IntroductiontoRingsIdealsModules

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Introduction to Rings, Ideals and Modules
Research · September 2020
DOI: 10.13140/RG.2.2.21094.86088
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INTRODUCTION TO RINGS, IDEALS AND MODULES
BALCHANDAR REDDY
CONTENTS
1. Rings 1
2. SubRings & Ideals 2
3. Modules 4
4. Quadratic Rings 5
References 8
1. Rings
a ring is one of the fundamental algebraic structures used in abstract algebra. It
consists of a set equipped with two binary operations that generalise the arithmetic
operations of addition and multiplication.
A ring is a set R equipped with two binary operations + and · satisfying the
following three sets of axioms, called the ring axioms.
1. R is an abelian group under addition, meaning that:
◦ (a + b) + c = a + (b + c) for all a, b, c in R (that is, +
is associative).
◦ a + b = b + a for all a, b in R (that is, + is commutative).
◦ There is an element 0 in R such that a + 0 = a for all a in R (that
is, 0 is the additive identity).
◦ For each a in R there exists −a in R such that a + (−a) = 0 (that is,
−a is the additive inverse of a).
2. R is a monoid under multiplication, meaning that:

BALCHANDAR REDDY 2
◦ (a · b) · c = a · (b · c) for all a, b, c in R (that is, · is associative).
◦ There is an element 1 in R such that a · 1 = a and 1 · a = a for
all a in R (that is, 1 is the multiplicative identity). [5]
3. Multiplication is distributive with respect to addition, meaning that:
◦ a ⋅ (b + c) = (a · b) + (a · c) for all a, b, c in R (left distributivity).
◦ (b + c) · a = (b · a) + (c · a) for all a, b, c in R (right
distributivity).
If a Ring has a multiplicative identity, 1 ∈ R, we call R as a “ring with identity”.
Examples:
1. Z 4 = { ¯0, ¯1, ¯2, ¯3 }, Z, Q, R, C.
2. M 2 (R) : set of 2*2 matrices with real entries, M N (R), U 2 (R).
Remarks:
1. Let R be a ring and x ∈ R. a) We say that x is a zero-divisor if x y = 0 for some
y ≠ 0. b) We say that x is a nilpotent if x n = 0 for some n ∈ N. c) We say that x is
a unit or invertible if x y = 1 for some y ∈ R.
2. In Z/6Z, the element 2 is a zero-divisor, as 2 · 3 = 0, while 5 is a unit, since 5 2
= 1. In Z/4Z the element 2 ≠ 0 is nilpotent, as 2 2 = 0.
3. The set of zero-divisors in R is not an ideal in general, e.g. 2 +3 = 5 ∈ Z/6Z.
Definitions:
1. A domain is a ring in which the only zero-divisor is 0.
2. A field is a ring in which every nonzero element is invertible.
2. SubRings & Ideals
A subset S of R is said to be a subring if it can be regarded as a ring with the
addition and the multiplication restricted from R to S. Equivalently, S is a subring
if it is not empty, and for any x, y in S, x y, x + y and - x are in S.
Examples:
1. 2 Z, the subset of even integers, is a subring of Z.
2. Z is a subring of the polynomial ring Z[ x].
A subset I of a ring R is called an ideal if
(i) I is a subring of R;
(ii) for all a ∈ I and r ∈ R, ar ∈ I and ra ∈ I
If I is an ideal of R we denote this fact by I = R.
Δ

