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I'd like to prove (i) implies (ii) where:

(i) Whenever $f: A \to B$ is injective and $A,B$ are finitely generated then $f \otimes \operatorname{id}: A \otimes P \to B \otimes P$ is injective.

(ii) If $f: M \to N$ is injective then $f \otimes \operatorname{id}: M \otimes P \to N \otimes P$ for arbitrary $R$-modules $M,N$.

Can you tell me if this is correct (I'm spelling out all the details because I want to be sure that I understand what I'm doing):

Let $M,N$ be arbitrary modules and let $f: M \to N$ be injective. Assume that $f \otimes \operatorname{id}$ is not injective so that there is an element $u = \sum_{i=1}^n m_i \otimes p_i$ in $M \otimes P$ such that $u \neq 0$ and $(f \otimes \operatorname{id}) (u) = \sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $N \otimes P$.

Let $M_0$ be the module generated by $m_1, \dots , m_n$.

$\sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $N \otimes P$ hence by proposition (2.13) in Atiyah-Macdonald we get finitely generated submodules $X \subset N$,$Y \subset P$ such that $\sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $X \otimes Y$. Let $S$ be the (finite) set generating $X$.

Let $Z$ be the module generated by $S$ and $f(m_1), \dots , f(m_n)$. Then $Z$ is finitely generated and contains $f(M_0)$. Since $f : M \to N$ is injective we hence have that $f\mid_{M_0} : M_0 \to Z$ is injective and well-defined. By (i) we get that $f\mid_{M_0} \otimes \operatorname{id}$ is injective.

We have $(f\mid_{M_0} \otimes \operatorname{id})(u) = \sum_{i=1}^n f(m_i) \otimes p_i = 0$ in $X \otimes Y \subset Z \otimes Y$. Hence by injectivity of $f\mid_{M_0} \otimes \operatorname{id} : M_0 \otimes Y \to Z \otimes Y$ we get that $u = 0$ in $M_0 \otimes Y \subset M \otimes P$ and hence in $M \otimes P$.

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The main idea of the proof is correct.

However, I don't understand why you want to assume that $f \otimes \operatorname{id}$ is not injective. Basically you take a $u \in M\otimes P$ such that $f \otimes \operatorname{id} (u)=0$ and then you show that $u=0$. So this is a straight-forward proof - no contraposition or contradiction is needed.

But this is only a minor thing. As I said above the remaining should be correct.

Note that an important point here is that zero stays zero if you "go up", i.e. $M'\subset M,N' \subset N$ and $u=0 \in M'\otimes N' \Rightarrow u=0 \in M\otimes N$. But if you "go down" with a non-zero-element then it may become zero, i.e. $M'\subset M,N' \subset N$ and $0 \neq u \in M'\otimes N'$ then not necessarily $0 \neq u \in M\otimes N$. (cf. Atiyah-Macdonald the remark ii) right before (2.13)).

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  • $\begingroup$ Awesome, thank you very much! $\endgroup$ – Rudy the Reindeer Jun 11 '12 at 13:47

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