Magma
A magma is a set $$M$$ with a binary operation $$\star$$ defined on $$M$$
Monoid
A monoid is a set $$X$$ with a binary operation $$\star$$ on $$X$$ such that $$\star$$ 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

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 $$R := \left\{ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} : a, b \in \mathbb{ Z } \right\}$$ 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 $$\left( R, \oplus, \otimes \right)$$ be a ring, then we say that an element $$x \in R \setminus \left\{ 0 _ R \right\}$$ is a zero divisor if there is some $$y \in R \setminus \left\{ 0 _ R \right\}$$ such that the following equation holds $x \otimes y = y \otimes x = 0 _ R$
The Integers form a Ring
$$\mathbb{ Z }$$ 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 $R [ x ] = \left\{ r_n x ^ n + r_ { n - 1 } x ^ { n - 1 } + \ldots + r _ 1 x ^ 1 + r ^ 0 : n \in \mathbb{ N } _ 0, r _ i \in R \right\}$ Along with addition, defined such that for any $$m \le n \in \mathbb{ N } _ 0$$, we have: $\sum _ { i = 0 } ^ n r _ i x ^ i + \sum _ { i = 0 } ^ m s _ i x ^ i := \sum _ { i = 0 } ^ n r _ i x ^ i + \sum _ { i = 0 } ^ n s _ i x ^ i := \sum _ { i = 0 } ^ n \left( r _ i + s _ i \right) x ^ i$ Where for every $$m < k \le n$$ we've set $$s _ k = 0$$, thus artifically padding the polynomial without changing it. As for multiplication, we define $\left( \sum _ { i = 0 } ^ n r _ i x ^ i \right) \left( \sum _ { j = 0 } ^ m s _ i x ^ i \right) := \sum _ { k = 0 } ^ { n + m } t _ k x ^ k$ 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.

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 $$p$$ is $$n$$, and it is denoted by $$\operatorname{ deg } \left( p \right)$$
Square Matrices form a Non-Commutative Ring
Let $$R$$ be a ring and then $$M _ { n \times n } \left( R \right)$$ 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 $$r \in R$$ $0 r = 0$
$$-r = \left( -1 \right) r$$
$$\left( -1 \right) \left( -r \right) = r$$
The set of One Element Forms a Ring
Suppose that $$R = \left\{ r \right\}$$, 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 $\operatorname{ deg } \left( f g \right) = \operatorname{ deg } \left( f \right) + \operatorname{ deg } \left( g \right)$

We first start by setting $$n := \operatorname{ deg } \left( f \right)$$ and $$m = \operatorname{ deg } \left( g \right)$$. 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 $$\operatorname{ deg } \left( \cdot \right)$$ 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 $$\operatorname{ deg } \left( fg \right) = \operatorname{ deg } \left( f \right) + \operatorname{ g }$$ 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 $$\left( R, \oplus, \otimes \right)$$ is a ring, and let $$\left( R, \oplus \right)$$ be it's additive group, then we say that $$I \subseteq R$$ is a left ideal of $$R$$ if
• $$\left( I, \oplus \right)$$ is a subgroup of $$\left( R, \oplus \right)$$
• For every $$r \in R$$ and $$x \in I$$ we have $$r \otimes x \in I$$
Right Ideal
Suppose that $$\left( R, \oplus, \otimes \right)$$ is a ring, and let $$\left( R, \oplus \right)$$ be it's additive group, then we say that $$I \subseteq R$$ is a left ideal of $$R$$ if
• $$\left( I, \oplus \right)$$ is a subgroup of $$\left( R, \oplus \right)$$
• For every $$r \in R$$ and $$x \in I$$ we have $$x \otimes r \in I$$
Binary Operation Applied to Sets
Suppose that $$\star$$ is a binary operation on a set $$X$$, and that $$A, B \subseteq X$$, then we define: $A \star B := \left\{ a \star b: a \in A, b \in B \right\}$

