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This is part of the exercise, I'm stuck with it.

$A$ is a commutative ring with unit.

1) Suppose we have an homomorphism $\phi : A^{m} \to A^{n}$ surjective. Is true that $m \geq n $ ?

2) Suppose we have an homomorphism $\phi : A^{m} \to A^{n}$ injective. Is true that $m \leq n $ ?

user26857
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WLOG
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  • Do you know how to do the exercise if $A^m$ and $A^n$ are $k$-vector spaces? – RghtHndSd May 04 '14 at 15:07
  • @rghthndsd: yes – WLOG May 04 '14 at 15:12
  • Follow the same methodology here. Namely for (1) choose elements of the domain that map to $(1, 0, 0, ...)$, (0, 1, 0, ...)$, etc. Show that they are independent. This depends on knowing that the rank of a free module is well-defined - is this ok? – RghtHndSd May 04 '14 at 15:17
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    Duplicate of 1) http://math.stackexchange.com/questions/20178/given-a-commutative-ring-r-and-an-epimorphism-rm-to-rn-is-then-m-geq-n; 2) http://math.stackexchange.com/questions/106786/am-hookrightarrow-an-implies-m-leq-n-for-a-ring-a-neq-0 – user26857 May 04 '14 at 18:13

1 Answers1

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Hints: The answer to both is yes. For (1) tensor by the residue field of a maximal ideal.

For (2) assume, by way of contradiction, that $m > n$. Consider $A^n \subseteq A^m$ and consider $\phi\colon A^m \to A^m$ as a map into $A^m$. Let $\pi\colon A^m \to A^m$ be a projection map such that $\pi\phi = 0$. Now use Proposition 2.4 on $\phi$ so show it satisfies some relation. Then use $\pi$ and the injectivity of $\phi$ to reduce that relation to $\phi = 0$.

Jim
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