Intereting Posts

Existence of valuation rings in an algebraic function field of one variable
Method for coming up with consecutive integers not relatively prime to $(100!)$
Square roots — positive and negative
Result due to Cohn, unique division ring whose unit group is a given group?
An application of the General Lebesgue Dominated convergence theorem
Factoring morphisms in abelian categories
Proving $\left( \sum_{n=-\infty}^{\infty} e^{-\pi n^2} \right)^2= 1 + 4 \sum_{n=0}^{\infty} \frac{(-1)^n}{e^{(2n+1)\pi} – 1}$
Show that A has the same cardinality as $A\setminus\{x,y,z\}$
Nonzero $f \in C()$ for which $\int_0^1 f(x)x^n dx = 0$ for all $n$
universal property in quotient topology
What are k-cycles?
Explanation of Lagrange Interpolating Polynomial
A finite Monoid $M$ is a group if and only if it has only one idempotent element
I have a problem understanding the proof of Rencontres numbers (Derangements)
Find equation of ellipse given two tangent lines at given points and a point on ellipse

My question refers to the proof of the second of the following lemma given in Cohen-Macaulay rings by Bruns and Herzog.

Lemma 1.3.4 (Bruns and Herzog): Let $(R,\mathfrak m,k)$ be a local ring, and $\phi:F \rightarrow G$ a homomorphism of finite $R$-modules. Suppose that $F$ is free, and let $M$ be an $R$-module with $m \in \operatorname{Ass}(M)$. Suppose that $\phi \otimes M$ is injective. Then

(a) $\phi \otimes k$ is injective

(b) if $G$ is a free $R$-module, then $\phi$ is injective, and $\phi(F)$ is a free direct summand of $G$.

I can see why (a) is true. But for (b), Bruns and Herzog write: “one notes that its conclusion is equivalent to the injectivity of $\phi \otimes k$. This is an easy consequence of Nakayama’s lemma.”

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Question 1: I find this statement confusing, since if the conclusion of (b) is equivalent to the injectivity of $\phi \otimes k$, which we already proved in (a), then why do we need the extra assumption that $G$ is free?

Question 2: What would be a proof of (b)?

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Let $(R,\mathfrak m,k)$ be a local ring, and $\phi:F \rightarrow G$ a homomorphism of finite free $R$-modules. Then $\phi \otimes k$ is injective iff $\phi$ is injective, and $\phi(F)$ is a free direct summand of $G$.

“$\Leftarrow$” This is not difficult and I leave it to you.

“$\Rightarrow$” Let $\{e_1,\dots,e_m\}\subset F$ be an $R$-basis. (In the following we denote $\phi \otimes k$ by $\overline{\phi}$, and $G\otimes k$ by $\overline G$.) Then $\overline{\phi}(\overline e_1),\dots,\overline{\phi}(\overline e_m)$ are linearly independent over $k$ and we can find $\{\overline g_{m+1},\dots,\overline g_n\}\subset\overline G$ such that $\{\overline{\phi}(\overline e_1),\dots,\overline{\phi}(\overline e_m),\overline g_{m+1},\dots,\overline g_n\}$ is a $k$-basis of $\overline G$. Nakayama’s Lemma shows that $\{\phi(e_1),\dots,\phi(e_m),g_{m+1},\dots,g_n\}$ is a minimal system of generators for $G$. Since $\hbox{rank}\ G=\dim_k\overline G=n$, it follows that $\{\phi(e_1),\dots,\phi(e_m),g_{m+1},\dots,g_n\}$ is an $R$-basis of $G$. This shows that $G=\phi(F)\oplus\langle g_{m+1},\dots,g_n\rangle$ and $\phi $ injective.

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