Rings

Magma

A magma is a set

M

with a binary operation

⋆

defined on

M

Monoid

A monoid is a set

X

with a binary operation

⋆

X

such that

⋆

is associative and it has an identity

Ring

A ring is a non-empty set

R

equipped with two binary operations

\oplus, \otimes

such that

$(R, \oplus)$ form an abelian group (We denote it's identity as $0_{R}$ )
$\otimes$ is associative
$\otimes$ distributes into $\oplus$

Note that we used the symbols $\oplus, \otimes$ because this definition generalizes some of the properties that we know true for $+, \cdot$ when working with numbers. Note that on paper, you will usually see $+, \cdot$ instead of the circled operations. They are used here to emphasize that these are generic operations.

Commutative Ring

A commutative ring is a ring

R

such that

\otimes

is commutative.

Ring with Unity

A ring with unity is a ring

R

that contains an identity element relative to

\otimes

, which we denote by

1_{R}

Crone

We employ the word Crone to specify a commutative ring with unity

The above word is derived from Commutative Ring with ONE

Two by Two Matrices with 0 In Bottom Left Form a Crone

Let

R := {(\begin{matrix} a & b \\ 0 & a \end{matrix}) : a, b \in Z}

along with the usual definition for matrix addition and matrix multiplication, show that this forms a crone.

In writing sometimes the original definition of ring is written as rng (without the i) to denote that there is no assumption on it having a multiplicative identity, and the above definition is just identified by ring (with the character i, representing that we assume there is an multiplicative identity. In speaking we cannot distinguish "rng" and "ring", so we prefer the names listed here.

Zero Divisor

Let

(R, \oplus, \otimes)

be a ring, then we say that an element

x \in R ∖ {0_{R}}

is a zero divisor if there is some

y \in R ∖ {0_{R}}

such that the following equation holds

x \otimes y = y \otimes x = 0_{R}

The Integers form a Ring

Z

forms a ring

Every Field is a Ring

The Integers Modulo N are a Ring

Polynomials Form a Ring with Ring Coefficients

Suppose that

R

is a ring and that the set

R [x] = {r_{n} x^{n} + r_{n - 1} x^{n - 1} + \dots + r_{1} x^{1} + r^{0} : n \in N_{0}, r_{i} \in R}

Along with addition, defined such that for any

m \leq n \in N_{0}

, we have:

\sum_{i = 0}^{n} r_{i} x^{i} + \sum_{i = 0}^{m} s_{i} x^{i} := \sum_{i = 0}^{n} r_{i} x^{i} + \sum_{i = 0}^{n} s_{i} x^{i} := \sum_{i = 0}^{n} (r_{i} + s_{i}) x^{i}

Where for every

m \leq k \leq n

we've set

s_{k} = 0

, thus artifically padding the polynomial without changing it. As for multiplication, we define

(\sum_{i = 0}^{n} r_{i} x^{i}) (\sum_{j = 0}^{m} s_{i} x^{i}) := \sum_{k = 0}^{n + m} t_{k} x^{k}

Wherein

t_{k} = \sum_{i + j = k} r_{i} s_{j}

Note that the definition of $t_{k}$ describes how many ways to get the same degree after the generalized expansion rule has been applied to the two. Additionally these objects are not yet functions, they are simply symbolic formal expressions.

Leading Coefficient

Let

R

be a ring and

p = \sum_{i = 0}^{n} r_{i} x^{i}

be a polynomial in

R [x]

such that

r_{n} \neq 0

, then

r_{i}

is called the leading coefficient of

p

Zero Polynomial

Let

R

be a ring and

p = \sum_{i = 0}^{n} r_{i} x^{i}

be a polynomial such that for every

i \in 0, . . ., n

we have

r_{i} = 0

, then we say that

p

is the zero polynomial.

Monic

Let

R

be a ring and

p \in R [x]

such that

r_{n} = 1

, then

p

is said to be monic.

