The first assertion follows from Cor. 1 of Prop. 7 of § 3, No. 4. Since B is flat over A, one can identify m_B^k = B.\rho(m_A^k) with B \otimes_A m_A^k for every integer k \geqslant 0, hence m_B/m_B^2 with B \otimes_A (m_A/m_A^2), or again with \kappa_B \otimes_{\kappa_A} (m_A/m_A^2). One therefore has

(2)

In fact, $\hat{A}$ is flat over A, and one has $m_{\hat{A}} = \hat{A}.m_A$ (III, § 3, No. 4, Theorem 3 and § 2, No. 12, Cor. 2 to Prop. 16).

If these conditions are fulfilled, every system of coordinates of A is a completely secant sequence for A.

(ii) \Rightarrow (i): for every secant subset has at most $\dim(A)$ elements (§ 3, No. 2, Theorem 1). (iii) \Rightarrow (ii): for every completely secant sequence is secant (§ 3, No. 2, corollary to Proposition 3). (iv) \Rightarrow (iii): let $(x_1, ..., x_r)$ be a sequence of elements of $m_A$ whose classes modulo $m_A^2$ form a basis $(\xi_1, ..., \xi_r)$ of $m_A/m_A^2$ over the field $\kappa_A$. If property (iv) is satisfied, $\mathrm{gr}(A)$ is the polynomial algebra $\kappa_A[\xi_1, ..., \xi_r]$ and the sequence $(x_1, ..., x_r)$ is completely secant (A, X, p. 160, Theorem 1). This also proves the last assertion of Theorem 1. (i) \Rightarrow (iv): put $r = [m_A/m_A^2 : \kappa_A]$. By formula (6) of § 4, No. 2, the Poincaré series of the graded vector space S over the field $\kappa_A$ is equal to

Suppose that the canonical homomorphism $\gamma : S \to \mathrm{gr}(A)$ is not bijective. Since $\gamma$ is surjective, there exists a homogeneous element $u$ of $S$, of degree $d > 0$, annihilated by $\gamma$. We then have

By Theorem 3 of § 4, No. 4 and Lemma 2 of § 4, No. 1, therefore $\dim(A) < r$, and A is not regular.

Finally, let us prove the equivalence of conditions (iv) to (vi). Now (iv) means that one has $H_{A,m_A} = (1 - T)^{-s}$ with $s = [m_A/m_A^2 : \kappa_A]$. Therefore conditions (iv), (v), (vi) mean that one has $H_{A,m_A} = (1 - T)^{-m}$, with respectively $m = [m_A/m_A^2 : \kappa_A]$, $m \geq 0$, $m = \dim(A)$. But if one has $H_{A,m_A} = (1 - T)^{-m}$, then $\dim(A) = m$ by § 4, No. 4, Theorem 3 and $[m_A/m_A^2 : \kappa_A] = m$ (since $(1 - T)^{-m} = 1 + mT + \cdots$). The equivalence of conditions (iv) to (vi) follows immediately from this.

Suppose that A is regular. Then $\mathrm{gr}(A)$ is isomorphic to a polynomial algebra in a finite number of indeterminates over a field (Theorem 1, (iv)). Hence $\mathrm{gr}(A)$ is a noetherian integrally closed ring (V, § 1, No. 3, Corollary 3 to Proposition 13), and therefore A is integrally closed (V, § 1, No. 4, Proposition 15).

We shall see in a later chapter that every regular noetherian local ring is factorial.

The stated condition is obviously necessary.

Suppose conversely that $\sigma$ induces isomorphisms of $\mathrm{gr}_0(B)$ onto $\mathrm{gr}_0(A)$ and of $\mathrm{gr}_1(B)$ onto $\mathrm{gr}_1(A)$. Since the ring $\mathrm{gr}(B)$ is generated by $\mathrm{gr}_0(B) \cup \mathrm{gr}_1(B)$ and $\mathrm{gr}(A)$ is the symmetric algebra of the vector space $\mathrm{gr}_1(A)$ over the field $\mathrm{gr}_0(A)$, the homomorphism $\mathrm{gr}(\sigma)$ is bijective. Hence $\sigma$ is bijective (III, § 2, No. 8, Corollary 3 to Theorem 1).

