Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV 95
and
Z/A ~ Im(f) = f (z) .
But Im f is a subring of K. Therefore, Im(f) is without zero divisor. It follows therefore that Z/A is
without zero divisor. Therefore either A = {0} or A is a prime ideal.
If A = {0}, then
ne = 0 ⇔ n = 0 .
If A is a prime ideal then we can find a prime number p such that
A = ker f = < p >
(ii)
Hence from (i) and (ii)
ne = 0 ⇔ p | n .
Thus we have seen that if K is a field, then one of the following two cases, holds
(i) ne = 0 ⇔ n = 0
(ii)
ne = 0 ⇔ p | n where p is a prime.
In the first case we say that the field K is of characteristic zero while in the second case, K is called a
field of characteristic p. Thus characteristic of a filed is zero or a prime number.
It is clear that a field of characteristic zero is infinite since in that case Z/A = Z and therefore Z ~ Im(f) .
Hence Im(f) and K are infinite
Example 1. The characteristic of the field Q of rational numbers is zero, since ne = 0 n = 0 (Θ e ≠ 0).
2. The characteristic of the field Z/ < p > is a prime number p.
Definition. Fields with non-zero characteristic are known as Modular Fields.
Definition. A field is said to be prime if it has no subfield other than itself.
Examples 1. If p is a prime, then Z/pZ is a prime field. Additive group Z/pZ . Hence Z/pZ is a prime
field.
2. Field Q of rational numbers is a prime field. To prove it let K be a subfield of Q. Then 1∈K. Since
K is an additive subgroup of Q, therefore 1+1 = 2∈K. Similarly 3∈K. Now K being a field, every nonzero
element of a K is invertible under multiplication. Therefore, n∈K, n ≠ 0 ∈ K. Then m∈K,
1
n
1 m ∈ K ∈ K. Hence K contains all rational numbers. Hence K = Q as a consequence of which Q
n n
is a prime field.
We have seen that the field Q of rational numbers and Z/pZ are prime fields. Now we shall prove that
upto isomorphism there are only two prime fields Q and Z/pZ.
Proof. Let K be any prime field and let e denote the unit element of the same. Since K is prime, the
subfield generated by e must coincide with K.
Consider the mapping f : Z → K defined by
f(n) = ne, n ∈ Z .
This mapping is a ring homomorphism. For,
f(n+m) = (n+m)e
= ne+me = f(n) + f(m)
f(nm) = (nm)e = (ne) (me) = f(n)f(m)