-bases of $M$ have the same cardinality.

I don't remember where I first saw this though... maybe someone else has a reference? I saw this first in the case that $A = \mathbb{Z}$ and $\mathfrak{m} = (2)$ for free abelian groups $M$, to show that the rank is well-defined.

$\endgroup$
2
  • $\begingroup$ This doesn't show that any linearly independent set of $m<n$ elements in $A^n$ is a subset of a linearly independent set of $n$ elements. $\endgroup$ Jul 25, 2010 at 18:04
  • $\begingroup$ Oh, shoot. I misread the question. Thanks for pointing that out, and sorry for my spam... $\endgroup$
    – felix
    Jul 25, 2010 at 18:12
-1
$\begingroup$

I have spotted the mistake in my proof. So here is the "wrong" proof:

Let $v_{1},\ldots,v_{m}$ be linearly independent elements in $A^{n}$, where $m\lt n$. Write them as $n$-tuples of elements in $A$, thereby forming an $n$-by-$m$ matrix. Linear independence of $v_{i}\ $s means that the rank of this matrix is $m$. So there is an $m$-by-$m$ minor with non-zero determinant. By exchanging rows if necessary, bring these $m$ rows to the top part of the matrix. Now add a colomn to the right side of the matrix whose entries are $0$ except at the $n+1$ th position, where the entry is $1$. Then the new $n$-by-$(m+1)$ matrix has rank $(m+1)$ and hence the $(m+1)$ columns, the first $m$ of which are the $v_{i}\ $s, are linearly independent.

The mistake was the notion of rank of a matrix. When the entries are not from integral domain, the proper definition should be the largest integer $m$ such that there is no nonzero element in the ring annihilating the determinant of every $m$-by-$m$ minor. In the above example, I can't conclude that the rank of the $n$-by-$(m+1)$ matrix is $(m+1)$.

With this, I can now exhale a sigh of relief and continue believing that this is not true in general. (By the way I also know that it is true for free modules of infinite rank)

$\endgroup$
4
  • $\begingroup$ continue believe it is not true in general? don't you mean continue believing that it is true in general? Anyway, see my answer, which settles the problem. $\endgroup$
    – KConrad
    Jul 25, 2010 at 19:14
  • $\begingroup$ I have a strong reason to believe that it is not true in general. See mathoverflow.net/questions/30066/… $\endgroup$
    – ashpool
    Jul 25, 2010 at 19:21
  • $\begingroup$ Ah, do you think it is not true because Lazarus did not include "finite free modules" in his list of cases where it is true? $\endgroup$
    – KConrad
    Jul 25, 2010 at 19:34
  • $\begingroup$ Yes. Also see books.google.com/… for non-commutative case. $\endgroup$
    – ashpool
    Jul 25, 2010 at 19:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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