David Eisenbud's book "The Geometry of Schemes" states the following:
Proposition IV-25: Let $A$ be a Noetherian ring and $x, y \in A$; let $B$ be the Rees algebra $$ B = A[xt, yt] \subset A[t].$$ If $x, y \in A$ is a regular sequence, then $$ B \cong A[X, Y] / (yX - xY)$$ via the map $X \mapsto xt$, $Y \mapsto yt$.
Proof. First we invert $x$ and set $X' = x^{-1}X \in A[x^{-1}][X, Y]$. The element $yX' - Y \in A[x^{-1}][X, Y] = A[x^{-1}][X', Y]$ generates the kernel of the map $$A[x^{-1}][X', Y] \longrightarrow A[x^{-1}][t],$$ $$X' \mapsto t,$$ $$Y \mapsto yt.$$ Since $(yX' - xY) = (yX - Y)$ in the ring $A[x^{-1}][X, Y]$, it suffices to show that $x$ is a nonzero divisor modulo $yX - xY$ in $A[X, Y]$.
In the second part of the proof he proves that $x$ is a nonzero divisor using Koszuls homology.
Could someone provide the skeleton of the proof ?
Why we invert $x$ in the beginning and why it suffices to show that $x$ is a nonzero divisor modulo $yX - xY$ in $A[X, Y]$ ?
$\bullet$ It seems we invert $x$ is to work with localizations. So we get an isomorphism on the level of the local rings which then somehow gives us the desired isomorphism.
