Remark 61.8.3. The construction $A \mapsto T(A)$ is functorial in the following sense: If $A \to A'$ is a ring map, then we can construct a commutative diagram

$\xymatrix{ A \ar[r] \ar[d] & T(A) \ar[d] \\ A' \ar[r] & T(A') }$

Namely, given $(A \to A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n))$ in $S(A)$ we can use the ring map $\varphi : A \to A'$ to obtain a corresponding element $(A' \to A'[x_1, \ldots , x_ n]/(f^\varphi _1, \ldots , f^\varphi _ n))$ of $S(A')$ where $f^\varphi$ means the polynomial obtained by applying $\varphi$ to the coefficients of the polynomial $f$. Moreover, there is a commutative diagram

$\xymatrix{ A \ar[r] \ar[d] & A[x_1, \ldots , x_ n]/(f_1, \ldots , f_ n) \ar[d] \\ A' \ar[r] & A'[x_1, \ldots , x_ n]/(f^\varphi _1, \ldots , f^\varphi _ n) }$

which is a in the category of rings. For $E \subset S(A)$ finite, set $E' = \varphi (E)$ and define $B_ E \to B_{E'}$ in the obvious manner. Taking the colimit gives the desired map $T(A) \to T(A')$, see Categories, Lemma 4.14.8.

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