As a consequence of Proposition 20.14, some polynomials f1, . . . , fn ∈ K[X] are relatively
prime iff there exist u1, . . . , un ∈ K[X] such that
u1f1 + · · · + unfn = 1.
The identity
u1f1 + · · · + unfn = 1
of part (2) of Proposition 20.14 is also called the Bezout identity.
We now consider the factorization of polynomials of a single variable into irreducible
factors.
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CHAPTER 20. POLYNOMIALS, IDEALS AND PID’S
20.5
Factorization and Irreducible Factors in K[X]
Definition 20.8. Given a field K, a polynomial p ∈ K[X] is irreducible or indecomposable
or prime if deg(p) ≥ 1 and if p is not divisible by any polynomial q ∈ K[X] such that
1 ≤ deg(q) < deg(p). Equivalently, p is irreducible if deg(p) ≥ 1 and if p = q1q2, then either
q1 ∈ K or q2 ∈ K (and of course, q1 = 0, q2 = 0).
Example 20.2. Every polynomial aX + b of degree 1 is irreducible. Over the field R, the
polynomial X2 + 1 is irreducible (why?), but X3 + 1 is not irreducible, since
X3 + 1 = (X + 1)(X2 − X + 1).
The polynomial X2 − X + 1 is irreducible over R (why?). It would seem that X4 + 1 is
irreducible over R, but in fact,
√
√
X4 + 1 = (X2 − 2X + 1)(X2 + 2X + 1).
However, in view of the above factorization, X4 + 1 is irreducible over Q.
It can be shown that the irreducible polynomials over R are the polynomials of degree 1,
or the polynomials of degree 2 of the form aX2 + bX + c, for which b2 − 4ac < 0 (i.e., those
having no real roots). This is not easy to prove! Over the complex numbers C, the only
irreducible polynomials are those of degree 1. This is a version of a fact often referred to as
the “Fundamental theorem of Algebra”, or, as the French sometimes say, as “d’Alembert’s
theorem”!
We already observed that for any two nonzero polynomials f, g ∈ K[X], f divides g iff
(g) ⊆ (f). In view of the definition of a maximal ideal given in Definition 20.4, we now prove
that a polynomial p ∈ K[X] is irreducible iff (p) is a maximal ideal in K[X].
Proposition 20.15. A polynomial p ∈ K[X] is irreducible iff (p) is a maximal ideal in
K[X].
Proof. Since K[X] is an integral domain, for all nonzero polynomials p, q ∈ K[X], deg(pq) =
deg(p) + deg(q), and thus, (p) = K[X] iff deg(p) ≥ 1. Assume that p ∈ K[X] is irreducible.
Since every ideal in K[X] is a principal ideal, every ideal in K[X] is of the form (q), for
some q ∈ K[X]. If (p) ⊆ (q), with deg(q) ≥ 1, then q divides p, and since p ∈ K[X] is
irreducible, this implies that p = λq for some λ = 0 in K, and so, (p) = (q). Thus, (p) is a
maximal ideal. Conversely, assume that (p) is a maximal ideal. Then, as we showed above,
deg(p) ≥ 1, and if q divides p, with deg(q) ≥ 1, then (p) ⊆ (q), and since (p) is a maximal
ideal, this implies that (p) = (q), which means that p = λq for some λ = 0 in K, and so, p
is irreducible.
Let p ∈ K[X] be irreducible. Then, for every nonzero polynomial g ∈ K[X], either p and
g are relatively prime, or p divides g. Indeed, if d is any gcd of p and g, if d is a constant, then
20.5. FACTORIZATION AND IRREDUCIBLE FACTORS IN K[X]
549
p and g are relatively prime, and if not, because p is irreducible, we have d = λp for some
λ = 0 in K, and thus, p divides g. As a consequence, if p, q ∈ K[X] are both irreducible,
then either p and q are relatively prime, or p = λq for some λ = 0 in K. In particular, if
p, q ∈ K[X] are both irreducible monic polynomials and p = q, then p and q are relatively
prime.
We now prove the (unique) factorization of polynomials into irreducible factors.
Theorem 20.16. Given any field K, for every nonzero polynomial
f = adXd + ad−1Xd−1 + · · · + a0
of degree d = deg(f ) ≥ 1 in K[X], there exists a unique set { p1, k1 , . . . , pm, km } such that
f = adpk1
1 · · · pkm
m ,
where the pi ∈ K[X] are distinct irreducible monic polynomials, the ki are (not necessarily
distinct) integers, and m ≥ 1, ki ≥ 1.
