Support and tensor product doubts Some questions:

*

*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 $f \circ g$.


*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?


*Let $M$ be an $A$-module let $\textrm{Supp}(M)=\{P \in \mathrm{Spec}(A) : M_{P} \neq 0\}$, 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:
$\textrm{Supp}(\mathbb{Z}) = \{0\} \cup \{(p) : p\textrm{ is prime}\}$ 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 $\textrm{Supp}(\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}) = \{(0)\} \cup \{(p): p \textrm{ is prime}\}=\textrm{Spec}(\mathbb{Z})$.
Is this OK?
Thanks in advance
EDIT. Sorry, in 3. I meant $\textrm{Supp} (\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z})$.
 A: (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 $P\neq (2)$.
A: $\DeclareMathOperator{\supp}{supp}$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 Exercice 3.19(vi) in Atiyah–Macdonald.
A: 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 ${\bf Z}$.
