Page 236 - DMTH403_ABSTRACT_ALGEBRA

Unit 24: Irreducibility and Field Extensions

But p | a and p | b . So we reach a contradiction. Therefore, our supposition is false. That is, our Notes
r
s
theorem is true.
Let us shift our attention to polynomials over Q now.

3 1 1
2
Consider any polynomial over Q, say f(x) = x + x + 3x + . If we take the l.c.m of the
3
2 5 3
denominators, is., of 2, 5, 1 and 3, i.e., 30 and multiply f(x) by it, what do we get? We get
30f(x) = 45x + 6x + 90x + 10  Z[x]
2
3
Using the same process, we can multiply any f(x)  Q[x] by a suitable integer d so that df(x),
 Z[X]. We will use this fact while relating irreducibility in Q[x] with irreducibility in Z[x].
Theorem 2: If f(x)  Z[x] is irreducible in Z[x], then it is irreducible in Q[x].
Proof: Let us suppose that f(x) is not irreducible over Q[x]. Then we should reach a contradiction.
So let f(x) = g(x) h(x) in Q[x], where neither g(x) nor h(x) is unit, i.e., deg g(x) > 0, deg h(x) > 0. Since
g(x)  Q[x].  m  Z such that mg(x)  Z[x]. Similarly,  n  Z such that nh(x)  Z[x]. Then,

mnf(x) = mg(x) nh(x) ... (1)
Thus, (1) gives us
mnrf (x) = stg (x)h (x) ...(2)
1
1
1
Since g (x) and h (x) are primitive, Theorem 1 says that g (x) h (x) is primitive. Thus, the content
1
1
1
1
of the right hand side polynomial in (2) is st. But the content of the left hand side polynomial in
(2) is mnr. Thus, (2) says that mnr = st.
Hence, using the cancellation law in (2), we get f,(x) = g,(x) h (x).
1
Therefore, f(x) = rf (x) = (rg (x)) h (x) in Z[x], where neither rp (x) nor h (x) is a unit. This
1
1
1
1
1
contradicts the fact that f(x) is irreducible in Z[x].
Thus, our supposition is false. Hence, f(x) must be irreducible in Q[x].
What this result says is that to check irreducibility of ii polynomial in Q[x], it is enough to check
it in Z[x]. And, for checking it in Z[x] we have the terrific Eisenstein’s criterion, that we mentioned
at the beginning.
Theorem 3 (Eisenstein’s Criterion): Let f(x) = a + a x + ... + a,,x  Z[x]. Suppose that for some
n
l
0
prime number p;
(i) P | a ,
n
(ii) p | a , p | a , ..., p | a , and
0
n-1
1
(iii) p | a .
2
0
Then f(x) is irreducible in Z[x] (and hence Q[x]).
Proof: Suppose f(x) is reducible in Z[x].
Let f(x) = g(x) h(x),
where g(x) = b + b x + ... + b,, x , m > 0 and
m
1
0
h(x) = c + c x + ... + c x , r > 0.
r
r
0
1

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