(E₁) $x$ is a root of a monic polynomial in the polynomial ring $\mathbf{A}[X]$. (E₂) The subalgebra $\mathbf{A}[x]$ of $\mathbf{R}$ is a finitely generated $\mathbf{A}$-module. (E₃) There exists a faithful module over the ring $\mathbf{A}[x]$ which is a finitely generated $\mathbf{A}$-module.

Let us show first that (E₁) implies (E₂). Let

be a monic polynomial in $\mathbf{A}[X]$ with $x$ as a root; for every integer $q \geq 0$ let $M_q$ be the sub-$\mathbf{A}$-module of $\mathbf{R}$ generated by $1, x, \ldots, x^n + q$. Then for all $q \geq 1$, whence, by induction on $q$,

(*) The results of this chapter and the next one depend on no Book other than Books I to VI, nor on Chapter I, § 4, nor on Chapter 111, § 5.

We conclude that $A[x]$ is equal to $M_0$ and is therefore a finitely generated $A$-module.

As the commutative ring $A[x]$ is a faithful module over itself, ($E_{II}$) implies ($E_{III}$).

Finally, the fact that ($E_{III}$) implies ($E_1$) will follow from the following more precise lemma:

Let $(u_i)_{1 \leq i \leq n}$ be a finite family of elements of $M$ such that $M = \sum_{i=1}^n Au_i$. For all $i$, there exists by hypothesis a finite family $(q_{ij})_{1 \leq j \leq n}$ of elements of $q$ such that

Therefore (Algebra, Chapter III, § 8) if $d$ is the determinant of the matrix $(q_{ij} - \delta_{ij}x)$ with elements in $A[x]$ ($\delta_{ij}$ denoting the Kronecker index), then $du_i = 0$ for all $i$ and hence $dM = 0$; as $M$ is assumed to be a faithful $A[x]$-module, necessarily $d = 0$. This means that $x$ is a root of the polynomial $\det(q_{ij} - \delta_{ij}X)$ in $A[X]$ which, up to a sign, is a monic polynomial whose coefficientsother than the dominant coefficient belong to $q$.

The condition is obviously necessary by virtue of property (E_{II}); it is also sufficient by virtue of (E_{III}), for $R'$ is a faithful $A[x]$-module (since it contains the unit element of $R$).

The condition is necessary by virtue of (E_{II}); it is sufficient for if $A[x]$ is a sub-A-module of a finitely generated A-module, it is itself a finitely generated A-module (Algebra, Chapter VIII, § 2, no. 3, Proposition 7).

The hypothesis that $A$ is Noetherian cannot be omitted from the statement (Exercise 2).

is commutative. If an element $x \in R$ is integral over $A$, then $g(x)$ is integral over $A'$.

If $x^n + a_1 x^{n-1} + \cdots + a_n = 0$ where $a_i \in A$ for $1 \leq i \leq n$, we deduce that

There exists a $K$-isomorphism $f$ of $K(x)$ onto $K(x')$ such that $f(x) = x'$ and the elements of $A$ are invariant under $f$.

It is obviously sufficient to prove the first assertion. The condition is necessary by Proposition 2. Conversely, if each of the $x_i$ is integral over $A$, the subalgebra $A[x_i]$ of $R_i$ is a finitely generated $A$-module and hence so is the subalgebra $\prod_{i=1}^n A[x_i]$ of $R$; as $A[x]$ is contained in this subalgebra, $x$ is integral over $A$ by Proposition 1.

We argue by induction on $n$, the proposition being just (E_{II}) for $n = 1$. The induction hypothesis implies that $B = A[x_1, \ldots, x_{n-1}]$ is a finitely generated A-module; as $x_n$ is integral over $B$, $B[x_n] = A[x_1, \ldots, x_n]$ is a finitely generated B-module and therefore also a finitely generated A-module (Algebra, Chapter II, § 1, no. 13, Proposition 25).

In fact, if $x, y$ are two elements of $R$ integral over $A$, it follows from Proposition 4 that $A[x, y]$ is a finitely generated A-module; as it contains $x + y$ and $xy$, the corollary follows from Proposition 1.

In a non-commutative algebra the sum and product of two integral elements over $A$ are not necessarily integral over $A$ (Exercise 4).