BALCHANDAR REDDY 3
Proposition. Let I be a non-empty subset of a ring R. Then I
all a, b ∈ I and r ∈ R, a − b ∈ I, ar ∈ I, ra ∈ I.
R if and only if for
Examples:
1. Let R be a non-zero ring. Then R has at least two ideals, namely R and {0}.
We often write 0 for {0}.
2. 2 Z is an ideal of Z.
3. Ring: Z(x) = a n x n + a n−1 x n−1 + . . . . + a 0 | a i ∈ Z.
Ideal: J = x . Z(x) = = a n x n + a n−1 x n−1 + . . . . a 0 x | a i ∈ Z
Is J the ideal of Z(x) ?
(a) J is a subgroup of Z(x) , as it satisfies all ring properties.
(b) f (x) ∈ Z(x), j(x) ∈ J
f (x) . j(x) ∈ J, as f (x) . j(x) = k . f (x) . j 1 (x) ∈ J
similarly, j(x) . f (x) ∈ J , as ring is commutative both f (x) . j(x) and
j(x) . f (x) are equal.
Left & Right ideals:
I is an ideal of Ring R if:
1. (I, + ) ≤ (R, + ) “I is an additive subgroup of R”
For any r ∈ R, x ∈ I:
2. r . x ∈ I
3. x . r ∈ I
Left ideal:
1. (I, + ) ≤ (R, + ) “I is an additive subgroup of R”
2. r . x ∈ I
Right ideal:
1. (I, + ) ≤ (R, + ) “I is an additive subgroup of R”
2. x . r ∈ I
An Ideal is a subring I ≤ R that is left ideal and right ideal.
Theorem: Intersection of any two ideals of a ring is also an ideal.
Δ Δ Δ
Proof: Given S 1 = R, S2 = R, is S1 ∩ S 2 = R ?
To proof whether a set is an ideal of a ring, we should check the following
conditions.
a) is S 1 ∩ S 2 non-empty?
0 ∈ S 1 , 0 ∈ S 2 , therefore it is obvious that 0 ∈ S 1 ∩ S 2 . Hence, S 1 ∩ S 2 is
not empty.
b) Is S 1 ∩ S 2 a subgroup of R?
Δ
=

BALCHANDAR REDDY 4
Δ
x ∈ S 1 , x ∈ S 2 and y ∈ S 1 , y ∈ S 2 . As S 1 = R it implies x − y ∈ S1 similarly
x − y ∈ S 2 . Hence, we conclude that S 1 ∩ S 2 a subgroup of R.
c) For some x ∈ S 1 ∩ S 2 , r ∈ R is xr, rx ∈ S 1 ∩ S 2 ?
xr, rx ∈ S 1 , xr, rx ∈ S 2 therefore xr ∈ S 1 ∩ S 2 and rx ∈ S 1 ∩ S 2 , This
concludes the proof.
Exercises:
Δ Δ Δ
1. Given S 1 = R, S2 = R, is S1 ⋅ S 2 = R ?
Δ Δ Δ
2. Given S 1 = R, S2 = R, is S1 + S 2 = R ?
Δ Δ Δ
3. Given S 1 = R, S2 = R, is S1 ∪ S 2 = R ?
Prime ideal:
A proper ideal P is a prime ideal if for any two elements x, y ∈ R such that xy ∈ p
either x ∈ p or y ∈ p.
The ideal pZ is prime if and only if p is a prime number.
Cosets:
Let I be an ideal of a ring R and x ∈ R . Then the set of elements
x + I := x + i |i ∈ I is the coset of x in R with respect to I.
R
We denote the set of all cosets of R with respect to I by .
I
3. Modules
a module is one of the fundamental algebraic structures used in abstract algebra
and it follows the properties,
1. Abelian group M with a ring of scalars R.
2. r . m is a scaled element.
3. Distributive properties:
1.
2.
r(m 1 + m 2 ) = r . m 1 + r . m 2
(r 1 + r 2 )m = r 1 m + r 2 m
4. Associative property:
(r 1 . r 2 ) . m = r 1 . (r 2 . m)
5. 1.m = m
Example:
1. Module M: any abelian group
Ring R: Z
Scalar multiplication:
Let r ∈ Z and a ∈ M
r > 0 = > r . a = a + a + . . . r terms.
r = 0 = > 0.a = 0 MZ
Every abelian group is a -module.

