How to prove $ ((S\times_A T)(1))_{(f\otimes g)}\cong S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}} T(1)_{(g)}$? I'm currently trying to solve Exercise 5.11 in chapter 2 of Hartshorne: Let $S,T$ be $\mathbb{Z}_{\geq 0}$-graded rings with $S_0=T_0=A$, and define their Cartesian product $S\times_A T$ to be
$$
S\times_A T=\bigoplus_{d\geq 0}S_d\otimes_A T_d.
$$
I already showed that $\operatorname{Proj}(S\times_A T)\cong \operatorname{Proj}(S)\times_{\operatorname{Spec}(A)}\operatorname{Proj}(T)$. Now I am trying to show that
$$
\mathcal{O}_{\operatorname{Proj}(S\times_A T)}\cong p_1^{*}\mathcal{O}_{\operatorname{Proj}(S)}\otimes p_2^{*}\mathcal{O}_{\operatorname{Proj}(T)}
$$
where $p_1$ and $p_2$ are the structure maps making $\operatorname{Proj}(S\times_A T)$ the product of $\operatorname{Proj}(S)$ and $\operatorname{Proj}(T)$. I'm stuck at the following very specific part of the proof:
Is it true that
$$
((S\times_A T)(1))_{(f\otimes g)}\cong S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}} T(1)_{(g)}
$$
For all $f\in S_d$ and $g\in T_d$ where $d\in\mathbb{Z}_{\geq 1}$ is arbitrary? I constructed a map going from right to left by mapping $s/f^n\otimes[(s'\otimes t')/(f\otimes g)^n]\otimes t/g^n$ to $((ss')\otimes (tt'))/(f\otimes g)^{2n}$ and showed that it is surjective. However, I'm struggling to prove that the map is injective.
Edit: As suggested by KReiser, I tried to directly construct an inverse. However, I'm only able to do this under the extra assumption that $S$ and $T$ are generated by $S_1$ resp. $T_1$ as $A$-algebras, so that we can assume that $f$ and $g$ are of degree $1$. Here's why.
To justify the well-definedness of the map
$$
((S\times_A T)(1))_{(f\otimes g)}\to S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}}T(1)_{(g)}\\
(s\otimes t)/(f\otimes g)^n\mapsto (s/f^n)\otimes 1\otimes (t/g^n)
$$
we would like to start by defining it on the homogeneous pieces of $(S\times_A T)(1)$ and then pass to the localisation. However, for this to work, we would have to relax the target to $S(1)_{f}\otimes_{S_{(f)}}(S\times_A T)_{f\otimes g}\otimes_{T_{(g)}}T(1)_{g}$, and only afterwards show that the image is contained in $S(1)_{(f)}\otimes_{S_{(f)}}(S\times_A T)_{(f\otimes g)}\otimes_{T_{(g)}}T(1)_{(g)}$, because otherwise, the map
$$
s\otimes t\in (S\times_A T)(1)\mapsto (s/1)\otimes 1\otimes (t/1)
$$
wouldn't be well-defined. So at this point we have a map of abelian groups $(S\times_A T)(1)\to S(1)_{f}\otimes_{S_{(f)}}(S\times_A T)_{f\otimes g}\otimes_{T_{(g)}}T(1)_{g}$. Now to pass to the localisation on the source, as we have only a morphism of abelian groups at this point, we would have to justify by hand why it is well defined. That is, we want to show that if $(s\otimes t)/(f\otimes g)^n$ is equal to $0$ inside $(S\times_A T)(1)_{f\otimes g}$, then $(s/1)\otimes 1\otimes (t/1)$ is $0$ inside $S(1)_{f}\otimes_{S_{(f)}}(S\times_A T)_{f\otimes g}\otimes_{T_{(g)}}T(1)_{g}$. If we now suppose that $f$ and $g$ are of degree $1$, then this can be done: if $e$ is the degree of $s$ and $t$, then
$$
(s/1)\otimes 1\otimes (t/1)=((s/f^{e+1})sf^{e+1})\otimes 1\otimes ((t/g^{e+1})tg^{e+1})=\\
=(sf^{e+1}/1)\otimes ((s\otimes t)/(f\otimes g)^{e+1})\otimes (tg^{e+1}/1)=0
$$
because of the middle term. However, I don't see how to do this without assuming that $f$ and $g$ are of degree $1$.
 A: How about constructing an inverse directly? If $(s\otimes t)/(f\otimes g)^d$ is an element of $(S\times_A T)(1)_{(f\otimes g)}$, send it to $s/f^d \otimes 1 \otimes t/g^d$. To see that this is mutually inverse with your map, we just need to do a little calculation. The first composition is relatively straightforward:
$$(s\otimes t)/(f\otimes g)^d \mapsto s/f^d \otimes 1 \otimes t/g^d = s/f^d \otimes [(f\otimes g)^d/(f\otimes g)^d] \otimes t/g^d \\ \mapsto (sf^d\otimes tg^d)/(f\otimes g)^{2d} = (s\otimes t)/(f\otimes g)^d.$$
The second direction requires a small tensor manipulation:
$$s/f^n \otimes [(s'\otimes t')/(f\otimes g)^n] \otimes t/g^n \mapsto (ss'\otimes tt')/(f\otimes g)^{2n} \\\mapsto ss'/f^{2n} \otimes 1 \otimes tt'/g^{2n} = s/f^n \otimes [(s'/f^n \otimes 1)\cdot (1\otimes t'/g^n)] \otimes t/g^n$$
where we used that $s'/f^n$ and $t'/g^n$ are actually elements of $S_{(f)}$ and $T_{(g)}$ so they can be pulled across the tensor product.
The lesson here is that sometimes it can be easier to write down an inverse map than to check bijectivity directly.

Edit: Since your troubles aren't quite resolved, let's write down a more explicit construction and check we don't need $f,g$ of degree one, just that they're the same degree.
Start with the map $S_d\times T_d \to S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_g$ defined by $(s,t)\mapsto s\otimes 1 \otimes t$. This is $A$-bilinear, as $A$ sits naturally inside the degree zero parts of $S_{(f)}$, $T_{(g)}$, and $(S\times_A T)_{f\otimes g}$, and the map clearly respects addition. So we get that our map factors through $S_d\otimes_A T_d$. Taking direct sums, we get a map $S\times_A T\to S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_g$ defined on pure tensors by $s\otimes t\mapsto s\otimes 1 \otimes t$.
Next, we check that the action of $f\otimes g$ on $S\times_A T$ is invertible on the target: $f\otimes 1 \otimes g$ has inverse $1/f\otimes 1 \otimes 1/g$ in the target, so our map factors through $(S\times_A T)_{f\otimes g}$ and we get a map $$(S\times_A T)_{f\otimes g} \to S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_g.$$
It remains to check that the homogeneous elements of total degree one in $(S\times_A T)_{f\otimes g}$ map to homogeneous elements of the appropriate degree in $S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{f\otimes g} \otimes_{T_{(g)}} T(1)_{g}$. If $(s\otimes t)/(f\otimes g)^n$ is of homogeneous degree one, then both $s/f^n$ and $t/g^n$ are too, so $s/f^n\otimes 1 \otimes t/g^n$ lies in $S(1)_f \otimes_{S_{(f)}} (S\times_A T)_{(f\otimes g)} \otimes_{T_{(g)}} T(1)_{(g)}$ and we've won.
