§ 2. Zeros of polynomials

1. Roots of a polynomial in one indeterminate. Multiplicity

Let $g \in A[(X_i)_{i \in I}]$ and let $E$ be a unital associative $A$ -algebra. Let $x = (x_i)_{i \in I}$ , be a family of pairwise permutable elements of $E$ . We shall say that $x$ is a zero of $g$ in $E^l$ if $g(x) = 0$ . If $f$ is a polynomial in a single indeterminate, a zero of $f$ in $E$ is also called a root of $f$ in $E$ .

Proposition 1

— Let $f \in A[X]$ and $\alpha \in A$ . The remainder of the division of $f$ by $X - \alpha$ is $f(\alpha)$ . For $\alpha$ to be a root of $f$ it is necessary and sufficient that $X - \alpha$ should be a divisor of $f$ in $A[X]$ .

For if $u,\ v \in \mathbf{A}[X]$ are such that $f = (X - \alpha)\ u + v,\ \deg v < 1$ , then $v$ is a scalar and $f(a) = (a - a)\ u(\alpha) + v = v$ . This proves the first assertion and the second follows from it.

Proposition 2

— Let $f \in \mathbf{A}[X]$ , $a \in \mathbf{A}$ , and let $h$ be an integer $\geq 0$ . The following conditions are equivalent: (i) $f$ is divisible by $(X - \alpha)^h$ but not by $(X - \alpha)^{h-1}$ ; (ii) *there exists $g \in \mathbf{A}[X]$ such that $f = (X - \alpha)^h g$ and $g(a) \neq 0$ . (i) $\Rightarrow$ (ii) follows at once from Prop. 1. (ii) $\Rightarrow$ (i) : Suppose that $f = (X - \alpha)^h g$ , where $g$ does not admit $a$ as a root. Then $f$ is divisible by $(X - \alpha)^h$ ; if $g_1 \in \mathbf{A}[X]$ existed such that $f = (X - \alpha)^{h+1} g_1$ , then since $(X - \alpha)^n$ is not a divisor of zero in $\mathbf{A}[X]$ (IV, p. 9, Prop. 7), we have $g = (X - a)\ g_1$ and so $g(\alpha) = 0$ , which is absurd.

Proposition 3

— Let $f$ be a non-zero element of $\mathbf{A}[X]$ and $a \in \mathbf{A}$ . There exists just one integer $h \geq 0$ satisfying the conditions (i) and (ii) of Prop. 2. This is clear for condition (i), bearing in mind the fact that iff is divisible by $(X - \alpha)^h$ , then $\deg f \geq h$ (IV, p. 9, Prop. 7).

Definition 1

— With the above notation we say that $a$ is of order $h$ , or multiplicity $h$ relative to $f$ . If $h > 0$ we also say that $a$ is a root of order $h$ or multiplicity $h$ off. A root of order 1 is called a simple root, a root of order 2 a double root,... A root of order $> 1$ is said to be multiple.

Remark 1

If $f = 0$ we agree to say that $a$ has order $\geq h$ relative to $f$ , whatever $a \in \mathbf{A}$ and the integer $h \geq 0$ . For any $f \in \mathbf{A}[X]$ and $a \in \mathbf{A}$ , to say that $a$ has order $\geq h$ relative to $f$ means that $(X - \alpha)^h$ divides $f$ .

Remark 2

Let $B$ be a commutative ring containing $\mathbf{A}$ as subring. Let $f \in \mathbf{A}[X]$ be non-zero and $a \in \mathbf{A}$ . The order of $a$ relative to $f$ is the same, whether we consider $f$ as element of $B[X]$ or as element of $\mathbf{A}[X]$ . This is clear from condition (ii) of Prop. 2.

Proposition 4

— Let $f$ and $g$ be non-zero elements of $\mathbf{A}[X]$ . Let $a \in \mathbf{A}$ , and let the orders of $a$ relative to $f$ and $g$ be $p$ and $q$ respectively. (i) The order of $a$ relative to $f + g$ is $\geq \inf(p, q)$ . It is equal to $\inf(p, q)$ if $p \neq q$ . (ii) The order of $a$ relative to $fg$ is $\geq p + q$ . It is equal to $p + q$ if $\mathbf{A}$ is an integral domain.