BALCHANDAR REDDY 5
R 3
2. Group: M = = {(x, y, z)|x, y, z ∈ R} under +
Scalar Ring: R = { M 3*3 }
r . m = matrix multiplication
4. Quadratic Rings
Quadratic integers are a generalisation of the integers to quadratic fields. They
are the algebraic integers of degree two.
Square free integer:
In mathematics, a square-free integer is an integer which is divisible by
no perfect square other than 1. That is, its prime factorization has exactly one
factor for each prime that appears in it. For example, 10 = 2 ⋅ 5 is square-free,
but 18 = 2 ⋅ 3 ⋅ 3 is not, because 18 is divisible by 9= 3 2 . The smallest positive
square-free numbers are
1,2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 26, 29, 30, 31, 33,.. ..
Pell's equation:
equation of the form x 2 − ny 2 = 1 is a pell's equation for which we study
the solutions in Z. If n is a perfect square, there will be only trivial solutions. We
consider the pell’s equation where n is not a perfect square. To understand this
equation thoroughly over Z, we need to work with another number system
Z[ n] = a + b n;
a, b ∈ Z
The reason we want to work specifically with the ring Z[ n] is of course so we
can factor Pell's equation:
x 2 − ny 2 = (x + n y)(x − n y)
The general Pell's equation and Z[ n]
let z ∈ C and z = x + iy, then we have z ¯z = x 2 + y 2 (where ¯z is the
complex conjugate). Drawing z as a vector in the complex plane, we see that z ¯z is
the square of the length of this vector, i.e., z ¯z = |z | 2 . Define the norm of z to be
N(z) = z ¯z
similarly,
N(z 1 z 2 ) = z 1 z 2 ¯z 1 z¯
2 = z 1 ¯z 1 z 2 z¯
2 = N(z 1 )N(z 2 )
i.e, the norm is multiplicative.
Similarly we define the norm to α = x + y n ∈ Z[ n] , norm of α to be
N(α) = αᾱ = (x + y n)(x − y n) = x 2 + ny 2

BALCHANDAR REDDY 6
Note: N(α) = N(ᾱ)
Gaussian Integers
So the solutions to pell’s equation are precisely the elements of norm 1 in Z[ n] .
Similarly if one wants to study the equation of the form
x 2 + ny 2 = k
It makes sense to look at the imaginary quadratic ring Z[ −n] . The imaginary
quadratic rings can be treated in the same basic way as the real quadratic rings
theoretically, however their flavour is quite different. For example x 2 + ny 2 = 1
has only finitely many solutions, but x 2 − ny 2 = 1 has infinitely many (for n nonsquare).
The Gaussian integers are the ring
Z[i] = a + bi;
a, b ∈ Z
For α = a + bi , the conjugate of α is α = a − bi , and the norm is
N(α) = αᾱ = a 2 + b 2 .
α c = |α |
Note that if we draw as a vector in the complex plane, denotes the
length of this vector, so the norm of α is just the square of the length of the vector
, i.e., .
α N(α) = c 2 c 2 c
The reason we want the norm to be the square of the length, instead of just the
length is because is always an integer, but rarely is. With this definition, the
norm is a map from the ring Z[i] into the ring of integers Z, N : Z[i] → Z.
Divisibility and primes in Z[i]
Observation: If β |α then N(β )|N(α) in Z.
1. α = 4 + i is a gaussian prime.
N(α) = 4 2 + 1 2 = 17 which is prime in Z, as N(α) is a prime, α is gaussian.
Other way to look at it is, N(α) = N(β )N(γ) , so the only possibilities are
N(β ) = 1,N(γ) = 17 and N(β ) = 17,N(γ) = 1 . Assuming the former, β is a
unit, so it has an inverse β −1 ∈ Z[i] . In other words, the only divisors of α are the
units and the unit times α.
2. α = 2 is not a gaussian prime.
The only reason α couldn’t be a prime in Z[i] is because it is the product of two
e l e m e n t s n o r m 2 . S o , t h e r e a r e 4 e l e m e n t s o f n o r m 2 :
1 + i, 1 − i, − 1 + i, − 1 − i.
We observe that