Proof. First, we prove the existence of such a factorization by induction on d = deg(f ).
Clearly, it is enough to prove the result for monic polynomials f of degree d = deg(f ) ≥ 1.
If d = 1, then f = X + a0, which is an irreducible monic polynomial.
Assume d ≥ 2, and assume the induction hypothesis for all monic polynomials of degree
< d. Consider the set S of all monic polynomials g such that deg(g) ≥ 1 and g divides
f . Since f ∈ S, the set S is nonempty, and thus, S contains some monic polynomial p1 of
minimal degree. Since deg(p1) ≥ 1, the monic polynomial p1 must be irreducible. Otherwise
we would have p1 = g1g2, for some monic polynomials g1, g2 such that deg(p1) > deg(g1) ≥ 1
and deg(p1) > deg(g2) ≥ 1, and since p1 divide f, then g1 would divide f, contradicting
the minimality of the degree of p1. Thus, we have f = p1q, for some irreducible monic
polynomial p1, with q also monic. Since deg(p1) ≥ 1, we have deg(q) < deg(f), and we can
apply the induction hypothesis to q. Thus, we obtain a factorization of the desired form.
We now prove uniqueness. Assume that
f = adpk1
1 · · · pkm
m ,
and
f = adqh1
1 · · · qhn
n .
Thus, we have
adpk1
1 · · · pkm
m = adqh1
1 · · · qhn
n .
We prove that m = n, pi = qi and hi = ki, for all i, with 1 ≤ i ≤ n.
The proof proceeds by induction on h1 + · · · + hn.
If h1 + · · · + hn = 1, then n = 1 and h1 = 1. Then, since K[X] is an integral domain, we
have
pk1
1 · · · pkm
m = q1,
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CHAPTER 20. POLYNOMIALS, IDEALS AND PID’S
and since q1 and the pi are irreducible monic, we must have m = 1 and p1 = q1.
If h1 + · · · + hn ≥ 2, since K[X] is an integral domain and since h1 ≥ 1, we have
pk1
1 · · · pkm
m = q1q,
with
q = qh1−1
1
· · · qhn
n ,
where (h1 − 1) + · · · + hn ≥ 1 (and qh1−1
1
= 1 if h1 = 1). Now, if q1 is not equal to any of
the pi, by a previous remark, q1 and pi are relatively prime, and by Proposition 20.13, q1
and pk1
1 · · · pkm
m
are relatively prime. But this contradicts the fact that q1 divides pk1
1 · · · pkm
m .
Thus, q1 is equal to one of the pi. Without loss of generality, we can assume that q1 = p1.
Then, since K[X] is an integral domain, we have
pk1−1
1
· · · pkm
m = qh1−1
1
· · · qhn
n ,
where pk1−1
1
= 1 if k1 = 1, and qh1−1
1
= 1 if h1 = 1. Now, (h1 − 1) + · · · + hn < h1 + · · · + hn,
and we can apply the induction hypothesis to conclude that m = n, pi = qi and hi = ki,
with 1 ≤ i ≤ n.
The above considerations about unique factorization into irreducible factors can be ex-
tended almost without changes to more general rings known as Euclidean domains. In such
rings, some abstract version of the division theorem is assumed to hold.
Definition 20.9. A Euclidean domain (or Euclidean ring) is an integral domain A such
that there exists a function ϕ : A → N with the following property: For all a, b ∈ A with
b = 0, there are some q, r ∈ A such that
a = bq + r
and ϕ(r) < ϕ(b).
Note that the pair (q, r) is not necessarily unique.
Actually, unique factorization holds in principal ideal domains (PID’s), see Theorem
21.12. As shown below, every Euclidean domain is a PID, and thus, unique factorization
holds for Euclidean domains.
Proposition 20.17. Every Euclidean domain A is a PID.
Proof. Let I be a nonnull ideal in A. Then, the set
{ϕ(a) | a ∈ I}
is nonempty, and thus, has a smallest element m. Let b be any (nonnull) element of I such
that m = ϕ(b). We claim that I = (b). Given any a ∈ I, we can write
a = bq + r
20.5. FACTORIZATION AND IRREDUCIBLE FACTORS IN K[X]
551
for some q, r ∈ A, with ϕ(r) < ϕ(b). Since b ∈ I and I is an ideal, we also have bq ∈ I,
and since a, bq ∈ I and I is an ideal, then r ∈ I with ϕ(r) < ϕ(b) = m, contradicting the
minimality of m. Thus, r = 0 and a ∈ (b). But then,
I ⊆ (b),
and since b ∈ I, we get
I = (b),
and A is a PID.