For we have $f(X) = (X - \alpha)^p f_1(X)$ , $g(X) = (X - \alpha)^q g_1(X)$ with $f_1(\alpha) \neq 0$ , $g_1(a) \neq 0$ . Suppose for example that $p \leq q$ ; then we have

f(X) + g(X) = (X - \alpha)^p (f_1(X) + (X - \alpha)^{q-p} g_1(X)) .

and if $p < q$ , $a$ is not a root of $f_1(X) + (X - \alpha)^{q-p} g_1(X)$ ; this proves (i). On the other hand, we have $f(X)g(X) = (X - \alpha)^{p+q} f_1(X)g_1(X)$ and $f_1(\alpha)g_1(\alpha) \neq 0$ if $A$ is an integral domain; this proves (ii).

Proposition 5

Suppose that $A$ is an integral domain. Let $f$ be a non-zero element of $A[X]$ , and $\alpha_1, \ldots, \alpha_p$ pairwise distinct roots of $f$ in $A$ , of orders $k_1, \ldots, k_r$ . We have

f(X) = (X - \alpha_1)^{k_1}(X - \alpha_2)^{k_2} \ldots (X - \alpha_p)^{k_p} g(X)

where $g \in A[X]$ and $\alpha_1, \ldots, \alpha_p$ are not roots of $g$ .

We proceed by induction on $p$ , the proposition being evident for $p = 1$ , by Def. 1. Suppose then that $f(X) = g_1(X)g_2(X)$ , where

g_1(X) = (X - \alpha_1)^{k_1} \ldots (X - \alpha_{p-1})^{k_{p-1}}, \quad g_2(X) \in A[X].

Since $A$ is an integral domain and $\alpha_p$ is distinct from $\alpha_1, \ldots, \alpha_{p-1}$ it follows that $\alpha_p$ is not a root of $g_1(X)$ , hence $\alpha_p$ is a root of order $k_p$ of $g_2(X)$ (Prop. 4, (ii)). It follows that $g_2(X)$ is divisible by $(X - \alpha_p)^{k_p}$ , and so

f(X) = (X - \alpha_1)^{k_1} \ldots (X - \alpha_p)^{k_p} g(X)

where $g(X) \in A[X]$ . Clearly $\alpha_1, \ldots, \alpha_p$ are not roots of $g$ .

Theorem 1

Let $A$ be an integral domain. Given a non-zero element $f$ of $A[X]$ , of degree $n$ , then the sum of the orders of all the roots off in $A$ is $\leq n$ .

This follows immediately from Prop. 5.

Corollary

Assume that $A$ is an integral domain and let $f, g \in A[X]$ , of degrees $\leq n$ . If there exist $n + 1$ pairwise distinct elements $x_1, \ldots, x_{n+1}$ of $A$ such that $f(x_i) = g(x_i)$ for $1 \leq i \leq n + 1$ , then $f = g$ .

It suffices to apply Th. 1 to $f - g$ .

Proposition 6 (Lagrange interpolation formula)

— *Let $K$ be a commutative field, $\alpha_1, \alpha_2, \ldots, a,$ distinct elements of $K$ and $\beta_1, \beta_2, \ldots, \beta_n$ any elements of $K$ . For $i = 1, 2, \ldots, n$ we put

f_i(X) = \prod_{j \in U(i)} (X - \alpha_j)/(\alpha_i - \alpha_j),

where $U(i)$ is the set of integers $j$ such that $j \neq i$ and $1 \leq j \leq n$ . Then $\beta_1 f_1 + \ldots + \beta_n f_n$ is the unique element $f$ of $K[X]$ such that $\deg f < n$ and $f(\alpha_i) = \beta_i$ for $1 \leq i \leq n$ .

The uniqueness off follows from the Cor. to Th. 1. Let $f = \beta_1 f_1 + \ldots + \beta_n f_n$ , then since $f_i$ has degree $n - 1$ , we have $\deg f < n$ . On the other hand, $f_i(\alpha_j) = 0$ for $j \neq i$ and $f_i(\alpha_i) = 1$ , hence $f(\alpha_i) = \beta_i$ for $1 \leq i \leq n$ .

Corollary

— Suppose that $\mathbf{A}$ is an integral domain. Let $f \in \mathbf{A}[X]$ , of degree $< n$ , and let $K$ be a subring of $\mathbf{A}$ which is a field. If there exist $n$ distinct elements $\alpha_1, \ldots, a_n$ of $\mathbf{A}$ such that $\alpha_i \in K$ and $f(\alpha_i) \in K$ for $i = 1, \ldots, n$ , then $f \in K[X]$ .

