Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

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ADVANCED ABSTRACT ALGEBRA
rt + M = (r+M) (t+M) ∈ A* (because A* is an ideal of R/M) .
rt ∈ A.
R being commutative tr also belongs to A.
Hence A is an ideal of R.
If a ∈ M, then
a+M = M ∈ R/M
(since M is the zero element of R/M)
a+M ∈ A* (since (1+M) (a+m) ∈ A* ,
a ∈ A A* being ideal of R/M)
Therefore
M ⊂ A ⊂ R .
Let us suppose that A* ≠ {0} then there exists an element r+M of A* such that
r +M ≠ M
But r+M ∈ A* r ∈ A ,
r+M ≠ M r ∉ M A ≠ M.
Thus we have proved that if A* ≠ {0}, then
M ⊂ A ⊂ R
Since M is maximal therefore, A = R. If r ∈ R then r ∈ A which implies that r +M ∈ A*. It follows
therefore, that A* = R/M.
We have proved therefore, that R/M has only two ideals {0} and R/M and hence R/M is a field.
Conversely, let R/M is a field. Then R/M has only two ideals {0} and R/M itself. Hence
A* = {0}
or
A* = R/M.
If A* = {0} then A* = M
Therefore,
A = {r | r +M ∈ A*}
= {r | r+M = M}
= {r | r ∈ M}
= M
If A* = R/M then
A = {r | r +M ∈ R/M}
(Θ M is zero element of R/M)
= {r | r ∈ M}
= R .
Therefore, R has only two ideals M and R. Hence M is a maximal ideal.
Imbedding of a ring and an integral domain.
Definition. If a ring R is isomorphic to a subring T of a ring S then R is called imbedded in S. The ring
S is called extension or over ring of R.
Theorem. Every ring R can be imbedded in a ring S with unit element.
Proof. Let S be a set defined by
S = Z × R = {(m,a) | m ∈ Z, a ∈ R}.