Ring Theory & Polynomial Rings

[Rings] [Polynomials]

Rings Theory

[Definition] [Properties] [Example]

Definition of a Ring

A ring is an algebraic system with two operations. A nonemty set R is considered a ring if in R there are two operations, + and · such that:

a) closure: a,b in R implies a + b in R and a · b in R

b) commutativity: a + b = b + a for a,b in R

c) associativity: (a + b) + c = a + (b + c) and (a · b) · c = a · (b · c) for a,b,c in R

d) identity: There exists 0 in R s.t. a + 0 = a for every a in R

e) additive inverse: Given a in R there exists a,b in R s.t. a + b = 0

f) distributive laws: a · (b + c) = a · b + a · c and (b+c) · a = b · a + c · a for a,b,c in R

(Notice that R is an abelian group under +.)

If there exists an element 1 in R s.t (a · 1) = (1 · a) for a in R then R is considered a ring with unit.

A ring with unit is called a division ring if there exists a multiplicative inverse in R.

If R is commutative under multiplication (a · b = b · a for a,b in R) then R is called a commutative ring.

R is considered an integral domain if it is a commutative ring where a · b = 0 only in a = 0 or b = 0.

Finally, if a ring R is a commutative division ring then R is called a Field. To learn more about fields click here.

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Properties

1) a0 = 0a = 0

0 = 0 + 0 so a0 = a(0 + 0) = a0 + a0, a0 - a0 = a0 + a0 - a0, 0 = a0

2) a(-b) = (-a)b = -(ab)

ab + a(-b) = a(b + -b) = a(0) = 0, so a(-b) = -(ab)

ab + (-a)b = (a + -a)b = (0)b = 0, so (-a)b = -(ab)

3) (-a)(-b) = ab

(-a)(-b) = -((-a)(-b)) = ab (We are in an abelian group.)

4) If 1 R, then (-1)a = -a

(-1)a + a = (-1 + 1)a = 0a = 0

5) (a +b)^2 = a^2 + b^2 + ab + ba for a,b in R

(a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + ab + ba

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Example

Integers

The set of integers forms a group under addition so we are half way to showing that it is a ring. Looking deeper, the set is also closed under multiplication, holds true for the distributive laws and the associative property under multiplication. Why?

The natural number x can be described by the ordered pair (a,b) where a > b such that x = a - b. If a < b then a - b is a negative number and if a = b then a - b = 0. The set of natural numbers contains neither, so we must come up with a new set.

In the set of natural numbers times the set of natural numbers we can introduce an equivalence relation ~ by (a,b) = (a*,b*) iff a + b* = a* + b. An integer is an equivalence class of ~. Let x = [(a,b)] and y = [(c,d)] where x, y are in the set of integers and a,b,c,d in natural numbers s.t. x + y := [(a + b,b + d)].

Closure under multiplication:

x·y = [(a,b)]·[(c,d)] = [(ac + bd, bc + ad)] = ac + bd - bc - ad = ac-bc+bd-ad and since we know that the set of integers is closed under addition and the set of natural numbers is closed under mulitplication we know that x·y in the set of integers.

Distributive laws:

Let x = [(a,b)], y = [(c,d)], and z = [(e,f)]

x(y+z) = xy + xz

x(y+z) = x·[(c+e, d+f)] = [(a,b)][(c+e,d+f)] = [(ac+ae+bd+bf, bc+be+ad+cf)] = [a(c+e) +b(d+f), b(c+e) +a(d+f)]=[a(c+e)-b(c+e), a(d+f)-b(d+f)] = [(a-b)·(c,d),(a-b)(d+f)]=(a,b)·(c,d)+(a,b)(e,f) = x·y+x·z

(y+z)x = yx+zx can be proved in similar fashion.

Associativity:

(x·y)·z = x·(y·z)

(x·y)·z = [(ac+bd,bc+ad)][(e,f)] = [(ace+bde+bcf+adf, bce+ade+acf+bdf)]=[(a,b)][(ce+df,de+cf)] = x·(y·z)

So the set of integers forms a ring. Since 1 is in the integers and 1 · a = a = a · 1 for all integers in a the set has a unit. That's not all. The ring is also commutative under multiplication:

Let x = [(a,b)] and y = [(c,d)] where a,b,c,d are natural numbers. x · y = [(ac + bd, bc + ad)] = [(ac + bd, ad + bc)] = [(ca + db, cb + da)] = y · x

We also know that in the integers a · b can't equal zero if a or b doesn't equal zero.

