Lemma 3

a) There exists a unique homomorphism $v : B \to A$ extending $u$ and mapping $T_i$ to $t_i$ for $1 \leq i \leq n$.

b) In order that $v$ be surjective, it is necessary and sufficient that the sequence $(u(p_1), ..., u(p_m), t_1, ..., t_n)$ generate the ideal $m_A$ of $A$, or again that the classes of these elements modulo $m_A^2$ generate $m_A/m_A^2$ as a vector space over the field $\kappa_A$.

c) In order that $v$ make $A$ into a finite $B$-algebra, it is necessary and sufficient that the sequence $(u(p_1), ..., u(p_m), t_1, ..., t_n)$ generate an ideal of definition of (the $m_A$-adic topology of) $A$.

Let $n$ denote the ideal of the ring $B$ generated by $T_1, ..., T_n$. Every homomorphism $v$ of $B$ into $A$ extending $u$ and such that $v(T_i) = t_i$ maps $n$ into $m_A$, and is therefore continuous when $B$ is endowed with the $n$-adic topology. The existence and uniqueness of $v$ then result from A, IV, p. 26, Prop. 4.

The ring $B = C[[T_1, ..., T_n]]$ is a complete noetherian local ring (III, § 2, No. 10, Corollary 6 to Theorem 2 and No. 6, Proposition 6), whose maximal ideal $m_B$ is generated by $p_1, ..., p_m, T_1, ..., T_n$. Hence $v(m_B) \subset m_A$ and $v$ defines a homomorphism gr$(v)$ of $\operatorname{gr}(B) = \bigoplus_{n=0}^\infty m_B^n/m_B^{n+1}$ into $\operatorname{gr}(A) = \bigoplus_{n=0}^\infty m_A^n/m_A^{n+1}$. Now the ring $\operatorname{gr}(A)$ is generated by $A/m_A = \kappa_A$ and $m_A/m_A^2$, gr$(v)$ induces an isomorphism of $\kappa_B = \kappa_C$ onto $\kappa_A$, and the classes modulo $m_B^2$ of the elements $p_1, ..., p_m, T_1, ..., T_n$ generate $m_B/m_B^2$ as a vector space over $\kappa_B$; moreover $v$ is surjective if and only if gr$(v)$ is surjective (III, § 2, No. 8, Corollary 2 to Theorem 1). This proves $b$.

The ideal of $A$ generated by the sequence $(u(p_1), ..., u(p_m), t_1, ..., t_n)$ is none other than $v(m_B)$ $A$. Since $m_A$ contains $v(m_B)$, $A$ is a Zariski ring for the $v(m_B)$ $A$-adic topology. The ring $A/v(m_B)$ $A$ is artinian if and only if its length as an $A$-module is finite. But since every simple module over $A$ is annihilated by $m_A$ and since, by hypothesis, $A/m_A$ and $B/m_B$ are isomorphic, this occurs if and only if the dimension over the field $B/m_B$ of the vector space $A/v(m_B)$ $A$ is finite. By IV, § 2, No. 5, Corollary 2 of Proposition 9, we see therefore that $v(m_B)$ $A$ is an ideal of definition of $A$ if and only if the dimension of $A/v(m_B)$ $A$ over $B/m_B$ is finite. This is indeed the case if $A$ is a finite $B$-algebra.

Suppose that $v(m_B)$ $A$ is an ideal of definition of $A$. The $m_B$-adic topology of the $B$-module $A$ then coincides with the $m_A$-adic topology of the ring $A$, and is therefore separated. Since $A/v(m_B)$ $A$ is a finitely generated module over $B/m_B$, $A$ is a finitely generated $B$-module (III, § 2, No. 3, Example 3 and No. 9, cor. 1 of Prop. 12). This proves $c$.