Degree

Let

R

be a ring and

p = \sum_{i = 0}^{n} r_{i} x^{i}

be a polynomial in

R [x]

such that

r_{n} \neq 0

, then we say that the degree of

p

n

, and it is denoted by

\deg (p)

Square Matrices form a Non-Commutative Ring

Let

R

be a ring and then

M_{n \times n} (R)

form a non-commutative ring

Note that a vector space forms an abelian group under addition, but there is no clear multiplication, using the cross product doesn't work.

Zero Multiplication in a Ring

Let

R

be a ring, then for any

r \in R

0 r = 0

Additive Inverse equals Multiplication by One's Additive Inverse

- r = (- 1) r

Additive Inverse times Additive Inverse of One Yields Original

(- 1) (- r) = r

The set of One Element Forms a Ring

Suppose that

R = {r}

, with addition and multiplication defined in the only way it can be, then

R

forms a ring, which is called the zero ring

Degree of Product equals Sum of Degrees

Suppose that

R

is a domain and

f, g

are non-zero polynomials in

R [x]

, then

\deg (f g) = \deg (f) + \deg (g)

We first start by setting $n := \deg (f)$ and $m = \deg (g)$ . Note that the highest power that could be produced by this product would be $x^{n + m}$ .

Suppose that $f = \sum_{i = 0}^{n} r_{i} x^{i}$ and $g = \sum_{i = 0}^{m} s_{i} x^{i}$ , by the definition of $\deg (\cdot)$ we know $r_{n}, s_{m}$ are non-zero, therefore since $R$ is a domain we know $r_{n} s_{n} \neq 0_{R}$ , which shows that the coefficient on $x^{n + m}$ is non zero, thus we can conclude that $\deg (f g) = \deg (f) + g$ as needed.

Polynomials Formed from a Domain Form a Domain

Suppose that

R

is a domain, then so is

R [x]

a direct sum of infinitely many nonzero rings has no unity I'm implicitly using the trivial fact that in any ring R with 1: R = {0} iff 1 = 0 (in fact, 0*x = x*0 = 0 in all rings, even those lacking 1) (since 1*x = x for all x, and 0*x = 0 for all x, in any ring with 1) Did you know that in any ring with 1, the axiom that addition is commutative is redundant with the rest of the axioms? That is, you can prove it, assuming only the other axioms.

Left Ideal

Suppose that

(R, \oplus, \otimes)

is a ring, and let

(R, \oplus)

be it's additive group, then we say that

I \subseteq R

is a left ideal of

R

$(I, \oplus)$ is a subgroup of $(R, \oplus)$
For every $r \in R$ and $x \in I$ we have $r \otimes x \in I$

Right Ideal

Suppose that

(R, \oplus, \otimes)

is a ring, and let

(R, \oplus)

be it's additive group, then we say that

I \subseteq R

is a left ideal of

R

$(I, \oplus)$ is a subgroup of $(R, \oplus)$
For every $r \in R$ and $x \in I$ we have $x \otimes r \in I$

Binary Operation Applied to Sets

Suppose that

⋆

is a binary operation on a set

X

, and that

A, B \subseteq X

, then we define:

A ⋆ B := {a ⋆ b : a \in A, b \in B}

Notice that $r, x$ have changed order

Subring

Suppose that

(R, \oplus, \otimes)

is a ring, and that

S \subseteq R

, if

(S, \oplus, \otimes)

is a ring, then we say that

S

is a subring

Subring with Unity

Suppose that

R

is a ring with unity then we say

S \subseteq R

is a subring with unity if

S

is a subring of

R

and

1_{R} \in S

Z

is a Subring with Unity of

Q

As per title.

We know that

Z \subseteq Q

, and let

+, \cdot

be multiplication as defined in

Q

, then for any

a, b \in Z

we can see that

a + b \in Q

this is because really we have

\frac{a}{1} + \frac{b}{1} = \frac{a + b}{1}

, similarly

a \cdot b \in Z

and we know that

1_{Q} \in Z

Subring Criterion

Suppose

(R, \oplus, \otimes)

is a ring and that

S \subseteq R

, if

$S$ is closed under $\oplus$ and $\otimes$
$S$ has inverses with respect to $\oplus$
$1_{R} \in S$

then

S

is a subring

Subfield

Suppose that

S

is a subcrone of

R

, then if

S

is a field, then we say that

S

is a subfield.