So the set of integers is an example of an integral domain.

Now that I have gone over the basics, I want to focus more of my attention on a specific type of ring, a polynomial ring.

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Polynomial Rings

[Definitions] [Properties] [Proof]

Definitions

F[x]

Where F is a field, F[x] is the ring of polynomials in x over F.

Degree of a Polynomial

If p(x) = a_0 + (a_1)x + ... + (a_n)x^n and a_n =/ 0, then the degree of p(x), deg p(x), is n. (deg(x^3 + 3x^2 + 1) = 3)

Monic Polynomial

f(x) is in F[x] is a monic polynomial if the coefficient of its highest power is 1.

Principal Ideal Domain

An integral domain R is a principal ideal domain if every ideal I in R is of the form I = {xa| x R} for some a in I.

Irreducible

The polynomial p(x) in F[x] is irreducible if

(1) p(x) is of positive degree

(2) given any polynomial f(x) in F[x], either p(x)|f(x) or p(x) is relatively prime to f(x)

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Properties

1. Equality

p(x) = a_0 + (a_1)x + ... + (a_n)x^n and q(x) = b_0 + (b_1)x + ... + (b_n)x^n are equal iff , a_0 = b_0, a_1 = b_1, ..., a_n = b_n (their corresponding coefficients are equal).

If q(x) = c_0 + (c_1)x + ... + (c_n)x^n, where m <n and c_(m+1) = ... = c_(n-1) = c_n = 0, then q(x) can be written, q(x) = c_0 + (c_1)x + ... + (c_m)x^m.

2. Addition

If p(x) = a_0 + (a_1)x + ... + (a_n)x^n and q(x) = b_0 + (b_1)x + ... + (b_n)x^n then p(x) + q(x) = c_0 + (c_1)x + (c_2)x^2 + ... + (c_n)x^n where c_0 = a_0 + b_0, c_1 = a_1 + b_1, ..., c_n = a_n + b_n

3. Multiplication

If p(x) = a_0 + (a_1)x + ... + (a_n)x^n and q(x) = b_0 + (b_1)x + ... + (b_n)x^n, then p(x)q(x) = c_0 + (c_1)x + ... + (c_n)x^n where c_i = (a_i)(b_0) + (a_(i-1))(b_1) + ... + (a_0)(b_i)

4. Division

Suppose f(x) and g(x) are in F[x], with g(x) 0. g(x) divides f(x), g(x)|f(x), if f(x) = a(x)g(x) for some a(x) in F[x].

Greatest Common Divisor

Given two polynomials f(x) and g(x) in F[x], (not both zero) the greatest common divisor of f(x) and g(x) is the monic polynomial d(x) s.t.

(1) d(x)|f(x) and d(x)|g(x)

(2) If h(x)|f(x) and h(x)|g(x), then h(x)|d(x)

The set of polynomials forms a ring in x over F

Proof:

From the above properties, we know that the set of polynomials is closed and associative under addition and mulitplication and commutative under addition. The additive identity of a polynomial is 0. The additive inverse of a polynomial p(x) is q(x) where q(x) = -(p(x)). Now all we have left to show is that the distributive laws hold. Let d(x) be a polynomial where d(x) = p(x)·(q(x) + r(x)) such that p(x), q(x) and r(x) are polynomials. Divide d(x) by p(x). Then p(x)|p(x) or p(x)|(q(x) + r(x)). p(x)|p(x), so p(x)|d(x) = q(x) + r(x). Let s(x) be a polynomial where s(x) = p(x)q(x) + p(x)r(x). Divide s(x) by p(x). Then p(x)|(p(x)q(x)) and p(x)|p(x)r(x). p(x)|(p(x)q(x)) = q(x) and p(x)|p(x)r(x) = r(x) so p(x)|s(x) = q(x) + r(x). Therefore p(x)|d(x) = p(x)|s(x) so d(x) = s(x) and thus p(x)·(q(x) + r(x)) = p(x)q(x) + p(x)r(x) and the distributive laws hold.

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