Example 91.5.9. Let $A \to B$ be a ring map. In this example we will construct an “explicit” resolution $P_\bullet$ of $B$ over $A$ of length $2$. To do this we follow the procedure of the proof of Proposition 91.5.2, see also the discussion in Remark 91.5.5.

We choose a surjection $P_0 = A[u_ i] \to B$ where $u_ i$ is a set of variables. Choose generators $f_ t \in P_0$, $t \in T$ of the ideal $\mathop{\mathrm{Ker}}(P_0 \to B)$. We choose $P_1 = A[u_ i, x_ t]$ with face maps $d_0$ and $d_1$ the unique $A$-algebra maps with $d_ j(u_ i) = u_ i$ and $d_0(x_ t) = 0$ and $d_1(x_ t) = f_ t$. The map $s_0 : P_0 \to P_1$ is the unique $A$-algebra map with $s_0(u_ i) = u_ i$. It is clear that

$P_1 \xrightarrow {d_0 - d_1} P_0 \to B \to 0$

is exact, in particular the map $(d_0, d_1) : P_1 \to P_0 \times _ B P_0$ is surjective. Thus, if $P_\bullet$ denotes the $1$-truncated simplicial $A$-algebra given by $P_0$, $P_1$, $d_0$, $d_1$, and $s_0$, then the augmentation $\text{cosk}_1(P_\bullet ) \to B$ is a trivial Kan fibration. The next step of the procedure in the proof of Proposition 91.5.2 is to choose a polynomial algebra $P_2$ and a surjection

$P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2$

Recall that

$\text{cosk}_1(P_\bullet )_2 = \{ (g_0, g_1, g_2) \in P_1^3 \mid d_0(g_0) = d_0(g_1), d_1(g_0) = d_0(g_2), d_1(g_1) = d_1(g_2)\}$

Thinking of $g_ i \in P_1$ as a polynomial in $x_ t$ the conditions are

$g_0(0) = g_1(0),\quad g_0(f_ t) = g_2(0),\quad g_1(f_ t) = g_2(f_ t)$

Thus $\text{cosk}_1(P_\bullet )_2$ contains the elements $y_ t = (x_ t, x_ t, f_ t)$ and $z_ t = (0, x_ t, x_ t)$. Every element $G$ in $\text{cosk}_1(P_\bullet )_2$ is of the form $G = H + (0, 0, g)$ where $H$ is in the image of $A[u_ i, y_ t, z_ t] \to \text{cosk}_1(P_\bullet )_2$. Here $g \in P_1$ is a polynomial with vanishing constant term such that $g(f_ t) = 0$ in $P_0$. Observe that

1. $g = x_ t x_{t'} - f_ t x_{t'}$ and

2. $g = \sum r_ t x_ t$ with $r_ t \in P_0$ if $\sum r_ t f_ t = 0$ in $P_0$

are elements of $P_1$ of the desired form. Let

$Rel = \mathop{\mathrm{Ker}}(\bigoplus \nolimits _{t \in T} P_0 \longrightarrow P_0),\quad (r_ t) \longmapsto \sum r_ tf_ t$

We set $P_2 = A[u_ i, y_ t, z_ t, v_ r, w_{t, t'}]$ where $r = (r_ t) \in Rel$, with map

$P_2 \longrightarrow \text{cosk}_1(P_\bullet )_2$

given by $y_ t \mapsto (x_ t, x_ t, f_ t)$, $z_ t \mapsto (0, x_ t, x_ t)$, $v_ r \mapsto (0, 0, \sum r_ t x_ t)$, and $w_{t, t'} \mapsto (0, 0, x_ t x_{t'} - f_ t x_{t'})$. A calculation (omitted) shows that this map is surjective. Our choice of the map displayed above determines the maps $d_0, d_1, d_2 : P_2 \to P_1$. Finally, the procedure in the proof of Proposition 91.5.2 tells us to choose the maps $s_0, s_1 : P_1 \to P_2$ lifting the two maps $P_1 \to \text{cosk}_1(P_\bullet )_2$. It is clear that we can take $s_ i$ to be the unique $A$-algebra maps determined by $s_0(x_ t) = y_ t$ and $s_1(x_ t) = z_ t$.

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