As a corollary of Proposition 20.17, the ring Z is a Euclidean domain (using the function
ϕ(a) = |a|) and thus, a PID. If K is a field, the function ϕ on K[X] defined such that
0
if f = 0,
ϕ(f ) =
deg(f ) + 1 if f = 0,
shows that K[X] is a Euclidean domain.
Example 20.3. A more interesting example of a Euclidean domain is the ring Z[i] of Gaus-
sian integers, i.e., the subring of C consisting of all complex numbers of the form a + ib,
where a, b ∈ Z. Using the function ϕ defined such that
ϕ(a + ib) = a2 + b2,
we leave it as an interesting exercise to prove that Z[i] is a Euclidean domain.
Not every PID is a Euclidean ring.
Remark: Given any integer, d ∈ Z, such that d = 0, 1 and d does not have any square factor
√
greater than one, the quadratic field , Q( d), is the field consisting of all complex numbers
√
√
of the form a + ib −d if d < 0, and of all the real numbers of the form a + b d if d > 0,
√
with a, b ∈ Q. The subring of Q( d) consisting of all elements as above for which a, b ∈ Z
√
√
is denoted by Z[ d]. We define the ring of integers of the field Q( d) as the subring of
√
Q(
d) consisting of the following elements:
√
(1) If d ≡ 2 (mod 4) or d ≡ 3 (mod 4), then all elements of the form a + ib −d (if d < 0)
√
or all elements of the form a + b d (if d > 0), with a, b ∈ Z;
√
(2) If d ≡ 1 (mod 4), then all elements of the form (a + ib −d)/2 (if d < 0) or all elements
√
of the form (a + b d)/2 (if d > 0), with a, b ∈ Z and with a, b either both even or both
odd.
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CHAPTER 20. POLYNOMIALS, IDEALS AND PID’S
√
Observe that when d ≡ 2 (mod 4) or d ≡ 3 (mod 4), the ring of integers of Q( d) is equal to
√
Z[
d]. For more on quadratic fields and their rings of integers, see Stark [96] (Chapter 8)
or Niven, Zuckerman and Montgomery [83] (Chapter 9). It can be shown that the rings of
√
integers, Z[ −d], where d = 19, 43, 67, 163, are PID’s, but not Euclidean rings.
√
Actually the rings of integers of Q( d) that are Euclidean domains are completely deter-
mined but the proof is quite difficult. It turns out that there are twenty one such rings corre-
sponding to the integers: −11, −7, −3, −2, −1, 2, 3, 5, 6, 7, 11, 13, 17, 19, 21, 29, 33, 37, 41, 57
and 73, see Stark [96] (Chapter 8).
It is possible to characterize a larger class of rings (in terms of ideals), factorial rings (or
unique factorization domains), for which unique factorization holds (see Section 21.1). We
now consider zeros (or roots) of polynomials.
20.6
Roots of Polynomials
We go back to the general case of an arbitrary ring for a little while.
Definition 20.10. Given a ring A and any polynomial f ∈ A[X], we say that some α ∈ A
is a zero of f , or a root of f , if f (α) = 0. Similarly, given a polynomial f ∈ A[X1, . . . , Xn],
we say that (α1, . . . , αn) ∈ An is a a zero of f, or a root of f, if f(α1, . . . , αn) = 0.
When f ∈ A[X] is the null polynomial, every α ∈ A is trivially a zero of f. This case
being trivial, we usually assume that we are considering zeros of nonnull polynomials.
Example 20.4. Considering the polynomial f (X) = X2 − 1, both +1 and −1 are zeros of
f (X). Over the field of reals, the polynomial g(X) = X2 + 1 has no zeros. Over the field C
of complex numbers, g(X) = X2 + 1 has two roots i and −i, the square roots of −1, which
are “imaginary numbers.”
We have the following basic proposition showing the relationship between polynomial
division and roots.
Proposition 20.18. Let f ∈ A[X] be any polynomial and α ∈ A any element of A. If the
result of dividing f by X − α is f = (X − α)q + r, then r = 0 iff f(α) = 0, i.e., α is a root
of f iff r = 0.
