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Problem:

Suppose $R$ is a commutative ring and $f_1, f_2 \in R$ generate the unit ideal. Let $M$ be an $R$-module, and assume we are given elements $$m_i \in M[f_i^{-1}]$$ such that $$m_1 = m_2 \in M[f_1^{-1}f_2^{-1}].$$

Show that there exists $m \in M$ so that $m = m_i$ when you localize at $f_i$.

My attempt:

We write $m_i \in M[f_i^{-1}]$ as a fraction $$m_i = \frac{\tilde{m_i}}{f_i^{e_i}},$$ and the compatibility assumption becomes $$(f_1f_2)^N (f_2^{e_2} \tilde{m_1} - f_1^{e_1} \tilde{m_2}) = 0, N \gg 0$$ in $M$.

The assumption that $f_1$ and $f_2$ generate the unit ideal is expressed by $af_1+bf_2=1$. By replacing $f_i$ with as high a power of $f_i$ as necessary, we may assume without loss of generality that $e_1=e_2=N=1$; it is a standard trick that the Bezout equation still holds.

Let $m = a \tilde {m_1} + b \tilde {m_2}$. This definition is motivated by what it should be if $\tilde{m_i} = f_i m$.

So we wish to show that $$m = m_1 = \frac{\tilde{m_1}}{f_1} \in M[f_1^{-1}],$$ i.e., $$f_1^N(f_1m-\tilde{m_1}) = 0$$ for sufficiently large $N$. Substituting the definition of $m$ back in and simplifying using the Bezout equation gives $$f_1^N(-b f_2 \tilde{m_1} + bf_1 \tilde{m_2})=0.$$ I know from the above that $f_2 \tilde{m_1}-f_1 \tilde{m_2}$ is killed by $f_1f_2$, but I don't know that it's killed by a high power of $f_1$.

This appears to be the approach outlined in the solution. What am I missing?

2 Answers 2

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Since $af_1+bf_2=1$, we can find $c$ and $d$ such that $$cf_1^2+df_2^2=1$$

Instead of letting $m = a \tilde {m_1} + b \tilde {m_2}$, we set $$m = c f_1 \tilde {m_1} + d f_2 \tilde {m_2}$$

We need to show $$f_1^M(f_1m-\tilde{m_1}) = 0$$ for sufficiently large M.

But this is $$f_1^M(-d f_2^2 \tilde{m_1} + d f_1 f_2 \tilde{m_2})=0$$ $$f_1^M f_2(-d f_2 \tilde{m_1} + d f_1 \tilde{m_2})=0$$

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    Do you have any suggestion as to how this definition of $m$, in contrast to the one I inferred, should be motivated? – 2017-02-14
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    ... especially since I'm pretty sure the hint in the back of the book uses my definition, and not your refinement? I mean, it's a bit tricky to translate between my notation and the book's, so I could be mistaken, but that's what it looks like to me. – 2017-02-14
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    @DustanLevenstein The book says "we may suppose that $N$ is large enough" It's kind of you need to raise the power of $f_i$ twice. – 2017-02-14
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    It says "We may suppose that $N$ is large enough so that for each $i$ the element $f_i^N m_i$ is the image of some element $m_i' \in M$." In my notation, this is accomplished with $N=1$ and $m_i' = \tilde{m_i}$. – 2017-02-14
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    @DustanLevenstein The book actually takes another $M>N$ that is large enough. That's what I mean by considering the sufficiently large $N$ twice. – 2017-02-14
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    Are we looking at different versions of the book? I only see one mention of the condition on $N$, and it's the one I quoted above. – 2017-02-14
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    @DustanLevenstein The book only mentions $N$ once because it doesn't set $N=1$ at the beginning. The book sets $N$ large enough only at the very last step. If you set $N=1$ (large enough) at the beginning, then at last you need to set $N$ a little bit larger. – 2017-02-14
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    All I did was make a couple of WLOG assumptions - so my notation should just be a special case of the book's, should it not? In my notation, one can choose to forget about the $N$ at the beginning - the assumptions can be rewritten with $1$ in place of $e_1, e_2, N$, and $N$ is an unbound variable. So according to the book I'm just looking for an $N$ large enough so that $f_i^N m_i$ is the image of an element of $M$. By definition, this is accomplished with $N=1$. – 2017-02-14
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To motivate refining the definition of $m$ as WWK has done, we note that by part (a) of Eisenbud's exercise, $m$ must necessarily be unique. The original version $m = a \tilde{m_1} + b \tilde{m_2}$ may not be unique. This can be seen either on the geometry side of the picture or on the commutative algebra side:

From a sheaf-theoretic point of view, the $m_i$ define sections of a sheaf of modules on $\operatorname{Spec}(R[f_i^{-1}])$, and $\tilde{m_i}$ extends $f_i m_i$ to a global section on $\operatorname{Spec}(R) = \operatorname{Spec}(R[f_1^{-1}]) \cup \operatorname{Spec}(R[f_2^{-1}])$. This extension may include some junk on $\operatorname{Spec}(R) - \operatorname{Spec}(R[f_i^{-1}])$, so for a well-defined $m$, it makes sense to cancel the junk by multiplying by $f_i$ again.

For a more elementary way to understand this, we recall that $\tilde{m_i}$ is only well-defined up to the annihilator of $f_i^N$, for $N \gg 0$, acting on $M$. My choice to restrict the compatibility assumption on $m_1$ and $m_2$ to $$f_1f_2(f_2\tilde{m_1} - f_1 \tilde{m_2})=0$$ actually restricts what can be added to $\tilde{m_i}$ though; this equation isn't necessarily preserved by adding an annihilator for $f_i^N$ for $N \ge 2$. It is, however, preserved by adding an annihilator for $f_i$, so it is necessary (though, a priori, not sufficient) that our formula for $m$ be well-defined up to that. If we write $\operatorname{Ann}_i := \operatorname{Ann}_M(f_i)$ for this annihilator, then my original definition of $m$ yields the set $$a(\tilde{m_1} + \operatorname{Ann}_1)+b(\tilde{m_2}+\operatorname{Ann}_2),$$ which is only well-defined up to the additive subgroup $$a \operatorname{Ann}_1 + b \operatorname{Ann}_2,$$ and there's no reason to believe that this additive subgroup should be $\{0\}$. WWK's refinement yields the set $$cf_1(\tilde{m_1}+\operatorname{Ann}_1)+df_2(\tilde{m_2} + \operatorname{Ann}_2),$$ which is well-defined up to the additive subgroup $$cf_1 \operatorname{Ann}_1 + df_2 \operatorname{Ann}_2,$$ which is $\{0\}$. As I said, this doesn't a priori imply that $m$ is well-defined as a function of $m_1$ and $m_2$; it just gets us a step closer to believing that $m$ is probably well-defined. WWK's argument, of course, implies that $m$ is well-defined.