Page 208 - DMTH403_ABSTRACT_ALGEBRA

Unit 20: Principal Ideal Domains

Thus, we have shown that d = (a,b), and d= ax+by for some x.y  R. Notes
The fact that F[x] is a PID gives-us the following corollary to Theorem 4.

Corollary: Let F be a field. Then any two polynomials f(x) and g(x) in F[x] have a g.c.d which is
of the form a(x)f(x)+b(x)g(x) for some a(x), b(x)  F[x].

1 ( x)

For example, (c), (x–1) = (x – 2x + 6x – 5) + (x – 2x + 1).
2
2
3
5 5
Now you can use Theorem 4 to prove the following exercise about relatively prime elements in
a PID, i.e., pairs of elements whose g.c.d is 1.
Let us now discuss a concept related to that of a prime element of a domain.
Definition: Let R be an integral domain. We say that an element x  R is irreducible if
(i) x is not a unit, and
(ii) if x = ab with a,b  R, then a is a unit or b is a unit.

Thus, an element is irreducible if it cannot be factored in a non-trivial way, i.e., its only factors
are its associates and the units in the ring.
So, for example, the irreducible elements of Z are the prime numbers and their associates. This
means that an element in Z is prime iff it is irreducible.
Another domain in which we can find several examples is F[x], where F is a field. Let us look at
the irreducible elements in R[x] and C[x], i.e., the irreducible polynomials over R and C. Consider
the following important theorem about polynomials in C[x]. You have already come across this
in the Linear Algebra course.

Theorem 5 (Fundamental Theorem of Algebra): Any non-constant polynomial in C[x] has a
root in C.
Does this tell us anything about the irreducible polynomials over C? Yes. In fact, we can also
write it as:
Theorem 5: A polynomial is irreducible in C[x] iff it is linear.
A corollary to this result is:

Theorem 6: Any irreducible polynomial in R[x] has degree 1 or degree 2.
We will not prove these results here but we will use them often when discussing polynomials
over R or C. You can use them to solve the following exercise.

Let us now discuss the relationship between prime and irreducible elements in a PID.
Theorem 7: In a PID an element is prime iff it is irreducible.

Proof: Let R be a PID and x  R be irreducible. Let x | ab, where a, b  R. Suppose x I a.
Then (x,a) = 1, since the only factor of x is itself, up to units. Thus, xb, Thus, x is prime.

Task Let R be a domain and p  R be a prime element. Show that p is irreducible.
(Hint: Suppose p = ab. Then p | ab. If p | a, then show that b must be a unit.)

LOVELY PROFESSIONAL UNIVERSITY 201