Page 238 - DMTH403_ABSTRACT_ALGEBRA

Unit 24: Irreducibility and Field Extensions

Now apply Eisenstein’s criterion taking p as the prime. We find that f(x+l) is irreducible. Notes
Therefore, f(x) is irreducible.

So far we have used the fact that if f(x)  Z[x] is irreducible over Z, then it is also irreducible over
Q. Do you think we can have a similar relationship between irreducibility in Q[x] and R[xl? To

2
answer this, consider f(x) = x - 2. This is irreducible in Q[x], but f(x) = (x  2)(x  2) in R[x].
Thus, we cannot extend irreducibility over Q to irreducibility over W.
But, we can generalise the fact that irreducibility in Z[x] implies irreducibility in Q[x]. This is not
only true for Z and Q; it is true for any UFD R and its field of quotients F. Let us state this
relationship explicitly.
Theorem 4: Let R be a UFD with field of quotients F.

(i) If f(x)  R[x] is an irreducible primitive polynomial, then it is also irreducible in F[x].
(ii) (Eisenstein’s Criterion) Let f(x) = a + a x + ... + a, x  R[x] and p  R be a prime element
n
1
0
such that p | a,, p | a and p | a for 0  i < n. Then f(x) is irreducible in F[x].
2
0 i
The proof of this result is on the same lines as that of Theorems 2 and 3. We will not be doing it
here. But if you are interested, you should try and prove the result yourself.
Now, we have already pointed out that if F is a field and f(x) is irreducible over F, then F[x]/
<f(x)> is field. How is this field related to F? That is part of what we will discuss in the next
section.

24.2 Field Extensions

We shall discuss subfields and field extensions. To start with let us define these terms. By now
the definition may be quite obvious to you.
Definition: A non-empty subset S of a field F is called a subfield of F if it is a field with respect
to the operations on F. If S$F, then S is galled a proper subfield of F.
A field K is called a field extension of F if F is a subfield of K. Thus, Q is a subfield of R and R is
a field extension of Q. Similarly, C is a field extension of Q as well as of R.
Note that a non-empty subset S of a field F is a subfield of F if
(i) S is a subgroup of (F,+), and
(ii) the set of all non-zero elements of S forms a subgroup of the group of non-zero elements
of F under multiplication.
Theorem 5: A non-empty subset S of a field F is a subfield of F if and only if
(i) a  S, b  S  a – b  S, and

(ii) a  S , b  S , b  0  ab  S.
-1
Now, let us look at a particular field extension of a field F. Since F[x] is an integral domain, we
can obtain its field of quotients. We denote this field by F(x). Then F is a subfield of F(x). Thus,
F(x) is a field extension of F. Its elements are expressions of the form f,( x) where f(x), g(x)  F[x]
and g(x) # 0.

g(x)
There is another way of obtaining a field extension of a field F from F[x]. We can look at quotient
rings of F[x] by its maximal ideals. You know that an ideal is maximal in F[x] iff it is generated
by an irreducible polynomial over F. So, F[x]/<f(x)> is a field iff f(x) is irreducible over F.

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