Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $M \otimes P$.

share|improve this question
add comment

1 Answer 1

up vote 2 down vote accepted

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)).

share|improve this answer
    
Awesome, thank you very much! –  Matt N. Jun 11 '12 at 13:47
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.