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Some questions:

1) This is proposition $3.5$ , page $39$ of Atiyah's and Macdonald's book.

Let $M$ be an $A$-module. Then $S^{-1}A \otimes_{A} M \cong S^{-1}M$ as $S^{-1}A$-modules.

So the idea is to use the universal property of the tensor product. The mapping $S^{-1}A \times M \rightarrow S^{-1}M$ given by $(a/s,m) \mapsto am/s$ is $A$-bilinear so we get an $A$-linear map $f: S^{-1}A \otimes_{A} M \rightarrow S^{-1}M$ given by $f((a/s) \times m) = am/s$.

Then the rest of the proof shows the map is injective and surjective. My question is, can't we simply give the inverse? let $g: S^{-1}M \rightarrow S^{-1}A \otimes_{A} M$ given by $g(m/s) = 1/s \otimes m$.

Then we have:

$(g \circ f)((a/s) \otimes m)=g(f(a/s))=g(am/s)=1/s \otimes am = a/s \otimes m$.

Similarly for $f \circ g$.

2) This is exercise $4$ (same book) page $44$: let $f: A \rightarrow B$ be a ring homomorphism and let $S$ be a multiplicatively closed subset of $A$. Let $T=f(S)$. Show that $S^{-1}B \cong T^{-1}B$ as $S^{-1}A$-modules.

My question here: why $S^{-1}B$ makes sense? I thought that we always needed that the multiplicatively closed set is a subset of the ring $B$, here $S \subseteq A$, why we can take the localization then?

3) Let $M$ be an $A$-module let $\textrm{Supp}(M)=\{P \in Spec(A) : M_{P} \neq 0\}$, the support of $M$.

I want to compute $\textrm{Supp}(\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z})$ (viewed as a $\mathbb{Z}$-module).

I know that in general $\textrm{Supp}(M_{1} \oplus M_{2}) = \textrm{Supp}(M_{1}) \cup \textrm{Supp}(M_{2})$.

So two doubts here:

$\textrm{Supp}(\mathbb{Z}) = \{0\} \cup \{(p) : \textrm{p is prime}\}$ right? because if we localize at such prime ideals we don't get the trivial module.

On the other hand I think $\textrm{Spec}(\mathbb{Z}/2\mathbb{Z}) = \{(0)\}$ and if we localize $\mathbb{Z}/2\mathbb{Z}$ at $(0)$ we get again $\mathbb{Z}/2\mathbb{Z}$ because $\mathbb{Z}/2\mathbb{Z}$ is a field right?

So in conclusion $\textrm{Supp}(\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}) = \{(0)\} \cup \{(p): \textrm{ p is prime}\}=\textrm{Spec}(\mathbb{Z})$.

Is this OK?

Thanks in advance

\textbf{EDIT}: Sorry I meant $\textrm{Supp} (\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z})$.

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3 Answers

up vote 3 down vote accepted

$\DeclareMathOperator{\supp}{supp}$For (1), I think that this is a fine way to do it, although you should be careful when defining maps out of something like $S^{-1}M$. Unfortunately, most books omit the universal property of a localised module, but it's a good exercise to figure out what it is. Less elegantly, you could define a map of sets $S \times M \to S^{-1}A \otimes M$ and check that it respects the equivalence relation used in the construction.

I think you have mixed up the direct sum and the tensor product in your attempt to solve (3). I agree with your computation of $\supp(\mathbf{Z})$, but as Bruno says that turns out to be unnecessary for this problem. Also note that you are not trying to find the support of $\mathbf{Z}/2\mathbf{Z}$ as a module over itself; if you were, then I would agree with your answer.

A fact that you might like, which would be overkill for this problem: if $M, N$ are finitely generated modules over $A$, then it is true that $\supp(M \otimes N) = \supp(M) \cap \supp(N)$. See Ex. 3.19(iv) in Atiyah-Macdonald.

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thanks, just corrected it, can you please have a look? –  user6495 Jul 18 '11 at 3:23
That certainly changes things, since by your remark you can calculate $\operatorname{supp}(\mathbf{Z} \oplus M)$ without knowing which $\mathbf{Z}$-module $M$ is. –  Dylan Moreland Jul 18 '11 at 3:44
supp(Z) is just spec(Z) right? –  user6495 Jul 18 '11 at 16:06
@user6495 Yes. I don't think it's too hard to show that the support of any ring $A$ as a module over itself is all of $\operatorname{Spec} A$. –  Dylan Moreland Jul 18 '11 at 16:24
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(1) You can certainly do that as well. You just have to check that your map $g$ is well-defined, and that it's a morphism of $S^{-1}A$-modules (remember that the expression $m/s$ is not unique).

(2) Here $B$ is given the canonical structure of an $A$-module, $a\cdot x = f(a)x$.

(3) Since $\mathbb{Z} \otimes_\mathbb{Z} \mathbb{Z}/2\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z}$ as $\mathbb{Z}$-modules, you have $\text{Supp}(\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}) = \text{Supp}(\mathbb{Z}/2\mathbb{Z})$. For a prime $P \in \mbox{Spec }\mathbb{Z}$, the localization $(\mathbb{Z}/2\mathbb{Z})_P$ is $0$ if and only if $P\neq (2)$.

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thanks , I'm aware $B$ is an $A$-module via restriction of scalars, but why $S$ is a subset of $B$? don't we need that the multiplicative subset is a subset of the ring? –  user6495 Jul 18 '11 at 3:31
Here we're talking about the localization of $B$ as an $A$-module, with the $A$-module structure that I pointed out. –  Bruno Joyal Jul 18 '11 at 4:30
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For 3), ${\bf Z} \otimes {\bf Z}/2$ is the same thing as ${\bf Z}/2$, so it support is all of $Spec({\bf Z}/2)$, which is the 0 ideal as you say. But you also seem to be mixing up the 0 ideal in ${\bf Z}/2$ and the one in ${\bf Z}$.

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But the support of $\mathbb{Z}/2$, as a $\mathbb{Z}$-module, is not the same thing as the support of $\mathbb{Z}/2$ as a $\mathbb{Z}/2$-module... –  Bruno Joyal Jul 17 '11 at 21:25
It is the same if you identify $Spec({\bf Z}/2)$ as a subset of $Spec({\bf Z})$. –  Steven Sam Jul 18 '11 at 3:29
Better to mix up the zero ideal of $\mathbf{Z}/2\mathbf{Z}$ and the prime $(2)$ of $\mathbf{Z}$. –  Dylan Moreland Jul 18 '11 at 3:40
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