Page 245 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Let F be a field and f (x)  F[x]. Then a  P is a root of f(x) if and only if (x–a) | f(x)).
We can generalise this criterion to define a root of multiplicity m of a polynomial in F[x].

Definition: Let F be a field and f(x)  F[x]. We say that a  F is a root of multiplicity m (where m
is a positive integer) of

f(x) if (x – a) | f(x), but (x–a) m+1 | f(x).
m
For example, 3 is a root of multiplicity 2 of the polynomial (x–3) (x+2)  Q[x]; and (–2) is a root
2
of multiplicity 1 of this polynomial.
Now, is it easy to obtain all the roots of a given polynomial? Any linear polynomial ax+b  F[x]
will have only one root, namely, -a b. This is because ax+b = 0 iff x = -a b.
-1
-1
In the case of a quadratic polynomial ax + bx + c  F[x], you know that its two roots are obtained
2
by applying the quadratic formula
2
 b  b  4ac
2a
For polynomials of higher degree we may be able to obtain some roots by trial and error. For
example, consider f(x) = x – 2x + 1  R[x]. Then, we try out x = 1 and find f(1) = 0. So, we find that
5
1 is a zero of f(x). But this method doesn’t give us all the roots of f(x).
As we have just seen, it is not easy to find all the roots of a given polynomial. But, we can give
a definite result about the number of roots of a polynomial.
Theorem 1: Let f(x) be a non-zero polynomial of degree n over a field F:Then f(x) has at most n
roots in F.

Proof: If n = 0, then f(x) is a non-zero constant polynomial.
Thus, it has no roots, and hence, it has at most 0 ( = n) roots in F.
So, let, us assume that n  1. We will use the principle of induction on n. If deg f(x) = 1,
then

f(x) = a + a x, where a , a,  F and a,  0.
0
0
1
So f(x) has only one root, namely, (–a a ).
-1
1
0
Now assume that the theorem is true for all polynomials in F[x] of degree  n. We will show that
the number of roots of f(x)  n.
If f(x) has no root in F, then the number of roots of f(x) in F is 0 S n. So, suppose f(x) has a root
a  F.
Then f(x) = (x – a) g(x), where deg g(x) = n–1.
Hence, by the induction hypothesis g(x) has at most n–1 roots in F, say a ,....,a . Now,
n-1
1
a is a root of g(x)  g(a ) = 0  f(a ) = (a,–a) g(a ) = 0
i
i
i
i
 a a is a root of f(x)  i = 1, ..., n – 1.
i
Thus, each root of g(x) is a root of f(x).
Now, b  F is a root of f(x) iff f(b) = 0, i.e., iff (b – a) g(b) = 0, i.e., iff b – a = 0 or g(b) = 0, since F is
an integral domain. Thus, b is a root of f(x) iff b = a or b is a root of g(x). So, the only roots of
f(x) are a and a , ..., a . Thus, f(x) has at the most n roots, and so, the theorem is true for n.
n-1
1

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