This follows from the equivalence of (i) and (iv) in Theorem 1, and from Proposition 11 of III, § 2, No. 9.

(i) the ring $A$ is regular, and $x$ forms part of a system of coordinates of $A$; (ii) the ring $A/x$ is regular and $x$ is a secant sequence for $A$; (iii) the ring $A/x$ is regular and $x$ is a completely secant sequence for $A$.

Moreover, when these conditions are satisfied, $x$ is a prime ideal of $A$.

(iii) $\Rightarrow$ (ii): this follows from the corollary of Prop. 3 of § 3, No. 2.

(ii) $\Rightarrow$ (i): suppose that $x$ is a secant sequence for $A$ and that the noetherian local ring $A/x$ is regular. Let $(x_{r+1}, ..., x_d)$ be a sequence of elements of $A$ whose classes modulo $x$ form a system of coordinates of $A/x$. Then the sequence $(x_1, ..., x_d)$ generates the ideal $m_A$ of $A$, and we have

It follows that $A$ is regular and that $(x_1, ..., x_d)$ is a system of coordinates of $A$.

(i) $\Rightarrow$ (iii): if condition (i) is satisfied, the sequence $x$ is completely secant (No. 2, Th. 1), hence secant by the corollary of Prop. 3 of § 3, No. 2. We therefore have

(3)

moreover, the classes of $x_1, ..., x_r$ modulo $m_A^2$ are linearly independent over the field $\kappa_A$, and therefore we have

(4)

Formulas (3) and (4) show that $A/x$ is regular.

Every regular noetherian local ring is an integral domain by Cor. 1 of Th. 1 of No. 2. It follows that $x$ is prime if $A/x$ is regular.

(i) $A$ is regular, and $t$ does not belong to $m_A^2$; (ii) $A/tA$ is regular and $\dim(A/tA) < \dim(A)$; (iii) $A/tA$ is regular, and $t$ is not a divisor of 0 in $A$.

The condition is sufficient by Prop. 2.

Suppose $A/q$ regular, and let $x = (x_1, ..., x_r)$ be a sequence of elements of $q$ whose classes modulo $m_A^2$ form a basis of $(q + m_A^2)/m_A^2$ over the field $\kappa_A$. Let $x$ be the ideal of $A$ generated by $x$. Then $x \subset q$ and $x$ forms part of a system of coordinates of $A$, hence the noetherian local ring $A/x$ is regular (Prop. 2); moreover, the vector spaces $m_A/(q + m_A^2)$ and $m_A/(x + m_A^2)$ have the same dimension over $\kappa_A$. It follows that the regular noetherian local rings $A/q$ and $A/x$ have the same dimension. Since the ideals $q$ and $x$ are prime and one has $x \subset q$, one concludes finally that $q = x$.

Let $\varphi$ be the canonical homomorphism of $A$ into the field of fractions $k$ of $A/p$ and let $\varphi': A[T] \to k[T]$ be the extension of $\varphi$ such that $\varphi'(T) = T$. Suppose that one has $P = P_1P_2$ where $P_1$ and $P_2$ are two monic polynomials of $A[T]$ distinct from 1. Then one has $T^d = \varphi'(P_1)\varphi'(P_2)$ in $k[T]$, denoting by $d$ the degree of $P$. If $d_i$ is the degree of $P_i$, one has therefore $\varphi'(P_i) = T^{d_i}$, that is to say $P_i(T) \equiv T^{d_i} \mod pA[T]$, and in particular $P_i(0) \in p$. But then $P(0) = P_1(0).P_2(0)$ belongs to $p^2$, contrary to the hypotheses. This proves a).

Assertion b) results from a) and prop. 11 of V, § 1, no 3.

Let $A$ be a noetherian local ring and $P_1, ..., P_r$ monic polynomials in $A[T]$, of degree $\geq 2$. Let $q$ be the ideal of $A[T_1, ..., T_r]$ generated by $P_1(T_1), ..., P_r(T_r)$ and let $B$ be the quotient $A$-algebra $A[T_1, ..., T_r]/q$. For $1 \leq i \leq r$, let $d_i$ denote the degree of $P_i$, $t_i$ the class of $T_i$ modulo $q$, and $\gamma_i$ the class of $c_i = P_i(0)$ modulo $m_A^2$. Suppose that $P_i(T) \equiv T^{d_i} \mod m_A A[T]$ for $1 \leq i \leq r$.

