# Atiyah-MacDonald, exercise 2.11

Let $A$ be a commutative ring with $1$ not equal to $0$. (The ring A is not necessarily a domain, and is not necessarily Noetherian.) Assume we have an injective map of free $A$-modules $A^m \to A^n$. Must we have $m \le n$?

I believe the answer is yes. For instance, why is there no injective map from $A^2 \to A^1$? Say it's represented by a matrix $(a_1, a_2)$. Then clearly $(a_2, -a_1)$ is in the kernel. In the $A^{n+1} \to A^{n}$ case, we can look at the $n \times (n+1)$ matrix which represents it; call it $M$. Let $M_i$ denote the determinant of the matrix obtained by deleting the $i$-th column. Let $v$ be the vector $(M_1, -M_2, ..., (-1)^nM_{n+1})$. Then $v$ is in the kernel of our map, because the vector $Mv^T$ has $i$-th component the determinant of the $(n+1) \times (n+1)$ matrix attained from $M$ by repeating the $i$-th row twice.

That almost finishes the proof, except it is possible that $v$ is the zero vector. I would like to see either this argument finished, or, even better, a nicer proof.

Thank you!

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The answer mathoverflow.net/questions/30860/… was given by Robin Chapman in another thread, which has been closed as duplicate. I find Robin's answer very nice, and hope the other thread won't be deleted. – Pierre-Yves Gaillard Jul 31 '10 at 15:40

Let M be the $n\times m$ matrix representing the injection $A^m \to A^n$. Define Di(M) to be the ideal generated by the determinants of all i-by-i minors of M. Let r be the largest possible integer such that Dr(M) has no annihilator (i.e. there is no nonzero element a∈A such that aDr(M)=0); I think r is usually called the McCoy rank of M.

Assume that $m>n$. We shall get a contradiction.

Choose a nonzero a∈A such that aDr+1(M)=0. By assumption, $a$ does not annihilate Dr(M), so there is some r-by-r minor that is not killed by $a$; we may assume it is the upper-left r-by-r minor. Thus, $r \leq n$, so that $r+1 \leq n+1\leq m$. Let M1, ..., Mr+1 be the cofactors of the upper-left (r+1)-by-(r+1) minor obtained by expanding along the bottom row. (This is well-defined even if the $r+1$-th row of $M$ does not exist, because these cofactors use only the first $r$ rows and the first $r+1$ columns of $A$, and $A$ has both since $r \leq n$ and $r+1 \leq m$.) Note, in particular, we know Mr+1 is the determinant of the upper-left r-by-r minor, so aMr+1≠0.

The claim is that the vector (aM1,...,aMr+1,0,0,...) (which we've already shown is non-zero) is in the kernel of M. To see that, note that when you dot this vector with the k-th row of M, you get $a$ times the determinant of the matrix obtained by replacing the bottom row of the upper-left (r+1)-by-(r+1) minor with the first r+1 entries in the k-th row. If k≤r, this determinant is zero because a row is repeated, and if k>r, this determinant is the determinant of some (r+1)-by-(r+1) minor, so it is annihilated by $a$. But since A is injective, the kernel of M is 0, and we have a contradiction.

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Here is another solution using only the Cayley-Hamilton Theorem for finitely generated modules (Proposition 2.4. in Atiyah-Macdonald) which, even though looks quite innocent, is a very powerful statement.

Assume by contradiction that there is an injective map $\phi: A^m \to A^n$ with $m>n$. The first idea is that we regard $A^n$ as a submodule of $A^m$, say the submodule generated by the first $n$ coordinates. Then, by the Cayley-Hamilton Theorem, $\phi$ satisfies some polynomial equation $$\phi^k + a_{k-1} \phi^{k-1} + \cdots + a_1 \phi + a_0 = 0.$$ Using the injectivity of $\phi$ it is easy to see that if this polynomial has the minimal possible degree, then $a_0 \ne 0$. But then, applying this polynomial of $\phi$ to $(0,\ldots,0, 1)$, the last coordinate will be $a_0$ which is a contradiction as it should be zero.

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Wonderful !!!!! – Pierre-Yves Gaillard Dec 1 '10 at 9:16
+1 This is a proof from The Book. – benblumsmith Sep 28 '12 at 14:43
Is Cayley-Hamilton Theorem necessary here? Is there no easier way to show that the powers of $\phi$ are linearly dependent over $A$? – Alexey Muranov Jun 14 '14 at 22:15
@Pierre-YvesGaillard - I apologise for a rather elementary question: In applying proposition 2.4 from Atiyah-Macdonald, one requirement is $\phi(A^m) \subset IA^m$, where $I$ is an ideal of $A$. In this question, what do you take to be the ideal $I$? – Terrence J Apr 17 at 12:56
@TerrenceJ - You can take $I:=A$. Thanks for your interest! – Pierre-Yves Gaillard Apr 17 at 14:44

Dear CJD, if you are still interested in your problem, already solved three weeks ago by Anton, here is another point of view.