2. Differential criterion for the multiplicity of a root

Proposition 7

— Let $f \in \mathbf{A}[X]$ and let $a \in \mathbf{A}$ be a root off: For $a$ to be a simple root off it is necessary and sufficient that $a$ should not be a root of the derivative $Df$ of $f$ .

By hypothesis we have $f = (X - a)\ g$ , where $g \in \mathbf{A}[X]$ . For $a$ to be a simple root of $f$ it is necessary and sufficient that $g(\alpha) \neq 0$ . Now we have $Df = g + (X - \alpha)\ Dg$ , whence $(Df)(\alpha) = g(\alpha)$ .

More generally:

Proposition 8

— Let $f \in \mathbf{A}[X]$ and $a \in \mathbf{A}$ , and suppose that $a$ has order $k \geq 1$ relative to $f$ : Then $a$ has order $\geq k - 1$ relative to $Df$ . If $k . 1$ is cancellable in $\mathbf{A}$ , then $a$ has order $k - 1$ relative to $Df$ .

By hypothesis there exists $g \in \mathbf{A}[X]$ such that $f = (X - a)^k g$ and $g(a) \neq 0$ . Hence $Df = k(X - \alpha)^{k-1}g + (X - \alpha)^k Dg = (X - \alpha)^{k-1}(kg + (X - \alpha)\ Dg)$ , which establishes the first part of the proposition. The value of $kg + (X - \alpha)\ Dg$ for $X = \alpha$ is $kg(a)$ , and this is non-zero if $k . 1$ is cancellable in $\mathbf{A}$ ; this proves the second part of the proposition.

Let $k$ be an integer $> 0$ such that $k . 1 = 0$ in $\mathbf{A}$ . If $f(X) = X^k$ , then $0$ is a root of order $k$ of $f$ , and a root of arbitrarily high order of $Df$ .

Corollary

— Let $f \in \mathbf{A}[X]$ , $a \in \mathbf{A}$ and $p$ an integer $\geq 0$ , further, suppose that $p! . 1$ is cancellable in $\mathbf{A}$ . Then for $a$ to be a root of order $p$ of $f$ , it is necessary and sufficient that $a$ should be a root of $f, Df, ..., D^{p-1}f$ and not a root of $D^pf$ .

This follows from Prop. 8 by induction on $p$ .

3. Polynomial functions on an infinite integral domain

Proposition 9

— Assume that $A$ is an integral domain. Let $I$ be a set, $(\mathbf{H}_i)_{i \in I}$ a family of infinite subsets of $\mathbf{A}$ and $H = \prod_{i \in I} \mathbf{H}_i \subset A'$ . Iff is a non-zero element of $\mathbf{A}[(X_i)_{i \in I}]$ , and $H_f$ the set of all $x \in H$ such that $f(x) \neq 0$ , then $H$ and $H_f$ are equipotent.

a) First suppose that $I$ is finite and put $n = \mathrm{Card}\ I$ . The proposition is clear for $n = 0$ ; we shall prove it by induction on $n$ . Choose an element $i_0$ of $I$ and put $J = I - \{i_0\}$ , $B = \mathbf{A}[(X_i)_{i \in J}]$ . Since $f \neq 0$ , we can write $f = \sum_{k=0}^m g_k X_{i_0}^k$ , where $g_0, \ldots, g_m \in B$ and $g_m \neq 0$ . By the induction hypothesis the set K of all $x \in \prod_{i \in s} H_i$ such that $g_m(x) \neq 0$ is equipotent with $\prod_{i \in J} H_i$ . For $x \in K$ the polynomial

h(X_{i_0}) = \sum_{k=0}^m g_k(x) X_{i_0}^k \in A[X_{i_0}]

is non-zero. By Th. 1 (IV, p. 16) the set of

a \in H_{i_0}

such that

h(\alpha) \neq 0

is equipotent with

H_{i_0}

whence

\operatorname{Card} H \geq \operatorname{Card} H_f \geq (\operatorname{Card} K) \cdot (\operatorname{Card} H_{i_0}) = \operatorname{Card} H,

and so

\operatorname{Card} H = \operatorname{Card} H_f

b) In the general case there is a finite subset $I'$ of I such that $f \in A[(X_i)_{i \in I'}]$ . Let $H'_f$ be the set of all $x \in \prod_{i \in I'} H_i$ such that $f(x) \neq 0$ . Then

H_f = H'_f \times \left( \prod_{i \in I - I'} H_i \right),

and it suffices to apply the first part of the proof to

H'_f

Corollary 1

— *We keep the hypothesis and notation of Prop. 9. If l is non-empty, then $H_f$ is infinite.

