This problem addresses the same question that has been asked in $A^m\hookrightarrow A^n$ implies $m\leq n$ for a ring $A\neq 0$, which is exercise 2.11 of Atiyah and Macdonald's Introduction to Commutative Algebra. The reason I am posting this again is because although the other threads provided a complete answer, there are additional questions that arise from the problem that are of concern to me. Specifically, I would like to know whether the following is true:

Problem. Let $f:A^m\rightarrow A^n$ be a monomorphism of $A$-modules with $A$ a non-zero commutative ring. Let $I\subset A$ be a maximal ideal in $A$. Is the following map injective $$f\otimes 1:A^m\otimes_A(A/I)\rightarrow A^n\otimes_A(A/I),\enspace (f\otimes 1)(a\otimes \bar{b})=f(a)\otimes \bar{b}\ ? $$

Discussion\Solution: Manos has pointed out the error in my proof, see the EDIT section below for additional discussion I know that in general tensor products are right exact, and so if we replace monomorphism by epimorphism then then statement always holds. Unfortunately we cannot use exactness here since the tensor product is not in general left exact (if $A/I$ is a flat module, we can use exactness). The reason I am considering the above homomorphism is because it is a $A/I$-linear transformation of $A/I$ -vector spaces, and thus if the result holds then $m\leq n$ since no injective linear transformation sends a vector space of higher dimension to one of lower dimension.

Set $k=A/I$. For $r=m,n$ we have that $A^r\otimes_A k$ is a $k$-vector space with basis $\{e_i\otimes 1: 1\leq i \leq r\}$. I will denote elements in $k$ by $[a]$ with $a\in A$.

Suppose that $\sum_{i=1}^m[\lambda_i](e_i\otimes 1)\in Ker(f\otimes 1)$. Then: \begin{align} (f\otimes 1)\left(\sum_{i=1}^m[\lambda_i](e_i\otimes 1) \right)&=(f\otimes 1)\left(\sum_{i=1}^m(e_i\otimes [\lambda_i]) \right)\\ &=\sum_{i=1}^m(f(e_i)\otimes [\lambda_i])\enspace (*)\\ \end{align}
For each $i$, $f(e_i)\otimes [\lambda_i]=f(e_i)\otimes \lambda_i\cdot [1]=\lambda_i f(e_i)\otimes 1=f(\lambda_ie_i)\otimes 1$ since $f$ is a homomorphism of $A$-modules (and by how the $A$ module structure on $A/I$ is defined). Then we have: $$(*)=\sum_{i=1}^m(f(\lambda_ie_i)\otimes 1)=\left(\sum_{i=1}^m f(\lambda_ie_i) \right)\otimes 1 =f\left(\sum_{i=1}^m \lambda_ie_i \right)\otimes 1 $$ Again, this holds since $f$ is a homomorphism of $A$-modules. Since our initial assumption was that $\sum_{i=1}^m[\lambda_i](e_i\otimes 1)\in Ker(f\otimes 1)$, $$f\left(\sum_{i=1}^m \lambda_ie_i \right)\otimes 1 =0 $$

From this it would be nice to conclude that each $\lambda_i\in I$ which then implies that $\sum_{i=1}^m[\lambda_i](e_i\otimes 1)=0$. This is the part I am not sure about. Clearly by the injectivity of $f$, $$f\left(\sum_{i=1}^m \lambda_ie_i \right)=0\iff \lambda_i=0\enspace \forall i $$ So we can omit this case. Thus $f\left(\sum_{i=1}^m \lambda_ie_i \right)=a\neq 0 \in A^n$. The only other time I can see the tensor $a\otimes 1$ being zero is if $a\in I\cdot A^n$, which should then imply that $ \sum_{i=1}^m \lambda_ie_i \in I\cdot A^m\implies \lambda_i\in I$ for all $i$. But I am not sure about this last statement or that this is the only time an element of this $k$-vector space is $0$.

If someone could read through this and help me out that would be great!

