Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

102
ADVANCED ABSTRACT ALGEBRA
that is,
But
and
g(x) q(x) + r(x) = g(x) q 1 (x) + r 1 (x) ,
g(x) [ q(x) − q 1 (x)] = r 1 (x) − r(x)
deg g(x) [q(x) − q 1 (x)] ≥ n
deg [r 1 (x) − r(x)] < n .
Hence (v) is possible only when
and
g(x) [q(x) − q 1 (x)] = 0
r 1 (x) − r(x) = 0
That is, when
q(x) = q 1 (x) and r(x) = r 1 (x)
Hence q(x) and r(x) are unique.
With the help of this theorem we shall prove that a polynomial domain F[x] over a field F is a principal
ideal domain.
Theorem. A polynomial domain F[x] over a field F is a principal ideal domain.
Proof. Let S be any ideal of F[x] other than the zero ideal and let g(x) be a polynomial of lowest degree
belonging to S. If f(x) is an arbitrary polynomial of S, then by division algorithm there exist uniquely
two polynomials q(x) and r(x) belonging to F[x] such that
f(x) = g(x) q(x) + r(x)
where r(x) = 0 or deg r(x) < deg g(x) .
Thus
r(x) = f(x) − g(x) q(x) ∈ S .
Also, since g(x) is a polynomial of lowest degree belonging to S, we see that deg r(x) cannot be less than
of g(x). Thus r(x) = 0 and we have
f(x) = g(x) q(x)
Since f(x) is arbitrary polynomial belonging to S, therefore
S = (g(x))
Hence F[x] is principal ideal domain.
Example. Show that the polynomial ring I[x] over the ring I of integers is not a principal ideal ring.
To establish this we have to produce an ideal of I[x] which is not a principal ideal. In fact we shall show
that the ideal (x, q) of the ring I[x] generated by two elements x and q of I[x] is not a principal ideal.
Let if possible (x, q) be a principal ideal generated by a member f(x) of I[x] so that we have
(x, q) = (f(x))
Thus we have relations of the form
q = f(x) g(x)
x = f(x) h(x)
where g(x) and h(x) are members of I[x]. These imply
deg f(x) + deg g(x) = deg q = 0
(i)
deg f(x) + deg h(x) = deg x = 1
(ii)
From (ii) we get
(v)