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In the part (a) of Exercise 3.4, it suggests that we may use the relation: $${\text{Content}(fg)}\subset{\text{Content}(f)\text{Content}(g)}\subset{\text{rad}\left(\text{Content}(fg)\right)}$$ to deduce that if $\text{Content}(f)$ contains a nonzerodivisor of $R$, then $f$ is nonzerodivisor of $S=R[x_1,. . . ,x_r]$.

But I don't know how to do this, for example, if all the coefficients of g are in the nilpotent radic of R, assuming that fg=0, the relation above gives nothing. Is there some tips please?

user26857
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Tibeku
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    I also get stuck with that: if $; \text{Cont.}(g)\subset \sqrt0\le R;$ then I can't get a contradiction, so either this is a case Eisenbud didn't see or else there's some little detail I (we) cannot see. I am writing him right now. – DonAntonio Jul 28 '16 at 08:54
  • @DonAntonio Dedekind-Mertens' lemma and McCoy's lemma are equivalent to [$c(f)=1$ and $fg=0 \implies g=0$] which we have to prove here, but under the hypothesis of Gauss-Joyal's lemma which is arguable weaker than the first two. In other words, Eisenbud's claim is hard to believe (unless he has a proof that others missed, which I don't think). – user26857 Aug 04 '16 at 07:23
  • On the other side, Theorem 3 of this paper claims that [$c(f)=c(g)=(1)\implies c(fg)=(1)$] is equivalent to McCoy's lemma, and the first follows from $c(f)c(g)\subseteq\sqrt{c(fg)}$. But for some reasons I'm not yet convinced that Theorem 3 is correct. (The references given by the author of the paper are rather vague, so I have to check all the details. I'll do this as soon as I can.) – user26857 Aug 04 '16 at 07:42
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    Well, I already wrote to David about my doubts on this and he answered with this: "Surprising as the result seems, I think the proof sketched in the hints is correct. DE" . This is, without any doubt, the lamest, most insufficient answer I've ever received. Perhaps he's already too old, perhaps he is way too busy with other things, but that answer supports your, and already my as well, doubts about that claim in Eisenbud's book. Perhaps I'm missing something, yet that's no way to address a honest mathematical doubt. – DonAntonio Aug 04 '16 at 09:01
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    @DonAntonio As far as I can see, there is no proof sketched in the hints for this assertion. – user26857 Aug 04 '16 at 16:56
  • @DonAntonio: I think this is a proof for Eisenbud's claim. Suppose $f$ is a zerodivisor. Then from this question http://math.stackexchange.com/q/948010/178578 $af = 0$ for some $0 \ne a \in R$ i.e. every coefficient of $f$ is in $\operatorname{ann}(a)$. Then $\operatorname{Content}(f) \subseteq \operatorname{ann}(a)$, so $\operatorname{Content}(f)$ consists of zerodivisors. But I don't know how to deduce this from the relation on contents – Jay Aug 04 '16 at 21:42
  • @user26857 Thanks for your links very much. I find them really helpful for me to understand this question, although I have not yet read through the paper you gave, which seems somehow difficult for me. – Tibeku Aug 07 '16 at 06:56
  • @Jay It is indeed a proof, I am sorry that I didn't find the related questions before. Thank you. – Tibeku Aug 07 '16 at 07:33
  • @user26857 I can't see how to prove McCoy's lemma from [$c(f)=1$ and $fg=0 \Rightarrow g=0 $ ]. Could you give me a hint? -- In theorem 3 of the paper of Nasehpour (5) $\Rightarrow$ (1) is the essential part we need, right? But for this step he refers to Northcott, who is proving a general Dedekind-Mertens' lemma by relying on DM for polynomials. So it does not help us. – user60589 Aug 09 '16 at 11:04

1 Answers1

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$\DeclareMathOperator{\Ann}{Ann}$ $\DeclareMathOperator{\Ass}{Ass}$ $\DeclareMathOperator{\Content}{Content}$

I don't know if this is the actual proof which Eisenbud had in mind, especially because this proof has nothing to do with the first half of the problem, but here's a possible proof. We start with some preparations:

Lemma Let $R$ be a $\mathbb{Z}^r$-graded ring, and let $M$ be a graded $R$-module. Let $m\in M\setminus\{0\}$ and suppose $P=\Ann(m)$ is prime. Then $P$ is homogeneous and $P$ is the annihilator of a homogeneous element.

The proof of the above lemma is entirely analogous to the one for Proposition 3.12 in Eisenbud, if we endow $\mathbb{Z}^r$ a lexicographic ordering. As a corollary, we obtain:

Corollary Let $R$ be a $\mathbb{Z}^r$-graded ring. If $f\in R$ is a zerodivisor, then there is a nonzero homogeneous element $g\in R$ such that $fg=0$.

(Proof) By hypothesis, there is some nonzero $h\in R$ such that $fh=0$. By replacing $R$ with the subring of $R$ generated by the homogeneous parts of $f$ and $h$, we may assume that $R$ is finitely generated; in particular, $R$ is a Noetherian ring by the Hilbert basis theorem. Since $f$ is a zerodivisor, it is contained in some associated prime $P\in\Ass_{R}(R)$. By the above lemma, $P$ is the annihilator of a homogeneous element. The claim follows. $\blacksquare$

Now back to your question, suppose $f$ is a zero divisor. By the Corollary you can find a nonzero monomial in $S$ which kills $f$, and hence a nonzero element $r\in R$ (hmm, I should have used a symbol other than $R$ in the above corollary; anyhow) which kills $f$. But then $r\Content (f)=0$, so $\Content (f)$ contains no nonzerodivisor.

Ken
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