EDIT. I am really just trying to come up with a proof that doesn't use Cayley-Hamilton for modules or determinants/linear algebra for modules (we have not discussed either in this class). We have covered basically the first two Chapters of Atiyah and Macdonald, but we skipped the Cayley-Hamilton theorem for modules and have not done any computations involving the matrix theory of modules. As Martin Brandenburg has kindly mentioned, it is possible that this question was given to us with the assumption that Cayley-Hamilton would be covered, which seems to seem very possible. The proofs in http://www.mathoverflow.net/questions/136, http://www.mathoverflow.net/questions/30860 or http://www.math.stackexchange.com/questions/106786 are very well done and I in particular like the Cayley-Hamilton one, but I was hoping to come up with a proof that only used concepts that have been covered in class, such as the basic properties of tensor products, exact sequences, modules (excluding Cayley-Hamilton and matrix theory) and rings.

  • 2
    $\begingroup$ Tensor $2 : \mathbb{Z} \to \mathbb{Z}$ with $\mathbb{F}_2$, it gets zero. General advice: Think about examples and structural arguments before diving into long element calculations ... $\endgroup$ Feb 6, 2014 at 0:05
  • $\begingroup$ can you elaborate on your comment? I am not sure I understand what you are saying. Edit: Nevermind I see it. I don't see where the proof goes wrong though, that is what I am interested in finding out. $\endgroup$
    – CWsl2
    Feb 6, 2014 at 0:07
  • $\begingroup$ Okay thanks I see now that it just goes wrong in that last step. I am still interested in a proof that does not use Cayley-Hamilton or matrix theory though $\endgroup$
    – CWsl2
    Feb 6, 2014 at 0:19
  • 1
    $\begingroup$ It is also possible that your professor / assistant didn't realize before if the exercise is solvable with the methods you have covered so far. Some assistants (not sure if this is the right word in english; I mean the one who helps to create the exercises and works together with the professor) just copy exercises from books. $\endgroup$ Feb 6, 2014 at 0:38
  • 3
    $\begingroup$ Have you seen Keith Conrad's proof using exterior powers? It is Cor. 5.11 in math.uconn.edu/~kconrad/blurbs/linmultialg/extmod.pdf. The main ingredient is Thm. 5.10, which is a consequence of Thm. 4.2., which is just the usual basis of the exterior power of a free module. Notice that this is a coordinate-free version of the proof using minors and matrices. In my opinion this is the most elegant proof. $\endgroup$ Feb 6, 2014 at 0:49

3 Answers 3


Let me reproduce the proof given by Keith Conrad in his notes on exterior powers. I will present it in a self-contained way. We really only need the definition of the exterior power $$\Lambda^n M := M^{\otimes n} / \langle m_1 \otimes \dotsc \otimes m_n : m_i = m_j \text{ for some } i \neq j \rangle.$$

Theorem: Let $M,N$ be free $A$-modules, and $\phi : M \to N$ be an injective homomorphism. Then for every $n \geq 0$ also $\Lambda^n \phi : \Lambda^n M \to \Lambda^n N$ is an injective homomorphism.

This implies that $A^n$ does not embed into $A^m$ for $n>m$, since otherwise$^\textbf{1}$ $A \cong \Lambda^n A^n $ embeds into $\Lambda^n A^m = 0$.

Proof of the Theorem. Since free modules are flat, we see by induction that $\phi^{\otimes n} : M^{\otimes n} \to N^{\otimes n}$ is injective. The map $M^n \to M^{\otimes n}$ which maps $(m_1,\dotsc,m_n)$ to $\sum_{\sigma \in \mathfrak{S}_n} \mathrm{sgn}(\sigma) \cdot m_{\sigma(1)} \otimes \dotsc \otimes m_{\sigma(n)}$ is multilinear alternating (check this!), hence lifts to a linear map $\alpha_M : \Lambda^n M \to M^{\otimes n}$. The diagram $$\begin{array}{c} \Lambda^n M & \xrightarrow{\alpha_M} & M^{\otimes n} \\ \downarrow && \downarrow \\ \Lambda^n N & \xrightarrow{\alpha_N} & N^{\otimes n} \end{array}$$ is commutative. Therefore, it suffices to prove that $\alpha_M$ is injective (because then also $\phi^{\otimes n} \circ \alpha_M = \alpha_N \circ \Lambda^n \phi$ and therefore $\Lambda^n \phi$ is injective).

