We now return to looking again at the mathematical entities known as rings, fields and ideals.
1.
Definitions:
i)
Binary
operation:
Let S be an abstract set. By binary operation on S,
denoted by o one means a mapping on a function from S x
S to S. In other words, a binary
operation assigns to each pair of
elements in S: (a,b) an element a o b in S.
ii)
Ring
By a ring (S, +, · ) one means an abstract set
S which has two binary
operations defined on it. The operation + is called addition, and the operation
· is called multiplication. Moreover, the set (S, +, ·) is
required to satisfy the following axioms:
Let
a, b and c Î S
A1): a + (b + c) = (a + b) + c
A2) ∃ an
element 0 Î S with the property that 0 + a = a
A3) For each element a Î S there exists an element
(-a) Î S such that a + (-a) = 0
A4) a + b = b + a
A5) a · (b · c) = (a · b)
· c
iii)
Commutative Ring
By a
commutative ring (S, + · ) one means a ring (S, + · ) which satisfies, in addition to
the ring axioms the axiom:
A6) For
all a, b
Î S then a · b
= b · a
And there
also exist the properties:
a)Closure: If a,b Î S, then the sum a+b and the
product a·b are uniquely defined and belong to S.
b)Associative laws: For
all a,b,c Î S,
a+(b+c) = (a+b)+c and a·(b·c) = (a·b)·c.
c)Commutative laws:
For all a,b Î S, a+b
= b+a and a·b = b·a.
d)Distributive laws: For all a,b,c Î S,
a·(b+c) = a·b + a·c
and (a+b)·c = a·c + b·c
e)Additive identity: The set R contains an additive
identity element, denoted by 0, such that for all a Î S,
a+0
= a and 0+a = a.
f)Additive inverses: For each a Î S, the equations:
a+x
= 0 and x+a = 0
have a solution x Î S, called the additive inverse of a,
and denoted by (-a).
The commutative ring S is called a commutative ring with identity if it contains an element 1, assumed to be different from 0, such that for all a Î S,
a·1
= a and 1·a = a.
In this case, 1 is called a multiplicative
identity element
iv)Commutative Ring
with unity
By a
commutative ring with unity one means a commutative ring (S, + · ) which satisfies the axiom:
A7) There
exists an element 1 Î S with the property:
1 ·
a = a for all a Î S
v)Integral
domain
By an
integral domain ring (S, + · ) one means a commutative ring with unity which
satisfies the following axiom:
If a, b Î S and a · b
= 0 then either a = 0 or b = 0.
vi)Fields
By a
field (S, + · ) one means a commutative ring with unity which
satisfies the additional axiom:
A7) For
every non-zero element a Î S there exists an
element a - 1 Î
S such that:
a · a - 1 = 1. The element
a - 1 is called the
reciprocal or multiplicative inverse of a.
vii)
Ordered Fields
An
ordered field is a field (S, + · ) which contains a
subset P (called the positive cone in P) which has the following properties:
P is a
proper subset of S.
O.1) P is
closed under addition.
O.2) P is closed under multiplication
O.3)
If a
Î S then one and only one of the following three
conditions must hold:
a =
0, a
Î P, or - a Î P
viii)Order
relation on an ordered field
Let (F , + ·
) be an ordered field
and let P represent the positive cone in F. We define a relation ‘<’ on F, called ‘less than’ by the formula:
< =
{(a, b) : a, b Î F, b – a Î P }
Or,
equivalently:
a < b if an only if b – a Î P
We also
define the relation ‘greater than’ on F by
> as follows:
s > b
if and only if b < a
Definition. The field K is
said to be an extension field of the field (F + · ) if (F + · ) is a subset of K which is a field under the
operations of K.
Definition. Let K be an extension field of F and let u Î
K. If there exists a nonzero polynomial f(x) Î
F[x] such that f(u)=0, then u is
said to be algebraic over F. If there does not exist such a polynomial,
then u is said to be
transcendental over K.
Proposition. Let K be an
extension field of F , and let u Î
K be algebraic over F. Then there exists a unique irreducible
polynomial p(x) Î F[x] such
that p(u)=0. It is characterized as the monic polynomial of minimal degree that
has u as a root.
Furthermore, if f(x) is any polynomial in F[x] with f(u)=0, then p(x) | f(x).
Definition. Let F
be a field and let f(x) = a0
+ a1 x + · · · + an xn be a polynomial in F [x]
of degree n> 0. An extension field K
of F is called a splitting field for f(x) over F
if there exist elements r1, r2, . . . , rn Î
F such that
(i)
f(x) = an (x-r1) (x-r2)
· · · (x-rn), and
(ii)
K = F (r1,r2,...,rn).
In the above situation we usually say that f(x) splits
over the field F. The elements r1, r2, . . . , rn
are roots of f(x), and so K is obtained
by adjoining to F a complete set of
roots of f(x).
Definition: Subfield of a field. Let
(F + · ) be a
field. A subset T ⊂ F is called a subfield if T is closed under the operations of + and · , and T is a field under those operations.
Definition: Ideal:
Let (S, + · ) be a commutative ring, by an ideal I in S one
means the following:
i)I is a subring of S,
ii)If a Î
I , b Î S then a, b Î
I
Definition: Congruence modulo I.
Let S be a
commutative ring. Let I be an ideal in S. We write a ≡ b modulo I for the two elements a,
b Î S if
a Î [ b ]. Hence,
b Î [ a ].
Definition: S is a commutative ring and I is an ideal.
Then let S/ I denote a quotient ring
and the set of congruence classes modulo I.
Perhaps the most
critical aspect of ideals to note is that: “an
ideal is to a ring as a normal subgroup is to a group”. Also, in a similar manner to groups, one can
have left and right ideals.
Let a Î S,
a commutative ring. Then the set (a) defined by: (a) = {ax: , x Î S }
is an ideal.
Now, let S be a commutative ring and I be an ideal in S. For
each element a Î S let [a]
be the subset of S defined by:
[a] = I + a = {a + j:
j Î S}
Further, let s be a subring of a ring S satisfying:
[a] s ⊆
s
and s[a] ⊆ s for
all a
Î S is an ideal (or two-sided ideal of S). Also, a subring s of S satisfying
[a] s ⊆
s
for all a Î
S is a left ideal of S. One satisfying
s [a] ⊆ s for all
a Î S is a right ideal of S.
Problems for Math Mavens:
1) Let Z be the
integers. Then prove for each class:
[a]
mod n there exists an integer r Î
Z, for: 0 < r < n, r Î [n]
2) Let Z be the
integers. The ideal:
(5) = {5
j: j Î Z }
Show all the congruence classes with respect
to this ideal.
Hint: [a] =
{a + j: j Î [(5)} =
{a + 5j:
j Î Z }
3) Let S be a commutative ring, and let I be an
ideal in S. If A, B are subset of S, then use set notation to define: i) A · B, A
+ B
4) Take
S as the set of integers, Z. Let the ideal I = (2) so that S / I = Z 2 Thence or otherwise, find:
a) [0] b)
[1] c) S/ I =
Z 5
5)
Show every ring S
has two ideals: S itself and {0}.
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