Notice that $$r, x$$ have changed order

Subring
Suppose that $$\left( R, \oplus, \otimes \right)$$ is a ring, and that $$S \subseteq R$$, if $$\left( S, \oplus, \otimes \right)$$ 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 $$1 _ R \in S$$
$$\mathbb{ Z }$$ is a Subring with Unity of $$\mathbb{ Q }$$
As per title.
We know that $$\mathbb{ Z } \subseteq \mathbb{ Q }$$, and let $$+, \cdot$$ be multiplication as defined in $$\mathbb{ Q }$$, then for any $$a, b \in \mathbb{ Z }$$ we can see that $$a + b \in \mathbb{ Q }$$ this is because really we have $$\frac{a}{1} + \frac{b}{1} = \frac{a + b}{1}$$, similarly $$a \cdot b \in \mathbb{ Z }$$ and we know that $$1 _ \mathbb{ Q } \in \mathbb{ Z }$$.
Subring Criterion
Suppose $$\left( R, \oplus, \otimes \right)$$ 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 $$\left( r, s \right)$$ is related to $$\left( u, v \right)$$ iff $$r \otimes v = s \otimes u$$. We will notate an element $$\left( x, y \right) \in R \times R$$ as $$\frac{x}{y}$$, 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 $\operatorname{ frac } \left( R \right) := \left\{ \left[ \frac{r}{s} \right] : r, s \in R, s \neq 0 _ R \right\}$
The Domain Fractions Can be Extended to a Crone
We define $$\oplus$$ as follows: $\left[ \frac{r}{s} \right] \oplus \left[ \frac{u}{v} \right] := \left[ \frac{rv + us}{sv} \right]$ And for $$\otimes$$ we define $\left[ \frac{r}{s} \right] \otimes \left[ \frac{u}{v} \right] := \left[ \frac{ru}{sv} \right]$ Then we claim that $$\left( \operatorname{ frac } \left( R \right) , \oplus, \otimes \right)$$ 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 $$\frac{ru}{sv} \notin \operatorname{ frac } \left( R \right)$$ (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 $$\operatorname{ frac } \left( R \right)$$ 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
$$\mathbb{ C }$$ is a field extension of $$\mathbb{ R }$$
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 $$\left[ E : F \right]$$
Degree of $$\mathbb{ C }$$ over $$\mathbb{ R }$$
Prove that $$\left[ \mathbb{ C } : \mathbb{ R } \right] = 2$$
We can prove it by showing that the basis $$\mathcal{ B } := \left( 1, i \right)$$ works
The Kernel of a Ring Homomorphism is a Proper Ideal
Suppose that $$\phi : R \to S$$ is a ring homomorphism, then $$\operatorname{ ker } \left( \phi \right)$$ is a proper ideal of $$R$$
A Ring Homomorphism is an Injection iff It's Kernel is Zero
Suppose that $$\phi : R \to S$$ is a ring homomorphism, then $$\phi$$ is injective iff $$\operatorname{ ker } \left( \phi \right) = \left\{ 0 _ R \right\}$$
Field to Ring Homomorphism Implies Injective
Suppose that $$F$$ is a field and that $$S$$ is a crone, then if $$\phi : F \to S$$ 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 $$\phi$$ is a crone homomorphism, then we know that $$\phi \left( 1 _ F \right) = 1 _ S$$, now suppose that $$x \in F \setminus \left\{ 0 _ F \right\}$$, then we have some $$x ^ { -1 }$$ such that $$x x ^ { -1 } = 1 _ F$$, so that $$1 _ S = \phi \left( 1 _ F \right) = \phi \left( x x ^ { -1 } \right)$$ but $$\phi$$ is a crone homomorphism, so that $$\phi \left( x x ^ { -1 } \right) = \phi \left( x \right) \phi \left( x ^ { -1 } \right) = \phi \left( x \right) \left[ \phi \left( x \right) \right] ^ { -1 }$$.