BALCHANDAR REDDY 7
2 = (1 + i)(1 − i) = (−1 − i)(−1 + i)
3. The elements of norm 2 : 1 + i, 1 − i, − 1 + i, − 1 − i are all gaussian
primes, because their norm is prime in Z.
Let z = a + ib be a Gaussian integer, when is z prime? Equivalently, what are the
positive real integers which are Gaussian prime? Obviously, every positive real
composite is not a Gaussian prime. Therefore, the question reduces to asking if
every real prime is a Gaussian prime. A real prime p can fail to be a Gaussian
prime only if there is a non-zero, non-real Gaussian integer w that divides p, i.e.,
p = N(w) . Thus, a real prime fails to be a Gaussian prime only if it is sum of two
squares.
For instance, the first real prime 2 = 1 2 + 1 2 is not a Gaussian prime because
2 = (1 + i)(1 − i) . Is every odd prime expressible as sum of squares of integers?
For instance, 3 is not expressible as sum of squares of two integers. Therefore, 3
is a Gaussian prime. What are the odd primes which can be expressed as sum of
two squares?
Proposition: Let p ∈ N be prime. Then p is also prime in Z[i] if and only if p is
not the sum of two squares.
The idea is that p is a prime in Z[i] iff p has only the trivial factorisation in Z[i] .
Whereas if p = a 2 + b 2 , we have a non-trivial factorisation of p has a + bi and
a − bi, neither of which are non units(as a, b ≠ 0) in Z[i] .
Fermat’s theorem on sums of two squares. An odd prime p
expressible as p = a 2 + b 2 , for some a, b ∈ N, iff p ≡ 1(mod4) .
is uniquely
Proof: It is obvious that p is 1 or 3 (mod 4) and a 2 + b 2 is either 0, 1 or 2 (mod
4) .Hence, p should be 1 (mod 4) . Therefore, we conclude that any odd prime that
is 3 (mod 4) is a Gaussian prime. For e.g., 3, 7, 11, 19 are gaussian primes,
where as 2, 5, 13, 17 are not.
Unique factorisation domain:
Example: Let α = 6i = 2 ⋅ 3i. In Z[i] , 3 remains prime since it is not a sum of
two squares. This means 3i is also a prime in Z[i] , since i is just a unit. On the
other hand, 2 = (1 + i)(1 − i) and we know 1 + i,1 − i are prime in Z [i]
because their norms (both 2) are prime in N. Unique factorisation says that
6i = (1 + i)(1 − i)(3i)
is the only factorisation of 6 into primes of Z[i] .

BALCHANDAR REDDY 8
Class number 1 corresponds to UFP(Unique factorisation property). For instance
Z[ −5] do not satisfy UFP. The number 6 ∈ Z[ −5] has two different prime
decompositions, 6 = 2 ⋅ 3 = (1 + 5i)(1 − 5i) . In fact, it has been shown that there
a r e e x a c t l y n i n e n e g a t i v e i n t e g e r s , −1, − 2, − 3, −7, − 11, − 19,
−43, − 67, − 163 for which the UFP holds true. These are called Heegner
numbers. It is an open problem whether there are finite number of positive
integers for which UFP is true in Z[ m] .
REFERENCES
1. Bourbaki, Nicolas (1994). Elements of the history of mathematics. Translated by
Meldrum, John.
2. Dummit, D. S., and Foote, R. M., 2004. Abstract Algebra, 3rd ed.
3. COHN, P. M. 1991. Algebraic numbers and algebraic functions. Chapman and Hall
Mathematics Series. Chapman & Hall, London.
4. EDWARDS, H. M. 1977. Fermat’s last theorem, volume 50 of Graduate Texts in
Mathematics. Springer-Verlag, New York. A genetic introduction to algebraic number
theory.
5. NARKIEWICZ, W. 1990. Elementary and analytic theory of algebraic numbers.
SpringerVerlag, Berlin, second edition.
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