Atiyah Macdonald proposition 2.19 iv)->iii) I have two questions about Atiyah Macdonald proposition 2.19 iv)->iii):
1.

iv) If $f: M' \to M$ is injective and $M, M'$ are finitely generated, then $f \otimes 1: M' \otimes N \to M \otimes N$ is injective.

Is $M, M'$ a mistake for $N, M'$? If not, where is it used that $M$ is finitely generated?
2.

Since $M_0$ and $M_0'$ are finitely generated,$f_0 \otimes 1$ is injective

Why is $f_0 \otimes 1$ injective?
 A: Let me write the proof of $\rm iv) \Rightarrow iii)$ as is in the text (with a couple of typos fixed):

Let $f \colon M’ \to M$ be injective and let $u = \sum x_i’ \otimes y_i \in \ker(f \otimes 1)$, so that $\sum f(x_i’) \otimes y_i = 0$ in $M \otimes N$. Let $M_0’$ be the submodule of $M’$ generated by the $x_i’$ and let $u_0$ denote $\sum x_i’ \otimes y_i$ as an element of $M_0’ \otimes N$. By (2.13) there exists a finitely generated submodule $M_0$ of $M$ containing $f(M_0’)$ and such that $\sum f(x_i’) \otimes y_i = 0$ as an element of $M_0 \otimes N$. If $f_0 \colon M_0’ \to M_0$ is the restriction of $f$, this means that $(f_0 \otimes 1)(u_0)=0$. Since $M_0$ and $M_0’$ are finitely generated, $f_0 \otimes 1$ is injective and therefore $u_0=0$, hence $u=0$.

Now:

*

*First, read the statement $\rm iv)$ as “If $f$ any injective linear map between finitely generated $A$-modules, then $f \otimes 1$ is injective”. Obviously, with $1$ being the identity map $N \to N$.

*In the definition of $u$, the summation $\sum$ reads as $\sum_{i=1}^n$, where $n$ is some positive integer. In other words, there is just a finite number of $x_i’$. Therefore, $M_0’$ is finitely generated by definition.

*Thus, $f_0$ is an injective linear map between finitely generated $A$-modules, so, by $\rm iv)$, $f_0 \otimes 1$ is injective as well.

