Is the first part of Eisenbud's Proposition 15.15's proof o.k?

Question

In the chapter on Gröbner bases from Eisenbud's "Commutative Algebra" the following statement appears as Proposition 15.15 (page 344):

Let $F$ be a free $S$ module with basis and monomial order compatible with a given monomial order on $S$. If $M\subset F$ is any submodule and $h_{1},\cdots,h_{u} \in S$ are such that $in(h_{1}),\cdots, in(h_{u})$ is a regular sequence on $F/in(M)$, then $h_{1},\cdots, h_{u}$ is a regular sequence on $F/M$ and $in(M+(h_{1},\cdots,h_{u})F)=in(M)+\sum_{i=1}^{u}in(h_{i})F$.

Here $S$ denotes the ring of polynomials in $r$ variables over an (infinite) field $k$. For more notation refer to the book.

Now, it is clear that it suffices to prove the statement for $u=1$. So writing $h_{1}=h$ Eisenbud argues:

"Suppose $hf\in M$ for some $f\in F$. We must prove that $f\in M$, and we may do induction on the size of $in(f)$. We have $in(hf)=in(h)in(f)\in in(M)$ [there is a typo in the book here], so by our hypothesis $in(f)\in in(M)$. Thus $h(f-in(f))\in M$, and by our induction $f-in(f)\in M$, so we are done."

My question is: why do one has that $h(f-in(f))\in M$? This is clearly equivalent to $h (in(f))\in M$, but I don't see why this holds. The last assertion presents a similar problem (it seems for me that the assumption $in(M)\subset M$ is behind all of this, perhaps the orders on $S$ and/or $F$ must fulfill some additional property).

The thing seems too elementary that I don't dare to look for counter examples or alternative proofs (do you know some?) without knowing more opinions, and I don't feel like ignoring it. Can somebody give her/his opinion please?

Answer 1

Indeed, $h(f-in(f))\in M$ does not hold in general. As an example, take $S=F=k[x,y], M=\langle x+y\rangle, h=y$ and any monomial order. Then $h \cdot (x+y) \in M$, but neither $hx$ nor $hy$ is in $M$.

The proof is easily fixed, though. After ".., so by our hypothesis, $in(f) \in in(M)$" continue as follows. Let $m \in M$ such that $in(m)=in(f)$. Then $h(f-m) \in M$, and $f-m$ has a smaller initial term, and we conclude by induction on $in(f)$.