Proposition 19

Tag 064Z, permanent. Cite it as [Bourbaki, Tag 064Z].

For a mapping g of M into M' to be an A[X]-homomorphism of Mu into M'_u, it is necessary and sufficient that g be an A-homomorphism of M into M' such that $g \circ u = u' \circ g$ . When this is so, if $\bar{g}$ is the A[X]-homomorphism of M[X] into M'[X] equal to $1_{A[x]} \otimes g$ (II, § 5, no. 1), the diagram*

\begin{array}{cccccc} M[X] & \xrightarrow{\psi} & M[X] & \xrightarrow{\phi} & M_u & \longrightarrow 0 \\ \downarrow \bar{g} & & \downarrow \bar{g} & & \downarrow g & \\ M'[X] & \xrightarrow{\psi'} & M'[X] & \xrightarrow{\phi'} & M'_{u'} & \longrightarrow 0 \end{array}

is commutative.

The condition $g \circ u = u' \circ g$ is obviously necessary by (43) for $g$ to be an A[X]-homomorphism; it is sufficient, for it implies by induction that $g \circ u^n = {u'}^n \circ g$ for every integer $n > 0$ . On the other hand, for all $x \in M$ and all $p \in A[X]$ ,

\phi'(\bar{g}(p \otimes x)) = \phi'(p \otimes g(x)) = p(u')(g(x)) = g(p(u)(x)) = g(\phi(p \otimes x))

and

\bar{u}'(g(p \otimes x)) = \bar{u}'(p \otimes g(x)) = p \otimes u'(g(x)) = p \otimes g(u(x)) = \bar{g}(\bar{u}(p \otimes x))

which proves the commutativity of diagram (48).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Cites

II, § 5, no. 1