Corollary 2

— *Suppose that A is an infinite integral domain or that A is an algebra over an infinite field. For every $f \in A[(X_i)_{i \in I}]$ , let $\tilde{f}: A^I \to A$ be the polynomial function defined by f (IV, p. 4). Then the mapping $f \mapsto \tilde{f}$ is injective.

When A is an infinite integral domain, the corollary follows at once from Prop. 9. Suppose that A is an algebra over an infinite field k. Let $f = \sum_{\nu \in \mathbf{N}^{(I)}} \alpha_\nu X^\nu$ be a non-zero element of $A[(X_i)_{i \in I}]$ ; then there exists $\nu_0 \in \mathbf{N}^{(I)}$ such that $\alpha_{\nu_0} \neq 0$ , and a k-linear form $\varphi$ on A such that $\varphi(\alpha_{\nu_0}) \neq 0$ . Let $g = \sum_{\nu \in \mathbf{N}^{(I)}} \varphi(a_\nu) X^\nu \in k[(X_i)_{i \in I}]$ ; we have $g \neq 0$ , hence there exists $x \in k^I$ such that $g(x) \neq 0$ . Then $\varphi(f(x)) = g(x) \neq 0$ , and so $f(x) \neq 0$ .

When A is an infinite integral domain or when A is an algebra over an infinite field, we shall usually identify f with $\tilde{f}$ .

Suppose that A is finite and let $f(X) = \prod_{a \in A} (X - a)$ , then $f \neq 0$ , but $\tilde{f} = 0$ . For other examples see IV, p. 88, exercises 7 and 8.

Theorem 2 (Principle of extension of algebraic identities)

— *Suppose that A is an infinite integral domain. Let $g_1, \ldots, g_m, f$ be elements of $A[(X_i)_{i \in I}]$ and assume the following hypotheses:

a) $g_1 \neq 0, \ldots, g_m \neq 0$ ; b) for all $x \in \mathbf{A}^1$ such that $g_1(x) \neq 0, \ldots, g_m(x) \neq 0$ , we have $f(x) = 0$ . Then $f = 0$ .

For if $f \neq 0$ , we have $fg_1 \ldots g_m \neq 0$ (IV, p. 9, Prop. 8), hence there exists $x \in \mathbf{A}^1$ such that $f(x)g_1(x) \ldots g_m(x) \neq 0$ (IV, p. 18, Cor. 2), which contradicts the hypothesis.

Scholium

— Let A be an integral domain and $f \in A[(X_i)_{i \in I}]$ . Th. 2 provides a convenient means of proving that $f = 0$ . It suffices to consider an infinite integral domain E containing A as subring; if we can show that $f((x_i)) = 0$ for all $(x_i) \in E^I$ (or even for those $(x_i) \in E^I$ at which a finite number of given non-zero polynomials do not vanish) then it follows that $f = 0$ . If A itself is not infinite, we can for example take E to be the ring $A[X]$ or its field of fractions.

Once we have proved the relation $f = 0$ , we can clearly deduce that $f((y_i)) = 0$ for all $(y_i) \in F^I$ where F is any unital associative and commutative A-algebra whatsoever; in particular F may be finite or non-integral.

In other words, the proof of the identity $f((x_i)) = 0$ when the $x_i$ run over an infinite integral domain containing A as subring (with the possible restriction that $g_k((x_i)) \neq 0$ for $1 \leq k \leq m$ , where the $g_k$ are non-zero polynomials) implies the same identity when the $x_i$ run over any unital associative and commutative A-algebra.

In particular let $f \in Z[(X_i)_{i \in I}]$ . If $f((x_i)) = 0$ when the $x_i$ run over $\mathbf{Z}$ (with the possible restriction that $g_k((x_i)) \neq 0$ for $1 \leq k \leq m$ , where the $g_k$ are non-zero elements of $\mathbf{Z}[(X_i)]$ ), then we have the same identity when the $x_i$ run over an arbitrary commutative ring.

Exercises

See the exercises for § 2.