Proof. We have f = (X − α)q + r, with deg(r) < 1 = deg(X − α). Thus, r is a constant in
K, and since f (α) = (α − α)q(α) + r, we get f(α) = r, and the proposition is trivial.
We now consider the issue of multiplicity of a root.
Proposition 20.19. Let f ∈ A[X] be any nonnull polynomial and h ≥ 0 any integer. The
following conditions are equivalent.
(1) f is divisible by (X − α)h but not by (X − α)h+1.
20.6. ROOTS OF POLYNOMIALS
553
(2) There is some g ∈ A[X] such that f = (X − α)hg and g(α) = 0.
Proof. Assume (1). Then, we have f = (X − α)hg for some g ∈ A[X]. If we had g(α) = 0,
by Proposition 20.18, g would be divisible by (X − α), and then f would be divisible by
(X − α)h+1, contradicting (1).
Assume (2), that is, f = (X − α)hg and g(α) = 0. If f is divisible by (X − α)h+1, then
we have f = (X − α)h+1g1, for some g1 ∈ A[X]. Then, we have
(X − α)hg = (X − α)h+1g1,
and thus
(X − α)h(g − (X − α)g1) = 0,
and since the leading coefficient of (X − α)h is 1 (show this by induction), by Proposition
20.1, (X − α)h is not a zero divisor, and we get g − (X − α)g1 = 0, i.e., g = (X − α)g1, and
so g(α) = 0, contrary to the hypothesis.
As a consequence of Proposition 20.19, for every nonnull polynomial f ∈ A[X] and every
α ∈ A, there is a unique integer h ≥ 0 such that f is divisible by (X − α)h but not by
(X − α)h+1. Indeed, since f is divisible by (X − α)h, we have h ≤ deg(f). When h = 0, α
is not a root of f , i.e., f (α) = 0. The interesting case is when α is a root of f .
Definition 20.11. Given a ring A and any nonnull polynomial f ∈ A[X], given any α ∈ A,
the unique h ≥ 0 such that f is divisible by (X − α)h but not by (X − α)h+1 is called the
order, or multiplicity, of α. We have h = 0 iff α is not a root of f , and when α is a root of f ,
if h = 1, we call α a simple root, if h = 2, a double root, and generally, a root of multiplicity
h ≥ 2 is called a multiple root.
Observe that Proposition 20.19 (2) implies that if A ⊆ B, where A and B are rings, for
every nonnull polynomial f ∈ A[X], if α ∈ A is a root of f, then the multiplicity of α with
respect to f ∈ A[X] and the multiplicity of α with respect to f considered as a polynomial
in B[X], is the same.
We now show that if the ring A is an integral domain, the number of roots of a nonzero
polynomial is at most its degree.
Proposition 20.20. Let f, g ∈ A[X] be nonnull polynomials, let α ∈ A, and let h ≥ 0 and
k ≥ 0 be the multiplicities of α with respect to f and g. The following properties hold.
(1) If l is the multiplicity of α with respect to (f + g), then l ≥ min(h, k). If h = k, then
l = min(h, k).
(2) If m is the multiplicity of α with respect to f g, then m ≥ h + k. If A is an integral
domain, then m = h + k.
554
CHAPTER 20. POLYNOMIALS, IDEALS AND PID’S
Proof. (1) We have f (X) = (X − α)hf1(X), g(X) = (X − α)kg1(X), with f1(α) = 0 and
g1(α) = 0. Clearly, l ≥ min(h, k). If h = k, assume h < k. Then, we have
f (X) + g(X) = (X − α)hf1(X) + (X − α)kg1(X) = (X − α)h(f1(X) + (X − α)k−hg1(X)),
and since (f1(X) + (X − α)k−hg1(X))(α) = f1(α) = 0, we have l = h = min(h, k).
(2) We have
f (X)g(X) = (X − α)h+kf1(X)g1(X),
with f1(α) = 0 and g1(α) = 0. Clearly, m ≥ h + k. If A is an integral domain, then
f1(α)g1(α) = 0, and so m = h + k.
Proposition 20.21. Let A be an integral domain. Let f be any nonnull polynomial f ∈ A[X]
and let α1, . . . , αm ∈ A be m ≥ 1 distinct roots of f of respective multiplicities k1, . . . , km.