Let $M:A^m \to A^n$ be injective. Let $B=\mathbb Z [\ldots,m_{ij},\ldots]$ be the subring of $A$ generated by all the entries of the matrix $M$ ; this $B$ is a noetherian ring and we have (by restriction) an injective linear map $M:B^m \to B^n$. In other words we may assume that $A$ is noetherian. Now we localize at a prime $\mathfrak p$ of $B$ of height zero and we get an injective map (localization is exact and thus preserves injections) $L:C^m \to C^n$ (the ring $C$ is the ring $B$ localized at $\mathfrak p$).

Ah, but now $C$ is noetherian of dimension zero, hence artinian and we can talk about lengths. Since lengths are additive in exact sequences (Atiyah-MacDonald, Proposition 6.9) we get $m.length(C) + length(coker L)= n.length(C)$, hence $m\leq n$.

Friendly greetings, Georges.

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I like this proof a lot. I guess it doesn't logically make sense as the solution to the Atiyah-MacDonald exercise, since it uses results that come later in the book, but this proof definitely requires less book-keeping than Anton's argument using determinants. Thanks for the post! – CJD Oct 28 '09 at 7:24
This is a response to the answer given by Georges Elencwajg posted on Oct 25. His answer is really nice, but there is a small correction to this answer. In the construction, $B=Z[{m_{ij}}]$ to be replaced with $B=Z(A)[{m_{ij}}]$, where $Z(A)$ is the prime sub ring of A. – N. Kumar Dec 30 '09 at 13:12
I wrote explicitly that B is the subring of A generated by the entries of the matrix M. Any subring of A must, of course, contain the prime subring of A. So your ring is exactly the one I described. – Georges Elencwajg Jan 2 '10 at 9:14
Sorry to have overlooked at it. – N. Kumar Nov 21 '12 at 14:45

I posted this question on a different site a couple of years ago. Eventually I found that a book of T.Y. Lam has a very nice treatment. Here is the writeup I posted on the other site:

After paging through several algebra books, I found that T.Y. Lam's GTM Lectures on Rings and Modules has a beautiful treatment of this question.

The above property of a (possibly noncommutative) ring is called the "strong rank condition." It is indeed stronger than the corresponding statement for surjections ("the rank condition") which is stronger than the isomorphism version "Invariant basis number property". However, in fact it is the case that all commutative rings satisfy the strong rank condition. Lam gives two proofs [pp. 12-16], and I will now sketch both of them.

First proof:

Step 1: The result holds for (left-) Noetherian rings. For this we establish:

Lemma: Let $M$ and $N$ be (left-) $A$-modules, with $N$ nonzero. If the direct sum $M \oplus N$ can be embedded in $M$, then $M$ is not a Noetherian $A$-module.

Proof: By hypothesis $M$ has a submodule $M_1 \oplus N_1$, with $M_1 \cong M$ and $N_1 \cong N$. But we can also embed $M \oplus N$ in $M_1$, meaning that $M_1$ contains a submodule $M_2 \oplus N_2$ with $M_2 \cong M$ and $N_2 \cong N$. Continuing in this way we construct an ascending chain of submodules $N_1$, $N_1 \oplus N_2$,..., contradiction.

So if A is (left-) Noetherian, apply the Lemma with $M = A^n$ and $N = A^{m-n}$. $M$ is a Noetherian $A$-module, and we conclude that $A^m$ cannot be embedded in $A^n$.

Step 2: We do the case of a commutative, but not necessarily Noetherian, ring. First observe that, defining linear independent subsets in the usual way, the strong rank condition precisely asserts that any set of more than $n$ elements in $A^n$ is linearly dependent. Thus a ring $A$ satisfies the strong rank condition iff: for all $m > n$, any homogeneous linear system of $n$ linear equations and $m$ unknowns has a nonzero solution in $A$.

So, let $MX = 0$ be any homogeneous linear system with coefficient matrix $M = (m_{ij}), \ 1 \leq i \leq n, 1 \leq j \leq m$. We want to show that it has a nonzero solution in $A$. But the subring $A' = \mathbb{Z}[a_{ij}]$, being a quotient of a polynomial ring in finitely many variables over a Noetherian ring, is Noetherian (by the Hilbert basis theorem), so by Step 1 there is (even) a nonzero solution $(x_1,...,x_m) \in (A')^m$.

This makes one wonder if it is necessary to consider the Noetherian case separately, and it is not. Lam's second proof comes from Bourbaki's Algebra, Chapter III, §7.9, Prop. 12, page 519. [Thanks to Georges Elencwajg for tracking down the reference.] It uses the following elegant characterization of linear independence in free modules:

Theorem: A subset $\{u_1,...,u_m\}$ in $M = A^n$ is linearly independent iff: if $a \in A$ is such that $a \cdot (u_1 \wedge \ldots \wedge u_m) = 0$, then $a = 0$.

Here $u_1\wedge \ldots \wedge u_m$ is an element of the exterior power $\Lambda^m(M)$.

(I will omit the proof here; the relevant passage is reproduced on Google books.)

This gives the result right away: if $m > n$, $\Lambda^m(A^n) = 0$.

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Hi Peter, nice to find you here. I have just checked that the reference to the English edition of Bourbaki is: Algebra, Chapter III, §7.9, Prop.12, page 519. Atiyah-MacDonald's exercise 2.11 is a consequence of Bourbaki's exercise 16, page 641. Friendly greetings, Georges. – Georges Elencwajg Oct 26 '09 at 20:27