If $$\phi \left( x \right) = 0 _ S$$, 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 $$\phi \left( x \right) \neq 0$$, which says that $$\operatorname{ ker } \left( \phi \right) = \left\{ 0 _ F \right\}$$, therefore we obtain that $$\phi$$ 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 $$\phi : F \to S$$ is a crone homomorphism, then $$\operatorname{ im } \left( \phi \right)$$ is a subfield of $$S$$ and is isomorphic to $$F$$

To show that $$\operatorname{ im } \left( \phi \right)$$ 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 $$x, y \in \operatorname{ im } \left( \phi \right)$$, by definition this means that there is some $$a, b \in F$$ such that $$\phi \left( a \right) = x$$ and $$\phi \left( b \right) = y$$, we want to now show that $$x \oplus y \in \operatorname{ im } \left( \phi \right)$$, but note that $$\phi \left( a + b \right) = \phi \left( a \right) \oplus \phi \left( b \right) = x \oplus y$$ therefore $$a + b$$ maps to it under $$\phi$$ so we know that $$x \oplus y \in \operatorname{ im } \left( \phi \right)$$, similarly we can see that $$\phi \left( a \cdot b \right) = \phi \left( a \right) \otimes \phi \left( b \right) = x \otimes y$$ so that also $$x \otimes y \in \operatorname{ im } \left( \phi \right)$$, where we've used the fact that $$\phi$$ is a homomorphism a few times.

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

To show it's a subfield, we must now show that $$\operatorname{ im } \left( \phi \right)$$ is a field, so we must prove that it has multiplicative inverses, thus let $$x \in \operatorname{ im } \left( \phi \right)$$ 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 $$\phi \left( a \right) = x$$, 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 $$x \phi \left( a ^ { -1 } \right) = \phi \left( a \right) \phi \left( a \right) ^ { -1 } = \phi \left( a a ^ { -1 } \right) = \phi \left( 1 _ F \right) = 1 _ S$$. In other words $$\phi \left( a ^ { -1 } \right) \in \operatorname{ im } \left( \phi \right)$$ is an inverse for $$x$$, now we conclude that $$\operatorname{ im } \left( \phi \right)$$ is a subfield of $$S$$.

We will now show that $$\operatorname{ im } \left( \phi \right)$$ is isomorphic to $$F$$ through $$\phi$$, we have to show that it's a homomorphism, but we already know it is because of the fact that $$\phi$$ 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 $$p \left( x \right) = x ^ 3 - 2 x ^ 2 + 1$$ and $$j \left( x \right) = x ^ 2 - x - 3$$ in $$\mathbb{ Q } \left[ x \right]$$ and express it as a linear combination
We employ the euclidean algorithm:

From our first pass of division we conclude that $$p \left( x \right) = \left( x - 1 \right) j \left( x \right) + \left( 2x - 2 \right)$$. Denote $$q _ 1 \left( x \right) = x - 1$$ and $$r _ 1 \left( x \right) = 2x - 2$$. From the second iteration we conclude that $$j \left( x \right) = \frac{1}{2} x r _ 1 \left( x \right) - 3$$ and denote $$q _ 2 \left( x \right) = \frac{1}{2} x$$ 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{align} -3 &= j \left( x \right) - q _ 2 \left( x \right) r _ 1 \left( x \right) \\ &= j \left( x \right) - q _ 2 \left( x \right) \left[ p \left( x \right) - q _ 1 \left( x \right) j \left( x \right) \right] \\ &= j \left( x \right) - q _ 2 \left( x \right) p \left( x \right) - q _ 2 \left( x \right) q _ 1 \left( x \right) j \left( x \right) \\ &= j \left( x \right) \left[ 1 - q _ 2 \left( x \right) q _ 1 \left( x \right) \right] - q _ 2 \left( x \right) p \left( x \right) \end{align} Therefore $$1 = j \left( x \right) \left[ \frac{-1}{3} \left( 1 - q _ 2 \left( x \right) q _ 1 \left( x \right) \right) \right] + p \left( x \right) \frac{q _ 2 \left( x \right)}{3}$$, note that it can be expanded explicitly from this point.