Proposition 4. — a) The ring $B$ is local and noetherian, with maximal ideal

One has $\dim(A) = \dim(B)$ and $[\kappa_B : \kappa_A] = 1$. The monomials $t_1^{\alpha(1)} ... t_r^{\alpha(r)}$, with $0 \leq \alpha(i) < d_i$ for $1 \leq i \leq r$, form a basis of the $A$-module $B$.

b) Let $\lambda$ be the canonical homomorphism of $m_A/m_A^2$ into $m_B/m_B^2$. Then the kernel of $\lambda$ is the $\kappa_A$-vector space generated by $\gamma_1, ..., \gamma_r$. The classes of the elements $t_1, ..., t_r$ form a basis over $\kappa_A$ of the cokernel of $\lambda$.

c) For $B$ to be regular, it is necessary and sufficient that $A$ be regular and that $\gamma_1, ..., \gamma_r$ be linearly independent in the $\kappa_A$-vector space $m_A/m_A^2$.

The $A$-algebra $B$ is isomorphic to the tensor product $B_1 \otimes_A \cdots \otimes_A B_r$ with $B_i = A[T]/(P_i)$ for $1 \leq i \leq r$. It follows that the monomials $t_1^{\alpha(1)} ... t_r^{\alpha(r)}$, with $0 \leq \alpha(i) < d_i$ for $1 \leq i \leq r$, form a basis of the $A$-module $B$. In particular, $B$ is integral over $A$, hence $A$ and $B$ have the same dimension by Th. 1 of § 2, No. 3.

According to Cor. 3 of Prop. 9 of IV, § 2, No. 5, the ring $B$ is noetherian, and every maximal ideal of $B$ contains $B.m_A$. Moreover, in view of the assumption made on $P_1, ..., P_r$ and the relation $P_i(t_i) = 0$, one has $t_i^{d_i} \in B.m_A$ for $1 \leq i \leq r$. Hence every maximal ideal of $B$ contains $t_1, ..., t_r$, and therefore also the ideal $q' = B.m_A + Bt_1 + \cdots + Bt_r$. Now one has $m_A = A \cap q'$ and $B = A + q'$, hence $B/q'$ is isomorphic to $A/m_A$ and $q'$ is a maximal ideal of $B$; consequently, $B$ is local and one has $[\kappa_B : \kappa_A] = 1$. This proves a).

Put $r = m_A^2 + \sum_{i=1}^r Ac_i$, and denote by $\varphi$ the canonical homomorphism of $(A/m_A^2)[T_1, ..., T_r]$ onto $B/m_B^2$. Since one has $m_B = B.m_A + \sum_{i=1}^r Bt_i$, the kernel $n$ of $\varphi$ is the ideal generated by the classes $\overline{P}_i(T_i)$ of the polynomials $P_i(T_i)$ modulo $m_A^2.A[T_1, ..., T_r]$ and the monomials $T_iT_j$ and $xT_i$ for $1 \leq i, j \leq r$ and $x$ in $m_A/m_A^2$. According to the assumption made on $P_i$, namely $P_i(T) \equiv T^{d_i} \mod m_A.A[T]$, one can replace $\overline{P}_i(T_i)$ by $\gamma_i$ in this description of $n$; consequently, the ring $B/m_B^2$ is isomorphic to the quotient of $(A/r)[T_1, ..., T_r]$ by the graded ideal generated by the monomials $T_i T_j$ and $x T_i$ for $x$ in $m_A / r$. Let $\tau_i$ denote the class of $t_i$ modulo $m_B^2$; one therefore has

whence

Assertion b) follows at once from this.

According to formula (6) and the relation $[ \kappa_B : \kappa_A ] = 1$, one has

Now, the $\kappa_A$-vector space $r / m_A^2$ is generated by $\gamma_1, ..., \gamma_r$ and one has

Assertion c) follows at once from formulas (7) and (8).