Then, we have
f (X) = (X − α1)k1 · · · (X − αm)kmg(X),
where g ∈ A[X] and g(αi) = 0 for all i, 1 ≤ i ≤ m.
Proof. We proceed by induction on m. The case m = 1 is obvious in view of Definition 20.11
(which itself, is justified by Proposition 20.19). If m ≥ 2, by the induction hypothesis, we
have
f (X) = (X − α1)k1 · · · (X − αm−1)km−1g1(X),
where g1 ∈ A[X] and g1(αi) = 0, for 1 ≤ i ≤ m − 1. Since A is an integral domain and
αi = αj for i = j, since αm is a root of f , we have
0 = (αm − α1)k1 · · · (αm − αm−1)km−1g1(αm),
which implies that g1(αm) = 0. Now, by Proposition 20.20 (2), since αm is not a root of the
polynomial (X − α1)k1 · · · (X − αm−1)km−1 and since A is an integral domain, αm must be a
root of multiplicity km of g1, which means that
g1(X) = (X − αm)kmg(X),
with g(αm) = 0. Since g1(αi) = 0 for 1 ≤ i ≤ m − 1 and A is an integral domain, we must
also have g(αi) = 0, for 1 ≤ i ≤ m − 1. Thus, we have
f (X) = (X − α1)k1 · · · (X − αm)kmg(X),
where g ∈ A[X], and g(αi) = 0 for 1 ≤ i ≤ m.
As a consequence of Proposition 20.21, we get the following important result.
Theorem 20.22. Let A be an integral domain. For every nonnull polynomial f ∈ A[X], if
the degree of f is n = deg(f ) and k1, . . . , km are the multiplicities of all the distinct roots of
f (where m ≥ 0), then k1 + · · · + km ≤ n.
20.6. ROOTS OF POLYNOMIALS
555
Proof. Immediate from Proposition 20.21.
Since fields are integral domains, Theorem 20.22 holds for nonnull polynomials over fields
and in particular, for R and C. An important consequence of Theorem 20.22 is the following.
Proposition 20.23. Let A be an integral domain. For any two polynomials f, g ∈ A[X], if
deg(f ) ≤ n, deg(g) ≤ n, and if there are n + 1 distinct elements α1, α2, . . . , αn+1 ∈ A (with
αi = αj for i = j) such that f (αi) = g(αi) for all i, 1 ≤ i ≤ n + 1, then f = g.
Proof. Assume f = g, then, (f −g) is nonnull, and since f(αi) = g(αi) for all i, 1 ≤ i ≤ n+1,
the polynomial (f − g) has n + 1 distinct roots. Thus, (f − g) has n + 1 distinct roots and
is of degree at most n, which contradicts Theorem 20.22.
Proposition 20.23 is often used to show that polynomials coincide. We will use it to show
some interpolation formulae due to Lagrange and Hermite. But first, we characterize the
multiplicity of a root of a polynomial. For this, we need the notion of derivative familiar in
analysis. Actually, we can simply define this notion algebraically.
First, we need to rule out some undesirable behaviors. Given a field K, as we saw in
Example 2.4, we can define a homomorphism χ : Z → K given by
χ(n) = n · 1,
where 1 is the multiplicative identity of K. Recall that we define n · a by
n · a = a + · · · + a
n
if n ≥ 0 (with 0 · a = 0) and
n · a = −(−n) · a
if n < 0. We say that the field K is of characteristic zero if the homomorphism χ is injective.
Then, for any a ∈ K with a = 0, we have n · a = 0 for all n = 0
The fields Q, R, and C are of characteristic zero. In fact, it is easy to see that every
field of characteristic zero contains a subfield isomorphic to Q. Thus, finite fields can’t be of
characteristic zero.
Remark: If a field is not of characteristic zero, it is not hard to show that its characteristic,
that is, the smallest n ≥ 2 such that n·1 = 0, is a prime number p. The characteristic p of K
is the generator of the principal ideal pZ, the kernel of the homomorphism χ : Z → K. Thus,
every finite field is of characteristic some prime p. Infinite fields of nonzero characteristic
also exist.
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CHAPTER 20. POLYNOMIALS, IDEALS AND PID’S
Definition 20.12. Let A be a ring. The derivative f , or Df , or D1f , of a polynomial
f ∈ A[X] is defined inductively as follows:
f = 0,
if f = 0, the null polynomial,
f = 0,
if f = a, a = 0, a ∈ A,
f = nanXn−1 + (n − 1)an−1Xn−2 + · · · + 2a2X + a1,
if f = anXn + an−1Xn−1 + · · · + a0, with n = deg(f) ≥ 1.
