Lemma 52.4.2. Let $I = (f_1, \ldots , f_ r)$ be an ideal of a Noetherian ring $A$ with $\text{cd}(A, I) = 1$. Let $c \geq 1$ and $\varphi _ j : I^ c \to A$, $j = 1, \ldots , r$ be as in Lemma 52.4.1. Then there is a unique graded $A$-algebra map

$\Phi : \bigoplus \nolimits _{n \geq 0} I^{nc} \to A[T_1, \ldots , T_ r]$

with $\Phi (g) = \sum \varphi _ j(g) T_ j$ for $g \in I^ c$. Moreover, the composition of $\Phi$ with the map $A[T_1, \ldots , T_ r] \to \bigoplus _{n \geq 0} I^ n$, $T_ j \mapsto f_ j$ is the inclusion map $\bigoplus _{n \geq 0} I^{nc} \to \bigoplus _{n \geq 0} I^ n$.

Proof. For each $j$ and $m \geq c$ the restriction of $\varphi _ j$ to $I^ m$ is a map $\varphi _ j : I^ m \to I^{m - c}$. Given $j_1, \ldots , j_ n \in \{ 1, \ldots , r\}$ we claim that the composition

$\varphi _{j_1} \ldots \varphi _{j_ n} : I^{nc} \to I^{(n - 1)c} \to \ldots \to I^ c \to A$

is independent of the order of the indices $j_1, \ldots , j_ n$. Namely, if $g = g_1 \ldots g_ n$ with $g_ i \in I^ c$, then we see that

$(\varphi _{j_1} \ldots \varphi _{j_ n})(g) = \varphi _{j_1}(g_1) \ldots \varphi _{j_ n}(g_ n)$

is independent of the ordering as multiplication in $A$ is commutative. Thus we can define $\Phi$ by sending $g \in I^{nc}$ to

$\Phi (g) = \sum \nolimits _{e_1 + \ldots + e_ r = n} (\varphi _1^{e_1} \circ \ldots \circ \varphi _ r^{e_ r})(g) T_1^{e_1} \ldots T_ r^{e_ r}$

It is straightforward to prove that this is a graded $A$-algebra homomorphism with the desired property. Uniqueness is immediate as is the final property. This proves the lemma. $\square$

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