Advanced Abstract Algebra - Maharshi Dayanand University, Rohtak

UNIT-IV 81
Definition. Let S be a subring of a ring R. If
x ∈ S, a ∈ R ax ∈ S ,
then S is called left ideal of R.
If x ∈ S , a ∈ R xa ∈ S ,
then S is called right ideal of R.
Ideals and Quotient Rings
If x ∈ S, a ∈ R xa ∈ S and ax ∈ S then S is called two sided ideal or simply ideal of R.
* If R is a commutative ring then all the three notions are same since in that case ax = xa ∈ S .
** Every ring has two trivial ideals :
(i) R itself and is called unit ideal.
(ii) Zero ideal [0] consisting of zero element only.
Any other ideal except these two trivial ideals is called proper ideal.
Theorem.
The intersection of any two left ideals of a ring is again a left ideal of the ring.
Proof. Let S 1 and S 2 be two ideals of R. S 1 and S 2 being subring of R, S 1 ∩ S 2 is also a subring of R.
Again let x ∈ S 1 ∩ S 2 .
Let a ∈ R.
x ∈ S 1 , x∈ S 2 .
Then since S 1 and S 2 are left ideals,
a ∈ R, x ∈ S 1 ax ∈ S 1
a ∈ R , x ∈ S 2 ax ∈ S 2
ax ∈ S 1 ∩ S 2
S 1 ∩ S 2 is a left ideal.
Theorem :- Let K(T) be the kernel of a ring homomorphism T : R → S . Then K(T) is a two sided
ideal of R.
Proof. Let a, b ∈ K (T). Then
T(a) = T(b) = 0 .
Therefore,
T(a+b) = T(a) + T(b) = 0 + 0 = 0 (by ring
T(ab) = T(a).T(b) = 0.0 = 0 homomorphism)
which implies that a+b, ab ∈ K(T). Hence K(T) is a subring of R.
Now let a ∈ K(T) and r ∈ R . It suffices to prove that ar, ra ∈ K(T)
T(ar) = T(a).T(r)
= 0 . T(r) (Θ a ∈ K(T) T(a) = 0)
= 0
This implies that ar ∈ K(T). Similarly,
T(ra) = T(r) T(a) = T(r).0 = 0
Hence K(T) is an ideal of R.
Theorem. A field has no proper ideal.
ra ∈ K(T) .