Every element of $B$ belongs to a sub-A-algebra of $B$ generated by a finite subset of $E$.

Remark (3) It follows from Proposition 4 that every commutative A-algebra integral over $A$ is the union of a right directed family of finite subalgebras over $A$.

Consider any element $x' = \sum_{i=1}^n x_i \otimes a'_i$ of $R \otimes_A A'$, where the $x_i$ belong to $R$ and the $a'_i$ to $A'$; as $x_i \otimes a'_i = (x_i \otimes 1)a'_i$ and the $x_i \otimes 1$ are integral over $A'$ (Proposition 2), so is $x$.

$B \otimes_A C[A]$ is integral over $C[A]$ by Proposition 5 and hence so is the canonical image $C[B]$ of $B \otimes_A C[A]$ in $R$ (considered as an A-algebra) by Proposition 2.

It is sufficient to verify that every $x \in C$ is integral over $A$. By hypothesis there exists a monic polynomial $X^n + b_1 X^{n-1} + \cdots + b$, with coefficients in $B$ with $x$ as a root; then $x$ is integral over $B' = A[b_1, \ldots, b_n]$ and $B'[x]$ is therefore a finitely generated B-module. But as $B$ is integral over $A$, $B'$ is a finitely generated A-module (Proposition 4); we conclude that B'[x] is also a finitely generated A-module (Algebra, Chapter 11, § 1, no. 13, Proposition 25) and therefore x is integral over A.

$R \otimes_A R$ is integral over R (Proposition 5) and hence the conclusion follows from Proposition 6.

The integral closure of A' in R is integral over A by no. 1, Proposition 6; it is therefore equal to A'.

Let K be the field of fractions of A and B the integral closure of A. Clearly

K is the field of fractions of B and it is sufficient to apply Proposition 7 to $R = K$.

This follows immediately from Definition 3 and no. 1, Corollary 1 to Proposition 2.

It is sufficient to apply Proposition 8 taking $R$ and the $B_\lambda$ equal to the field of fractions $K$ of $A$ and noting that a subring of $K$ integrally closed in $K$ is a fortiori an integrally closed domain since its field of fractions is contained in $K$.

This is an immediate consequence of no. 1, Proposition 3.

Let $A$ be a principal ideal domain, $K$ its field of fractions and $x$ an element of $K$. There exist two relatively prime elements $a, b$ of $A$ such that $x = ab^{-1}$ (Algebra, Chapter VII, § 1, no. 2, Proposition 1 and Chapter VI, § 1, no. 11, Proposition 9 (DIV)). If $x$ is integral over $A$, it is a root of a polynomial

X^n + c_1 X^{n-1} + \ldots + c_n \text{ of } A[X]. Then $a^n = b(-c_1 a^{n-1} - \ldots - c_n b^{n-1})$, which proves that $b$ divides $a''$. Since $a$ and $b$ are relatively prime, this implies that $b$ is invertible in $A$ (Algebra, Chapter VI, § 1, no. 12, Corollary 1 to Proposition 11 (DIV)); hence $x \in A$.

We proceed by induction on the degree $n$ of P, the lemma being obvious for $n = 0$ and $n = 1$. Suppose therefore that $n > 1$. Let $a$ be the ideal of $R[X]$ generated by P and let $f$ be the canonical homomorphism of $R[X]$ onto $B = R[X]/a$. Since P is monic, for every polynomial $Q \in R[X]$, $\deg(PQ) = \deg(P) + \deg(Q)$, whence $a \cap R = 0$; the restriction off to R is therefore injective. Identifying R with the subring $f(R)$ of B by means of $f$ and writing $b = f(X)$, we see that $b$ is a root of P in B, P being considered as a polynomial in $B[X]$. Then there exists a monic polynomial Q in $B[X]$ of degree $n - 1$ such that $P(X) = (X - b)Q(X)$ (Algebra, Chapter IV, § 1, no. 4, Proposition 5). By the induction hypothesis there exists a ring $R' \supseteq B$ such that in $R'[X]$ the polynomial Q is a product of monic polynomials of degree 1; clearly in $R'[X]$ P is then a product of monic polynomials of degree 1.

