Subrings, Ideals, Quotient Rings & Euclidean Rings Ideas
The concept of an ideal of a ring is analogous to that of a normal subgroup of a group. Some of the subrings which we call ideals play a very important role as the normal subgroups
in group theory.
Definition. (Ideal). Let. (R, +,• ) be a ring. A non-empty subset U of R is called a two-sided ideal or ideal if (1) a,b ∈ U=>a-b∈U and (2) a ∈ U and r ∈ R=> ar,r a∈ U.
(Ideal). A subring U of a ring R is called a (two-sided) ideal of R if, for every r ∈ R and every a ∈ U, both ra and ar are in U.
Note
- A subring u of the ring R satisfying rU⊂ U and Ur⊂U for all r ∈ R is an ideal.
- The (2) condition of ideal is stronger than the (2) condition of a subring.
- The condition (1), namely, a,b∈ U => a -b∈U is called module property.
- If U is an ideal of the ring R,+, • ) then (U, +) is a normal subgroup of the commutative group (R, +). Hence zero element in R is zero element in U.
Definition. A non-empty subset U of a ring R is called a right ideal if.
(1) a,b∈U=>a-b ∈U and(2)a ∈U,r ∈R=>ar ∈U . A non-empty subset U of a ring R is called a left ideal if (1) a,b ∈U a-b ell and a ∈U,r∈R=>ra ∈U .
Note.
- An ideal is both a left and a right ideal.
- For commutative rings left ideals coincide with right ideals.
Theorem 1. If U is an ideal of a ring R with unity element and 1∈ U then U =R.
Proof: by the definition ideal U⊂R.
Also x ∈R=>x.1 eR =>x.1∈U for x∈R, 1 ∈u (Def. of ideal) =>x ∈U .
R⊂U and hence U ⊂ R
Theorems On Ideals In Subrings And Quotient Rings
Theorem 2. A field has no proper non-trivial ideals, (or) The ideals of a field f are only {0} and F itself
Proof. Let U be an ideal of F so that U ≠ {0}
We now prove that U = F
By the definition of ideal, U⊂ F …… (1)
Let a∈ U and a≠ 0.
For a (≠ 0) ∈F there exists a -1 ∈ F so that aa -1 = 1
x ∈F => x 1 ∈ F=>x . 1 ∈U for xeF, 1∈U
=> x∈U .-. F ∈U … (2)
From (1) and (2): U = F Hence ideals of F are either {o} orF.
Theorem 3. If R is a commutative ring and a ∈ R then Ra = {ra\r| ∈ R \ is an ideal of R.
Proof. For 0 ∈ R, 0a = 0∈Ra. Ra≠ Φand R⊂R.
Let x,y∈Ra. Then x =r 1 a, y = r2a where r1,r2 ∈ R :. x,y ∈ Ra=>x-y ∈ R
x. y = r1a-r2a=(r1-r2)a= ra where r =r1~r2 ∈ R.
Let x ∈ Ra and r ∈R.
x.r = (r1a)r ( :.x = r1a Where r∈R) =r1(ar) = r1(ra) (By R5 and R is commutative )
=(r1r) a-r’a where r’ = r1r ∈ R .
Since R is commutative, x.r=r.x.
∴ x∈Ra, r ∈R=>xr = rx ∈Ra ……. (2)
Hence from (1) and (2): Ra is an ideal of R.
Note. 1. If R is a commutative ring and a e R then aR = { ar \ r e R } is an ideal of R.
2. If R is a r in and a ∈ R then Ra is a left ideal and aR is a right ideal.
Theorem 4. A commutative ring R with unity element is a field if R have no proper ideals.
Proof. Since the Ring R has no proper nontrivial ideals, the ideals of R are {o} and R only.
To prove that R is a field we have to show that every a (≠ 0) ∈ R has a multiplicative inverse. We know that aR = {ar|r ∈ R} is an ideal of R.
Since a ≠ 0 .aR ≠ {0} and hence aR =R (By hypothesis)
1 ∈ R => 1 ∈ a R =>1 = ab for some b≠ 0 ∈ R . Since R is conimutatiye,1= ab= ba .
a ≠ 0 ∈ R has a multiplicative inverse b∈ R . Hence R is a field.
Note. If R is a ring with a unity element and R has no proper nontrivial ideals then R is a division ring.
Examples Of Ideals In Quotient Rings And Euclidean Domains
Theorem. 5. The intersection of two ideals of a ring R is an ideal of R.
Proof. Let U1, and U2 be two ideals of the ring R.
If 0 ∈ R is the zero element, then 0∈ U1 and 0 ∈ U2 .
:. 0∈U1∩U2 and hence U1∩ U2 ≠φ
Let a,b ∈ U1 ∩ U2 and r ∈ R. Then a b∈U and a,b ∈U2.
a,b ∈ U1,r ∈R and U1 is an ideal =>a-b ∈U1 and ar, ra ∈ U1……………………….1
a,b ∈U2,r ∈R and U2 is an ideal=> a-b ∈U2 and ar, ra ∈U2………………………….2
From (1) and (2): a-b ∈ U1 ∩ U2 and ar, ra ∈ U1∩U2
Hence U1∩ U2 is an ideal of R.
Remark: The union of two ideals of a ring R need not be an ideal of R
For the ring Z of integers, A = { 2n | n∈ Z) and B = {3n\n ∈ Z } are two ideals.
But, for 2,3 ∈ A∪B, 3-2 = 1 ≠ A∪B. A∪B is not an ideal of Z
Understanding The Properties Of Ideals In Algebraic Structures
Theorem 6. If U1, and U2 are two ideals of a ring R then U1∪U2is ideal of R if and only if U1⊂ U2 or U2⊂U1
Proof. Let U1 ∪ U2 be an ideal of R. We now prove that U1⊂ U2 or U2⊂U1
If possible, suppose that U1⊄ U2 or U2⊄U1
Since U1⊄ U2 there exists an element a ∈ U1 and a ∈ U2.
Since U2⊄U1 there exists an element b ∈ U2 and ∈ U1.
a ∈ U1 and b∈ U2 => a, b ∈ U1⊂ U2
a,b ∈ U1⊂ U2 and U1⊂ U2 is an ideal =>a-b ∈ U1 ∪ U2
=> a-b ∈ U1 or a-b ∈ U2
But a-b ∈ U1=>a- {a-b) = b ∈U1 …………………..1
a-b ∈ U2 => b+(a-b)=a ∈ U2 ………………………2
Both (1) and (2) contradict a ∉U2 and b∉U1
Our supposition is wrong.
Hence U1⊂ U2 or U2⊂U1
Conversely, let U1⊂ U2 or U2⊂U1
Then U1 ∪ U2 = U2 or U1and is an ideal.
Note. If S1, and S2 are two subrings of a ring prove that S1 u S2 is also a subring iff either S1 ⊆ S2 or S2 ⊆S1