For this choose an ordered basis $B$ of $M$. Then clearly $\{b_1 \wedge \dotsc \wedge b_n : b_1 < \dotsc < b_n \text{ in } B\}$ is a generating set$^\textbf{1}$ of $\Lambda^n M$. Let $\omega = \sum_{b_1 < \dotsc < b_n} \lambda(b_1,\dotsc,b_n) \cdot (b_1 \wedge \dotsc \wedge b_n)$ be an element in the kernel of $\alpha_M$. Then in $M^{\otimes n}$ we have $$0 = \sum_{\sigma \in \mathfrak{S}_n} \sum_{b_1 < \dotsc < b_n} \lambda(b_1,\dotsc,b_n) \cdot \mathrm{sgn}(\sigma) \cdot b_{\sigma(1)} \otimes \dotsc \otimes b_{\sigma(n)}.$$ Since $\{b_{\sigma(1)} \otimes \dotsc \otimes b_{\sigma(n)} : \sigma \in \mathfrak{S}_n, b_1 < \dotsc < b_n \}$ is a subset of the basis $\{b_1 \otimes \dotsc \otimes b_n : b_i \in B\}$ of $M^{\otimes n}$, we conclude that $\lambda(b_1,\dotsc,b_n) \cdot \mathrm{sgn}(\sigma)$ and hence $\lambda(b_1,\dotsc,b_n)$ vanishes for all $b_1<\dotsc<b_n$. Thus, $\omega=0$. $\square$

$^\textbf{1}$ This argument actually shows that $\{b_1 \wedge \dotsc \wedge b_n : b_1 < \dotsc < b_n \text{ in } B\}$ is a basis of $\Lambda^n M$. In particular $A \cong \Lambda^n A^n$ and $0 = \Lambda^n A^m$ for $n>m$.

  • $\begingroup$ Very nicely put- I think this is my favourite argument so far and you have made it more clear. $\endgroup$
    – CWsl2
    Feb 6, 2014 at 1:23
  • $\begingroup$ Beautiful argument Martin using exterior powers! $\endgroup$
    – user38268
    Feb 6, 2014 at 15:42

If $f(\sum_{i=1}^m \lambda_i e_i) \otimes 1 =0$, then indeed we must have that $f(\sum_{i=1}^m \lambda_i e_i) \in I A^n$. The error in your proof is that this last relation does not imply that $\lambda_i \in I$.


One can see the error also in another way. Since $A^n$ is projective, hence flat, we have $\def\Tor{\operatorname{Tor}}\Tor^A_1(A^n,X)=0$ for all modules $X$; similarly, $\Tor^A_1(X,A)=0$. If you tensor by $A/I$ the exact sequence $$ 0\to A^m\to A^n\to C\to 0 $$ where $C$ is the cokernel of your map, you get the exact sequence $$ 0=\Tor^A_1(A^n,A/I)\to\Tor^A_1(C,A/I)\to A^m\otimes A/I\to A^n\otimes A/I \to C\otimes A/I\to 0 $$ You can compute $\Tor^A_1(C,A/I)$ by tensoring by $C$ the exact sequence $0\to I\to A\to A/I\to 0$, which gives $$ 0=\Tor^A_1(C,A)\to\Tor^A_1(C,A/I)\to C\otimes I\to C\otimes A\to C\otimes A/I\to 0 $$ Now you see that $$ \Tor^A_1(C,A/I)=\ker(C\otimes I\to CI) $$ because the epimorphism $C\otimes A\to C\otimes A/I$ is the canonical projection $C\to C/CI$ under cover, so its kernel is isomorphic to $CI$. The morphism $C\otimes I\to CI$ is not necessarily injective.

  • $\begingroup$ Oh, when I started to read the answer, I was hoping to find an alternative proof using Tor functors. :) $\endgroup$ Feb 6, 2014 at 20:28
  • $\begingroup$ @MartinBrandenburg Only if one is able to find a suitable maximal ideal $I$ such that $\operatorname{Tor}^A_1(C,A/I)=0$. This might be possible. $\endgroup$
    – egreg
    Feb 6, 2014 at 20:35

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