By a double application of Lemma 2, we see that there exists a ring R' containing R and families of elements $(a_i)_{1 \leq i \leq m}, (b_j)_{1 \leq j \leq n}$ of R' such that in $R'[X]$

the coefficients of P Q belong to the integral closure $A'$ of A in R' and hence (no. 2, Proposition 7) the elements $a, (1 \leq i \leq m)$ and $b, (1 \leq j \leq n)$ belong to A. It follows that the coefficientsof P and Qare integral over A (no. 1, Corollary 1 to Proposition 4).

Let A be an integral domain, K its field of fractions and K' a K-algebra (not necessarily commutative). Given an element $x \in K'$ algebraic over K, the polynomials $P \in K[X]$ such that $P(x) = 0$ form an ideal $a \neq 0$ of $K[X]$, necessarily principal (Algebra, Chapter IV, § 1, no. 5, Proposition 7). There exists a unique monic polynomial which generates a; generalizing the terminology introduced in Algebra, Chapter V, § 3, no. 1, Definition 3, this monic polynomial will be called the minimal polynomial of x over K.

There exists by hypothesis (no. 1, Theorem 1) a monic polynomial $Q \in A[X]$ such that $Q(x) = 0$. As $P$ divides $Q$ in $K[X]$, it follows from Proposition 11 that the coefficients of $P$ are integral over $A$.

Let $A$ be a ring and $R$ a (commutative) $A$-algebra; the homomorphism $\phi : A \to R$ defining the $A$-algebra structure on $R$ can be extended uniquely to a homomorphism $A[X] \to R[X]$ of polynomial rings over $A$ and $R$, leaving $X$ invariant and hence $R[X]$ is given a canonical $A[X]$-algebra structure.

By considering the polynomials of $R[X_1, \ldots, X_n]$ as polynomials in $X$, with coefficients in $R[X_1, \ldots, X_{n-1}]$, we see immediately that it is reduced to proving the proposition for $n = 1$. Then let $P$ be a polynomial in $R[X]$; it follows immediately from no. 1, Proposition 5 that, if the coefficients of $P$ are in the integral closure $B$ of $A$ in $R$, the element $P$, which belongs to $B[X] = B \otimes_A A[X]$, is integral over $A[X]$. Conversely, suppose that $P$ is integral over $A[X]$ and let

be a monic polynomial with coefficients $F_i \in A[X]$ with $P$ as a root. Let $r$ be an integer strictly greater than all the degrees of the polynomials $P$ and $F_i$ ($1 \leq i \leq m$) and let us write

Then $P_1$ is a root of the polynomial

with coefficients in $A[X]$; we may therefore write

Now the choice of $r$ implies that $-P_1$ is a monic polynomial of $R[X]$ and so is $G_m(X) = Q(X^r)$, the degrees of the polynomials $F_k(X) X^{r(m-k)}$ being all $< rm$ for $k \geq 1$. We conclude first of all that the polynomial

of $R[X]$ is also monic; moreover, as the coefficients of $G_i$ belong to $A$, Proposition 11 shows that $P_1$ has coefficients integral over $A$ and the coefficients of $P$ are therefore certainly integral over $A$.

This follows from Proposition 12 and Definition 3 of no. 2.

Arguing by induction on $n$, the problem is immediately reduced to the case $n = 1$. Let $K$ be the field of fractions of $A$, which is also that of $A'$; if an element $P$ of the field of fractions $K(X)$ of $A[X]$ is integral over $A[X]$, it belongs to the polynomial ring $K[X]$, for the latter is a principal ideal domain (Algebra, Chapter IV, § 1, no. 5, Proposition 7) and hence integrally closed (Proposition 10); the corollary then follows from Proposition 13 applied to $\mathbf{R} = K$.

By induction on $n$, it is sufficient to prove that $A[X]$ (resp. $A[[X]]$) is completely integrally closed. Then let $P$ be an element of the field of fractions of $A[X]$ (resp. $A[[X]]$) and suppose that there exists a non-zero element $Q \in A[X]$ (resp. $Q \in A[[X]]$) such that $Q P^m \in A[X]$ (resp. $Q P'' \in A[[X]]$) for every integer $m \geq 0$. If $K$ is the field of fractions of $A$, $A[X]$ (resp. $A[[X]]$) is a subring of $K[X]$ (resp. $K[[X]]$) and $K[X]$ (resp. $K[[X]]$) is a principal ideal domain (Algebra, Chapter VII, § 1, no. 1) and hence integrally closed (no. 4, Proposition 10) and Noetherian (Algebra, Chapter VIII, § 2, no. 3) and therefore completely integrally closed; then we have already seen that