The case where P has degree $\geq 2$ follows from prop. 4, where one takes $r = 1$; when P has degree 1, that is to say is of the form $T - c$ with $c \in m_A$, the corollary is immediate.

Put $C = A_p[t]$ and let d denote the degree of P. By prop. 3 applied to the ring $A_p$, the Eisenstein polynomial P is irreducible in $K[T]$ and $(1, t, ..., t^{d-1})$ is a basis of L over K, hence of C over $A_p$. Since P is monic, the kernel of the canonical homomorphism of $A_p[T]$ onto C is equal to (P). By the corollary to prop. 4 above, C is therefore a regular noetherian local ring of the same dimension as $A_p$, the maximal ideal $m_C$ of C is generated by $p \cup \{ t \}$ and the field $\kappa_C$ is a trivial extension of the field of fractions of $A/p$. In order to prove prop. 5, it is therefore enough to show that there exists a unique prime ideal q of B above p, and that one has $C = B_q$.

Put S = A − p. One knows (V, § 1, n° 5, prop. 16) that the integral closure of

(VI, § 8, n° 1).

In fact, discrete valuation rings are regular noetherian local rings of dimension 1; putting d = [L : K], one has t^d \in m_A - m_A^2, whence d = e(B_q/A_p). One has [\kappa_B : \kappa_A] = 1, whence f(B_q/A_p) = 1.

B) $p \neq 0$. A Cohen subring of A is any subring V of A which is a $p$-ring such that $A = m_A + V$ (IX, § 2, no 2, def. 2). The ring V is local; its maximal ideal $m_V$ is generated by $p.1_V$; hence one has $m_A \cap V = m_V$ and the canonical injection of V into A defines by passage to the quotient an isomorphism of the field $\kappa_V$ onto the field $\kappa_A$. If $p.1_A = 0$, V is a field of characteristic $p$. Otherwise V is a discrete valuation ring whose field of fractions has characteristic zero (IX, § 2, no 1, cor. 1 to prop. 1). One proves (IX, no 2, th. 1) that A possesses Cohen subrings.

a) Suppose that $V$ is a field. Then the $V$-algebra $A$ is isomorphic to the algebra of formal series $V[[X_1, ..., X_n]]$.

b) Suppose that $V$ is a complete discrete valuation ring and that one has $m_V \not\subset m_A^2$. Then the $V$-algebra $A$ is isomorphic to the algebra of formal series $V[[X_1, ..., X_{n-1}]]$.

c) Suppose that $V$ is a complete discrete valuation ring and that one has $m_V \subset m_A^2$. Then there exists a special polynomial $P$ in $V[[X_1, ..., X_{n-1}]] [T]$ and a $V$-isomorphism of $A$ onto $V[[X_1, ..., X_{n-1}]] [T]/(P)$.

Assertion a) follows at once from Cor. 3 of Th. 1 of No. 2.

Let us prove b). Let $(x_1, ..., x_m)$ be a sequence of elements of $m_A$, and let $\varphi_0$ be the homomorphism of $V[X_1, ..., X_m]$ into $A$ which coincides with the identity on $V$ and sends $X_i$ to $x_i$ for $1 \leq i \leq m$. If $a$ is the ideal of $V[X_1, ..., X_m]$ generated by $X_1, ..., X_m$, one has $\varphi_0(a) \subset m_A$, hence $\varphi_0$ extends by continuity to a homomorphism $\varphi$ of $V_m = V[[X_1, ..., X_m]]$ into $A$. Let $\pi$ be a uniformizer of $V$. By Corollary 2 of Theorem 1 of No. 2, $\varphi$ is an isomorphism of $V_m$ onto $A$ if and only if $(\pi, x_1, ..., x_m)$ is a system of coordinates of $A$. But the canonical mapping of $m_V/m_V^2$ into $m_A/m_A^2$ is injective, since one has $m_V \not\subset m_A^2$ and $m_V/m_V^2$ is of rank 1 over $\kappa_V$. Hence $m_A/(m_V + m_A^2)$ is of rank $n - 1$ and $(\pi, x_1, ..., x_m)$ is a system of coordinates of $A$ if and only if the classes of the $x_i$ form a basis of $m_A/(m_V + m_A^2)$ over $\kappa_V$. Hence b).