Fraction Equivalence Class

Suppose that

R

is a domain, now consider the relation defined

R \times R

where we say that

(r, s)

is related to

(u, v)

iff

r \otimes v = s \otimes u

. We will notate an element

(x, y) \in R \times R

\frac{x}{y}

, thus we have equivalence classes which we will denote by

[\frac{x}{y}]

, where we note that

\frac{a}{b} \in [\frac{x}{y}]

iff

a \otimes y = x \otimes b

Domain Fractions

Suppose that

R

is a domain, then we define

frac (R) := {[\frac{r}{s}] : r, s \in R, s \neq 0_{R}}

The Domain Fractions Can be Extended to a Crone

We define

\oplus

as follows:

[\frac{r}{s}] \oplus [\frac{u}{v}] := [\frac{r v + u s}{s v}]

And for

\otimes

we define

[\frac{r}{s}] \otimes [\frac{u}{v}] := [\frac{r u}{s v}]

Then we claim that

(frac (R), \oplus, \otimes)

is a crone and we call it the fraction crone

In the definition of

\otimes

note that we require the product of

s v

to be non-zero for if it was zero then clearly

\frac{r u}{s v} \notin frac (R)

(meaning it's not closed under

\otimes

, this is where the assumption that

R

was a domain is being used.

The Fraction Crone is a Field

Suppose that

frac (R)

is the fraction crone, then it is also a field

Because we can explicitly construct multiplicative inverses.

Field Extension

Suppose that

F

is a subfield of some field

E

, then we say that

E

is a field extension of

F

The complex numbers are a field extension of the reals

C

is a field extension of

R

Field Extension Yields a Vector Space

Suppose that

E

is a field extension of

F

then

E

forms a vector space over

F

Degree of a Field Extension

Suppose that

E

is a field extension of

F

, and let

V_{E}

be the vector space over

F

, then if this vector space is finite dimension, we say call it's dimension the degree of $E$ over $F$ and denote it as

[E : F]

Degree of

C

over

R

Prove that

[C : R] = 2

We can prove it by showing that the basis

B := (1, i)

works

The Kernel of a Ring Homomorphism is a Proper Ideal

Suppose that

ϕ : R \to S

is a ring homomorphism, then

\ker (ϕ)

is a proper ideal of

R

A Ring Homomorphism is an Injection iff It's Kernel is Zero

Suppose that

ϕ : R \to S

is a ring homomorphism, then

ϕ

is injective iff

\ker (ϕ) = {0_{R}}

Field to Ring Homomorphism Implies Injective

Suppose that

F

is a field and that

S

is a crone, then if

ϕ : F \to S

is a crone homomorphism, then it is also injective

If $S$ is the zero crone, then this trivially holds. So now we assume not the trivial crone

Since $ϕ$ is a crone homomorphism, then we know that $ϕ (1_{F}) = 1_{S}$ , now suppose that $x \in F ∖ {0_{F}}$ , then we have some $x^{- 1}$ such that $x x^{- 1} = 1_{F}$ , so that $1_{S} = ϕ (1_{F}) = ϕ (x x^{- 1})$ but $ϕ$ is a crone homomorphism, so that $ϕ (x x^{- 1}) = ϕ (x) ϕ (x^{- 1}) = ϕ (x) {[ϕ (x)]}^{- 1}$ .

If $ϕ (x) = 0_{S}$ , then we obtain that $1_{S} = 0_{S}$ , which is a contradiction becaues $S$ was not trivial. Therefore for any $x \neq 0$ we know that $ϕ (x) \neq 0$ , which says that $\ker (ϕ) = {0_{F}}$ , therefore we obtain that $ϕ$ is an isomorphism.

