Page 206 - DMTH403_ABSTRACT_ALGEBRA

Unit 20: Principal Ideal Domains

20.2 Principal Ideal Domain (PID) Notes

In the previous section you have proved that every ideal of F[x] is principal, where F is a field.
There are several other integral domains, apart from Euclidean domains, which have this property.
We give such rings a very appropriate name.

Definition: We call an integral domain R a principal ideal domain (PID, in short) if every ideal
in R is a principal ideal.

Note Every Euclidean domain is a PID

Thus, Z is a PID. Can you think of another example of a PID? What about Q and Q[x]? In fact, by
Theorem 2 all Euclidean domains are PIDs. But, the converse is not true. That is, every principal
ideal domain is not a Euclidean domain.
b
For example, the ring of all complex numbers of the form a   1 i 19 ,  where a, b  Z, is a
2
principal ideal domain, but not a Euclidean domain.
Now let us look at an example of an integral domain that is not a PID.

Example: Show that Z[x] is not a PID,
Solution: You know that Z[x] is a domain, since Z is one. We will show that all its ideals are not
principal. Consider the ideal of Z[x] generated by 2 and x, i.e., < 2,x>. We want to show that
< 2, x >  <f(x)> for any f(x)  Z[X].
On the contrary, suppose that 3 f(x)  Z[x] such that < 2, x > = < f(x) >. Clearly, f(x)  0.
Also, 3 g(x), h(x)  Z[x] such that

2 = f(x) g(x) and x = f(x) h(x).
Thus, deg f(x) + deg g(x) = deg 2 = 0 ...(1)
and deg f(x)+deg h(x) = deg x = 1 ...(2)
(I) shows that deg f(x) = 0, i.e., f(x)  Z, say f(x) = n.
Then (2) shows that deg h(x) = 1. Let h(x) = ax+b with a,b  Z.

Then x =f(x) h(x) = n(ax+b).
Comparing the coefficients on either side of this equation, we see that na = 1 and nb = 0. Thus, n
is a unit in Z, that is, n = If I.

Therefore, 1  < f(x) > = < x,2 >. Thus, we can write
I = x (a +a x+ ...+a x ) + 2(b +b x+ .... +b x ), where a ,b  Z  i = 0, l,.. ...., r and j = 0, 1,...,s.
r
s
1
s
j
i
1
0
0
r
Now, on comparing the constant term on either side we see that 1 = 2b . This can’t be true, since
0
2 is not invertible in Z. So we reach a contradiction.
Thus, < x,2 > is not a principal ideal.
Thus, Z[x] is not a P.I.D.
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