Page 235 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes 24.1 Irreducibility in Q[x]

We introduced you to irreducible polynomials in F[x], where F is a field. We also stated the
Fundamental Theorem of Algebra, which said that a polynomial over C is irreducible iff it is
linear. You also learnt that if a polynomial over R is irreducible, it must have degree 1 or
degree 2. Thus, any polynomial over R of degree more than 2 is reducible. And, using the
quadratic formula, we know which quadratic polynomials over R are irreducible.

Now let us look at polynomials over Q. Again, as for any field F, a linear polynomial over Q is
irreducible. Also, by using the quadratic formula we can explicitly obtain the roots of any
quadratic polynomial over Q, and hence figure out whether it is irreducible or not. But, can you
tell whether 2x + 3x – 6x + 3x + 12 is irreducible over Q. This criterion was discovered by the
4
3
7
5
nineteenth century mathematician Ferdinand Eisenstein. In this section we will build up the
theory for proving this useful criterion.
Let us start with a definition.
Definition: Let f(x) = a, + a x + . .. + a x  Z[x]. We define the content of f[x] to be the g.c.d. of the
n
n
1
integers a , a ,,..., a,.
0
1
We say that f(x) is primitive if the content of f(x) is 1.
For example, the content of 3x + 6x + 12 is the g.c.d. of 3, 6 and 12, i.e., 3. Thus, this polynomial
2
is not primitive. But x + 3x + 4x – 5 is primitive, since the g.c.d of 1, 0, 0, 3, 4, –5 is 1.
2
5
We will now prove that the product of primitive polynomials is a primitive polynomial. This
result is well known as Gauss’ lemma.
Theorem 1: Let f(x) and g(x) be primitive polynomials. Then so is f(x) g(x).
Proof: Let f(x) = a + a x + ... + a x  Z[x] and
n
0
n
1
g(x) = b + b x + ... + b x  Z[x], where the
m
0
1
m
g.c.d. of a , a , ..., a, is 1 and the g.c.d. of b , b ..., b is 1. Now
1
0
1
m
0
f(x) g(x) = c + c x + ... + c m+n x m+n
1
0
where c, = a b + a b + ... + a b .
1 k-1
k 0
0 k
To prove the result we shall assume that it is false, and then reach a contradiction. So, suppose
that f(x) g(x) is not primitive. Then the g.c.d. of c , c ...., c m+n is greater than 1, and hence some
0
1
prime p must divide it. Thus, p | c  i = 0, 1, ..., m+n. Since f(x) is primitive, p does not divide
i
some a . Let r be the least integer such that p| a . Similarly, let s be the least integer such that
i
r
p| b . s
Now consider
c = a b + a b + ... + a b + ... + a b 0
0 r+s
r+s
I r+s-1
r+s
r s
= a b + (a b + a b + ... + a b + a b + ... + a b )
r s
r+1 s-1
r+s
s+1
r-1
I r+s-1
0 r+s
0
By our choice of r and s, p | a , p | a , ..., p | a , and p | b , p | b , ..., p | b . Also p | c ,
s-1
r-1
r+s
0
1
0
1
Therefore, p | c – (a b +... + a b + a b + ... + a b )
0
r+s
r+s
s+1
r-1
s-1
r+1
0 r+s
i.e., p | a b,.
r
 p ( a, or p | b , since p is a prime.
s
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