Field to Crone Homomorphism Implies the Image of the Field Is a Subfield Isomorphic to the Original Field

Suppose that

F

is a field and that

S

is a crone, then if

ϕ : F \to S

is a crone homomorphism, then

im (ϕ)

is a subfield of

S

and is isomorphic to

F

To show that $im (ϕ)$ is a subfield of the crone $S$ we must show that it is also a crone under $S$ 's operations $\oplus, \otimes$ . So suppose that $x, y \in im (ϕ)$ , by definition this means that there is some $a, b \in F$ such that $ϕ (a) = x$ and $ϕ (b) = y$ , we want to now show that $x \oplus y \in im (ϕ)$ , but note that $ϕ (a + b) = ϕ (a) \oplus ϕ (b) = x \oplus y$ therefore $a + b$ maps to it under $ϕ$ so we know that $x \oplus y \in im (ϕ)$ , similarly we can see that $ϕ (a \cdot b) = ϕ (a) \otimes ϕ (b) = x \otimes y$ so that also $x \otimes y \in im (ϕ)$ , where we've used the fact that $ϕ$ is a homomorphism a few times.

We'll now show that $im (ϕ)$ has inverses with respect to $\oplus$ , so let $x \in im (ϕ)$ so we have some $a \in F$ such that $ϕ (a) = x$ , consider the element $ϕ (- a) \in im (ϕ)$ , we can see that $ϕ (a) \oplus ϕ (- a) = ϕ (a + - a) = ϕ (0_{F}) = 0_{S}$ , thus $im (ϕ)$ has inverses with respect to $\oplus$ , finally note that since $ϕ$ was a crone homomorphism, we also know that $ϕ (1_{F}) = 1_{S}$ which shows that $1_{S} \in im (ϕ)$ . So we concluded that $im (ϕ)$ is a subcrone.

To show it's a subfield, we must now show that $im (ϕ)$ is a field, so we must prove that it has multiplicative inverses, thus let $x \in im (ϕ)$ where $x \neq O_{S}$ , we want to prove that it has a multiplicative inverse, but first we know that there is some $a \in F$ such that $ϕ (a) = x$ , also note that $a \neq 0$ or else we have a contradiction, therefore since $F$ was a field, there is some $a^{- 1} \in F$ such that $a a^{- 1} = 1_{F}$ , now we note that $x ϕ (a^{- 1}) = ϕ (a) ϕ {(a)}^{- 1} = ϕ (a a^{- 1}) = ϕ (1_{F}) = 1_{S}$ . In other words $ϕ (a^{- 1}) \in im (ϕ)$ is an inverse for $x$ , now we conclude that $im (ϕ)$ is a subfield of $S$ .

We will now show that $im (ϕ)$ is isomorphic to $F$ through $ϕ$ , we have to show that it's a homomorphism, but we already know it is because of the fact that $ϕ$ is a ring homomorphism, so we just have to show it's a bijection, but every function is in bijection to it's own image, therefore it's an isomorphism.

GCD of two Polynomials

Find the gcd of

p (x) = x^{3} - 2 x^{2} + 1

and

j (x) = x^{2} - x - 3

Q [x]

and express it as a linear combination

We employ the euclidean algorithm:

From our first pass of division we conclude that $p (x) = (x - 1) j (x) + (2 x - 2)$ . Denote $q_{1} (x) = x - 1$ and $r_{1} (x) = 2 x - 2$ . From the second iteration we conclude that $j (x) = \frac{1}{2} x r_{1} (x) - 3$ and denote $q_{2} (x) = \frac{1}{2} x$ we can see that $- 3$ is going to be our last non-zero remainder, therefore we know that their gcd will be one. We will find $- 3$ as a linear combination of the others and then normalize it.

We do it all symbolically: $\begin{aligned} - 3 & = j (x) - q_{2} (x) r_{1} (x) \\ = j (x) - q_{2} (x) [p (x) - q_{1} (x) j (x)] \\ = j (x) - q_{2} (x) p (x) - q_{2} (x) q_{1} (x) j (x) \\ = j (x) [1 - q_{2} (x) q_{1} (x)] - q_{2} (x) p (x) \end{aligned}$ Therefore $1 = j (x) [\frac{- 1}{3} (1 - q_{2} (x) q_{1} (x))] + p (x) \frac{q_{2} (x)}{3}$ , note that it can be expanded explicitly from this point.