If A = K is a field of characteristic zero, if deg(f ) ≥ 1, the leading coefficient nan of f is
nonzero, and thus, f is not the null polynomial. Thus, if A = K is a field of characteristic
zero, when n = deg(f ) ≥ 1, we have deg(f ) = n − 1.
For rings or for fields of characteristic p ≥ 2, we could have f = 0, for a polynomial f
of degree ≥ 1.
The following standard properties of derivatives are recalled without proof (prove them
as an exercise).
Given any two polynomials, f, g ∈ A[X], we have
(f + g) = f + g ,
(f g) = f g + f g .
For example, if f = (X − α)kg and k ≥ 1, we have
f = k(X − α)k−1g + (X − α)kg .
We can now give a criterion for the existence of simple roots. The first proposition holds for
any ring.
Proposition 20.24. Let A be any ring. For every nonnull polynomial f ∈ A[X], α ∈ A is
a simple root of f iff α is a root of f and α is not a root of f .
Proof. Since α ∈ A is a root of f, we have f = (X − α)g for some g ∈ A[X]. Now, α is a
simple root of f iff g(α) = 0. However, we have f = g + (X − α)g , and so f (α) = g(α).
Thus, α is a simple root of f iff f (α) = 0.
We can improve the previous proposition as follows.
Proposition 20.25. Let A be any ring. For every nonnull polynomial f ∈ A[X], let α ∈ A
be a root of multiplicity k ≥ 1 of f. Then, α is a root of multiplicity at least k − 1 of f . If
A is a field of characteristic zero, then α is a root of multiplicity k − 1 of f .
20.6. ROOTS OF POLYNOMIALS
557
Proof. Since α ∈ A is a root of multiplicity k of f, we have f = (X − α)kg for some g ∈ A[X]
and g(α) = 0. Since
f = k(X − α)k−1g + (X − α)kg = (X − α)k−1(kg + (X − α)g ),
it is clear that the multiplicity of α w.r.t. f is at least k−1. Now, (kg+(X−α)g )(α) = kg(α),
and if A is of characteristic zero, since g(α) = 0, then kg(α) = 0. Thus, α is a root of
multiplicity k − 1 of f .
As a consequence, we obtain the following test for the existence of a root of multiplicity
k for a polynomial f :
Given a field K of characteristic zero, for any nonnull polynomial f ∈ K[X], any α ∈ K
is a root of multiplicity k ≥ 1 of f iff α is a root of f, D1f, D2f, . . . , Dk−1f, but not a root of
Dkf .
We can now return to polynomial functions and tie up some loose ends. Given a ring A,
recall that every polynomial f ∈ A[X1, . . . , Xn] induces a function fA : An → A defined such
that fA(α1, . . . , αn) = f (α1, . . . , αn), for every (α1, . . . , αn) ∈ An. We now give a sufficient
condition for the mapping f → fA to be injective.
Proposition 20.26. Let A be an integral domain. For every polynomial f ∈ A[X1, . . . , Xn],
if A1, . . . , An are n infinite subsets of A such that f (α1, . . . , αn) = 0 for all (α1, . . . , αn) ∈
A1 ×· · ·×An, then f = 0, i.e., f is the null polynomial. As a consequence, if A is an infinite
integral domain, then the map f → fA is injective.
Proof. We proceed by induction on n. Assume n = 1. If f ∈ A[X1] is nonnull, let m = deg(f)
be its degree. Since A1 is infinite and f (α1) = 0 for all α1 ∈ A1, then f has an infinite number
of roots. But since f is of degree m, this contradicts Theorem 20.22. Thus, f = 0.
If n ≥ 2, we can view f ∈ A[X1, . . . , Xn] as a polynomial
f = gmXm
n + gm−1X m−1
n
+ · · · + g0,
where the coefficients gi are polynomials in A[X1, . . . , Xn−1]. Now, for every (α1, . . . , αn−1) ∈
A1 × · · · × An−1, f(α1, . . . , αn−1, Xn) determines a polynomial h(Xn) ∈ A[Xn], and since An
is infinite and h(αn) = f (α1, . . . , αn−1, αn)