0-Universal Properties

Functions to single object sets

Let $A = a$ be a single element set. For every set $X$ , there exists exactly one function
$f_{X} : X \to a$ .

$* p r o o f * :$ Trivial. the only function is the one that takes each $x \in X$ to $f_{X} (x) = a$ .

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Ring

A ring $R$ is a triple $(R, +, \cdot)$ where $(R, +)$ is an abelian group with identity $0 \in R$ and $(R, \cdot)$ is a monoid with identity $1 \in R$ .
Morover, there are two way distributive laws $a (b + c) = (b + c) a = a b + a c$ .
The cartesian product of $n$ rings $\prod_{i = 1}^{n} R_{i}$ with addition and multiplication being element-wise on the tuples, also forms a ring.
A ring is commutative if $(R, \cdot)$ is a commutative monoid.

Ring homomorphisms

let $(R, +_{R}, \cdot_{R}) \overset{ϕ}{⟶} (S, +_{s}, \cdot_{s})$ be a function from ring $R$ to ring $S$ . $ϕ$ is a ring homomorphism if:

$ϕ (x +_{R} y) = ϕ (x) +_{R} ϕ (y)$
$ϕ (x \cdot_{R} y) = ϕ (x) \cdot_{R} ϕ (y)$
$ϕ (1_{R}) = 1_{S}$ (this one needs to be defined, cant be extracted as a result cause multiplication on a ring is a monoid. to prove that $ϕ (O_{R}) = O_{s}$ one uses the inverse available in the addition group, we don't have that luxury in the multiplication monoid).

Polynomial rings

let $N^{m}$ be the monoid of natural number tuples, with addition (defined as element-wise addition on tuples). define the set of functions $P$ from $N^{m} \to R$ where $R$ is a commutative ring, with finite support:

P = {p : N^{m} \to R : | p (x) \neq 0 \in R | < | N |}

where we have addition and multiplication on $P$ :

(p +_{P} q) (x) := p (x) +_{R} q (x)

(p \cdot_{P} q) (x) = \sum_{u + v = x}^{R} p (x) \cdot_{R} q (x)

Closure of addition and multiplication in polynomial rings

For $p, q \in P$ , both $p + q$ and $p q$ are in $P$ .

$* p r o o f * :$
Let $S (p) = {x : p (x) \neq 0}$ , which is finite by definition.

For addition: If $(p + q) (x) \neq 0$ , then either $p (x) \neq 0$ or $q (x) \neq 0$ . Thus
$S (p + q) \subseteq S (p) \cup S (q),$
which is finite.
For multiplication: If $(p q) (x) \neq 0$ , then there exist $u, v$ with $u + v = x$ , $p (u) \neq 0$ , and $q (v) \neq 0$ . Hence
$S (p q) \subseteq S (p) + S (q) := {u + v : u \in S (p), v \in S (q)} .$
Since $S (p), S (q)$ are finite, so is $S (p) + S (q)$ . Therefore $p q \in P$ .
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Polynomial ring multiplication forms a commutative monoid

Multiplication in $P$ is associative, commutative, and has an identity element $1_{P}$ .

$* p r o o f * :$
Associativity:

((p q) r) (x) = \sum_{u + v = x} p q (u) r (v) = \sum_{u + v = x} \sum_{s + t = u} p (s) q (t) r (v) .

Rearranging the addition (abelian group) and using associativity in $R$ ,

((p q) r) (x) = \sum_{s + t + v = x} p (s) q (t) r (v) .

Similarly,

(p (q r)) (x) = \sum_{s + t + v = x} p (s) q (t) r (v),

so multiplication is associative.

Commutativity:

(p q) (x) = \sum_{u + v = x} p (u) q (v) = \sum_{u + v = x} q (v) p (u) = (q p) (x) .

Identity: Define $1_{P}$ by $1_{P} (0, \dots, 0) = 1_{R}$ and $1_{P} (x) = 0_{R}$ otherwise. Then

(1_{P} p) (x) = \sum_{u + v = x} 1_{P} (u) p (v),

but only the term $u = (0, \dots, 0)$ contributes, yielding $p (x)$ . Thus $1_{P}$ is the identity.
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Polynomial ring addition forms an abelian group

With addition defined as above (pointwise), $(P, +)$ is an abelian group.

$* p r o o f * :$

Associativity/commutativity: Directly inherited from $(R, +)$ .
Identity: The zero polynomial $0_{P}$ defined by $0_{P} (x) = 0_{R}$ for all $x$ is the additive identity.
Inverses: For $p \in P$ , define $(- p) (x) = - p (x)$ (additive inverse in $R$ ). Then $p + (- p) = 0_{P}$ .
Thus $(P, +)$ is an abelian group.
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The ring of polynomials is itself a ring

Let $R$ be a commutative ring, and $P_{R}$ be the polynomial ring over $R$ as defined in Polynomial rings . Then $P_{R}$ is itself a commutative ring.

$* p r o o f * :$
combine Closure of addition and multiplication in polynomial rings , Polynomial ring multiplication forms a commutative monoid , Polynomial ring addition forms an abelian group
$◻$ .

Note: Now you may be wondering where are my good old ring of polynomials as expressions $R [x_{1}, x_{2}, \dots x_{m}]$ well, there is a natural isomorphism from $P_{R}$ to $R [x_{1}, x_{2}, \dots x_{m}]$ and I'll leave it to the reader to construct this classical polynomial ring and to find the isomorphism. hint: start with an empty expression, and then for any $p \in P_{R}$ whenever $(t_{1}, t_{2}, . . t_{m}) \in S (p)$ (the support of p) "add" (as defined in the polynomial expression ring) the term $p (t_{1}, t_{2}, \dots t_{m}) x_{1}^{t_{1}} x_{2}^{t_{2}} \dots x_{m}^{t_{m}}$ to the expression. Since we have a finite support set, we will get a finite polynomial expression. Bijectivity comes from the idea that every $p \in P_{R}$ is fully defined by its support and the image of its support in the ring, and hence induces a unique expression in $R [x_{1}, x_{2}, \dots x_{m}]$ proving homomorphicity is straightforward but calculative in the sense of writing a good amount of algebraic equations, and that's not really my style.