$P \in K[X]$ (resp. $P \in K[[X]]$). Let $P = \sum_{k=0}^{\infty} a_k X^k$ ($a_k \in K$) and $Q = \sum_{k=0}^{\infty} b_k X^k$ ($b_k \in A$) and we argue by reductio ad absurdum by supposing that the $a_i$ do not all belong to $A$; then there is a least index $i$ such that $a_i \notin A$; if we write

it follows immediately from the hypothesis that also $Q(P - P_1)^m \in A[X]$ (resp. $Q(P - P_1)^m \in A[[X]]$) for all $m \geq 0$. Let $j$ be the least integer such that $b_j \neq 0$; clearly in $Q(P - P_1)^m$ the term of least degree with a coefficient $\neq 0$ is $b_j a_i^m X^{j + mi}$ and hence $b_j a_i^m \in A$ for all $m \geq 0$; but as $A$ is completely integrally closed this implies $a_i \in A$, contrary to the hypothesis.

Let $(A_n)_{n \in \mathbf{Z}}$ be the filtration defined on $A$; as $\bigcap_{n \in \mathbf{Z}} A_n$, is the closure of the ideal (0) (Chapter III, § 2, no. 5), the hypothesis implies first that the filtration $(A_n)$ is separated and, as $\mathrm{gr}(A)$ is an integral domain, so then is $A$ (Chapter III, § 2, no. 3, Corollary to Proposition 1). Let $x = b/a$ be an element of the field of fractions $K$ of $A$ ($a \in A, b \in A$) for which there exists an element $d \neq 0$ of $A$ such that $dx^n \in A$ for all $n \geq 0$. We must prove that $b \in A a$ and, as by hypothesis the ideal $A a$ is closed, it is sufficient to show that, for all $n \in \mathbf{Z}, b \in A a + A$. As the filtration of $A$ is exhaustive, there exists an integer $q \in \mathbf{Z}$ such that $b \in A a + A_q$. It will therefore suffice to prove that the relation $b \in A a + A_q$ implies $b \in A a + A_{q+1}$.

Suppose then that $b = a y + z$ where $y \in A, z \in A_{q+1}$. Then by hypothesis $dx^n \in A$ for all $n \geq 0$, whence we obtain immediately $d(x - y)^n \in A$ for all $n \geq 0$; in other words, $dz^n = a^n t_n$ where $t_n \in A$ for all $n \geq 0$. We can obviously limit our attention to the case where $z \neq 0$. Let $v$ denote the order function on $A$ (Chapter III, § 1, no. 2) and let us write $v(d) = n_1,\ v(z) = n_2 \geq m,\ v(a) = n_3$; let $d', z', a'$ be the respective images of $d, z, a$ in $A_{n_1}/A_{n_1+1},\ A_{n_2}/A_{n_2+1},\ A_{n_3}/A_{n_3+1}$. For all $n \geq 0,\ v(dz^n) = n_1 + nn_2$ (Chapter III, § 2, no. 3, Proposition 1) and hence the canonical image in $\mathrm{gr}(A)$ of $dz^n$ is $d'{z'}^n$; similarly it is seen that the canonical image in $\mathrm{gr}(A)$ of $a^n t_n$ is of the form ${d'}^n t'_n$ where $t'_n \in \mathrm{gr}(A)$ and, as $a' \neq 0$ we deduce that, for all $n \geq 0,\ d'(z'/a')^n \in \mathrm{gr}(A)$. The hypothesis that $\mathrm{gr}(A)$ is completely integrally closed therefore implies the existence of an $s' \in \mathrm{gr}(A)$ such that $z' = a's'$; decomposing $s'$ into a sum of homogeneous elements, it is further seen (since $z'$ and $a'$ are homogeneous) that $s'$ may be assumed to be homogeneous, that is the image of an element $s \in A$; then $v(as) = v(z) = n_2$ and $z \equiv as \pmod{A_{n_2+1}}$; as $n_2 \geq m, a fortiori$ $z \equiv as \pmod{A_{m+1}}$, hence $b \equiv a(y+s) \pmod{A_{m+1}}$.