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I already read solutions for injectivity at MathOverflow and MathStackexchange and understand some of the solution. (Those are brilliant!) What I didn't understand is the reason we have to find another way without using the vector space structure, since isomorphism and surjective cases are solved by such way.

Let me introduce my proof using the vector space structure.

Let $A$ be a (commutative and unital) ring and $\phi:A^{m} \to A^{n}$ is injective. To show that $m \leq n$, as we can do for surjective or isomorphism case, construct $1 \otimes \phi: (A/\mathfrak{m})\otimes_{A}A^{m} \to (A/\mathfrak{m})\otimes_{A}A^{n}.$ Then this map is still injective. Also, $(A/\mathfrak{m})\otimes_{A}A^{m} \cong (A/\mathfrak{m})^{m}$ and $(A/\mathfrak{m})\otimes_{A}A^{n} \cong (A/\mathfrak{m})^{n}$ as an $A$-module. Moreover, we can regard those $(A/\mathfrak{m})^{m}$ and $(A/\mathfrak{m})^{n}$ as $(A/\mathfrak{m})$-module, and $1\otimes j$ as $A/\mathfrak{m}$-module map. Hence it suffices to show that if a linear transformation between two vector spaces $V^{m} \to V^{n}$ is injective, then $m \leq n$. And this easily follows from the fact that image of injective linear transformation has dimension $m$, thus $m \leq n$.

Is there any problem about this argument?

user124697
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    "Then this map is still injective"? – Angina Seng May 14 '19 at 04:44
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    @LordSharktheUnknown I see... We don't know whether $A/\mathfrak{m}$ is flat $A$-module or not. Oh, I did a very basic mistake. Thank you for pointing out! – user124697 May 14 '19 at 04:51
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    In most standard examples, e.g., $A=\Bbb Z$, then $A/\frak m$ is usually not flat. – Angina Seng May 14 '19 at 04:52
  • What about $B= \phi(A^m)$ and $A/\mathfrak{m} \otimes_A B$ ? It is a sub vector space of $A/\mathfrak{m}\otimes_A C$ for some full rank submodule $C$ of $ A$ ? – reuns May 14 '19 at 04:57

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