Is the first part of Eisenbud's Proposition 15.15's proof o.k? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:03:20Z http://mathoverflow.net/feeds/question/55237 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/55237/is-the-first-part-of-eisenbuds-proposition-15-15s-proof-o-k Is the first part of Eisenbud's Proposition 15.15's proof o.k? David 2011-02-12T21:14:01Z 2011-02-12T22:42:02Z <p>In the chapter on Gröbner bases from Eisenbud's "Commutative Algebra" the following statement appears as Proposition 15.15 (page 344):</p> <p><em>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</em> $h_{1},\cdots,h_{u} \in S$ <em>are such that</em> $in(h_{1}),\cdots, in(h_{u})$ <em>is a regular sequence on</em> $F/in(M)$, <em>then</em> $h_{1},\cdots, h_{u}$ <em>is a regular sequence on</em> $F/M$ <em>and</em> $in(M+(h_{1},\cdots,h_{u})F)=in(M)+\sum_{i=1}^{u}in(h_{i})F$.</p> <p>Here $S$ denotes the ring of polynomials in $r$ variables over an (infinite) field $k$. For more notation refer to the book.</p> <p>Now, it is clear that it suffices to prove the statement for $u=1$. So writing $h_{1}=h$ Eisenbud argues:</p> <p>"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."</p> <p>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). </p> <p>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?</p> http://mathoverflow.net/questions/55237/is-the-first-part-of-eisenbuds-proposition-15-15s-proof-o-k/55248#55248 Answer by Guntram for Is the first part of Eisenbud's Proposition 15.15's proof o.k? Guntram 2011-02-12T22:42:02Z 2011-02-12T22:42:02Z <p>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 $M$.</p> <p>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 $in(f)$. </p>