Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV 87
M ⊂ M′ ⊂ R M′ = R or M′ = M.
Example. An ideal generated by a prime number is a maximal ideal of the ring of integers. But
the zero ideal of the ring of integers is not maximal.
Proof. Let p be any prime integer and let S be any ideal containing the principal ideal generated by p.
Now the ring of integers being principal ideal ring the ideal S is a principal ideal and it is generated by
the integer q. We have therefore
(p) ⊂ (q) ⊂ R
p ∈ (q)
p = kq, k ∈ R .
Since p is prime, p = kq either k = 1 or q= 1 .
Now k = 1
p = q
(p) = (q)
and q = 1 (q) = (1) = R (Since R is generated by 1) .
Hence (p) is maximal ideal.
Theorem. Every maximal ideal M of a commutative ring R with unity is a prime ideal.
Proof. It suffices to prove that if a, b ∈ R then
ab ∈ M a ∈ M or b ∈ M .
Let us suppose that a ∉ M. If we prove that b ∈ M then we are done. It can be seen that the set
N = {ra + m | r ∈ R, m ∈ M}
is an ideal of R.
Since 1 ∈ R, therefore a + m ∈ N. But a+m ∉ M since a ∉ M. Therefore
M ⊂ N ⊂ R , M ≠ N.
M being a maximal ideal asserts that N = R. Therefore 1 ∈ R 1 ∈ N. So we can find two elements r
∈ R, m ∈ M such that
1 = ra + m
b = r(ab) + mb, b ∈ R
Since M is an ideal of R, therefore
ab ∈ M, r ∈ R r(ab) ∈ M
and m ∈ M , b ∈ R mb ∈ M .
Therefore b ∈ M.
Hence M is a prime ideal.
Theorem. An ideal M of a commutative ring R with unity is maximal if and only if R/M is a field.
Proof. Let M be a maximal ideal of R. Since R is a commutative ring with unity, R/M is also a
commutative ring with unity element. Let A* be an ideal of R/M and
A = {r | r+M ∈ A*}
If r , s ∈ A , then r+M, s+M ∈ A*. Therefore
(r−s) + M = (r+M) − (s+M) ∈ A*
r−s ∈ A
If r ∈ A, t ∈ R, then
r+M ∈ A* and