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> [!definition] Definition. ([[polynomial]])
> Let $R$ be a [[ring]]. A **polynomial $f(x)$ in [[indeterminate]] $x$ and with coefficients in $R$** is a (finite) linear combination of nonnegative 'powers' of $x$ with coefficients in $R$: $f(x)=\sum_{i \geq 0} a_{i}x^{i}=a_{0} + a_{1}x + a_{2}x^{2} +\dots,$
> where all $a_{i}$ are elements of $R$ (the **coefficients** and we require $a_{i}=0$ for $i \gg 0$). Two polynomials are taken to be **equal** if all the coefficients are equal.
>
> The set of polynomials in $x$ over $R$ is denoted $R[x]$. Since all but finitely $a_{i}$ are assumed 0, one usually employs notation $f(x)=a_{0}+a_{1}x+\dots+a_{n}x^{n}$
> for $\sum_{i \geq 0}a_{i}x^{i}$, if $a_{i} =0$ for $i>n$.
>
The **degree** of $f(x)$ is $-\infty$ if $f(x)=0$, and otherwise is the largest integer $d$ for which $a_{d} \neq 0$. Polynomials of degree $0$ are called **constant**; together with $0$ itself they form a 'copy' of $R$ inside $R[x]$.
>
^definition
> [!definition] The Ring of Polynomials with Coefficients in $R.$
> $R[x]$ itself is a [[ring]] with addition defined $f(x)+g(x):=\sum_{i \geq 0}(a_{i}+b_{i})x^{i}$
> and multiplication $f(x) \cdot g(x) := \sum_{k \geq 0} \sum_{i+j=k}a_{i}b_{j} x^{i+j}.$
>
>
Polynomials in multiple indeterminates are obtained inductively as $R[x_{1},\dots,x_{n}]:=R[x_{1}]\dots[x_{n}]$.
>
>One often identifies (as abelian groups) polynomials of $\text{degree}<d$ as elements of the direct sum $R^{\oplus d}=\underbrace{R \oplus \dots \oplus R}_{d \text{ times}},$
>indeed, the function $\begin{align}
\psi: R^{\oplus d} \to& R[x] \\
(r_{0},r_{1},\dots,r_{d-1}) \xmapsto{\psi}& r_{0}+r_{1}x+\dots+r_{d-1}x^{d-1}
\end{align}$
>is clearly an [[injection|injective]] [[group homomorphism|homomorphism]] of [[abelian group|abelian groups]], hence an [[group embedding|isomorphism onto its image]].
^definition-2
> [!note] Remark.
> The notion of degree behaves as we'd like when $R[](direct%20sum%20of%20vector%20spaces.md)gral domain]]: in this case, we have for nonzero $f(x),g(x) \in R[x]$ that $\text{deg}\big( f(x) \cdot g(x) \big)=\text{deg}\big( f(x) \big) + \text{deg}\big( g(x) \big).$
>
>> [!proof]- Proof of Remark.
> Let $d_{f}$ and $d_{g}$ be the degrees of $f,g$ respectively. Write $d=d_{f}+d_{g}$. Take the product definition $f(x) \cdot g(x)= \sum_{k \geq 0} \sum_{i+j=k} a_{i}b_{j}x^{i+j}.$
The goal is to show that the $k=d$ term, $\sum_{i+j=d}a_{i}b_{j}x^{i+j}$, is nonzero, while all terms with $k>d$ do equal zero. Take the entry in the sum where $i=d_{f}$ (also enforcing $j=d_{g}$). By hypothesis, the coefficient $a_{d_{f}} \neq 0$ and $b_{d_{g}} \neq 0$. Because $R$ is an [[integral domain]], their product must be nonzero. Thus, the $k=d$ term is indeed nonzero. Moreover, if we let $K>k$ be arbitrary, every term in the sum $\sum_{i+j=K}a_{i}b_{j}x^{i+j}$ must be zero, because whenever $i<d_{f}$ we have $j > d_{j}$ (thus $b_{j}=0$, hence $a_{i}b_{j}=0$) and whenever $j < d_{g}$ we have $i > d_{i}$ (thus $a_{i}=0$, hence $a_{i}b_{j}=0$).
^note
> [!basicproperties]
>
>- If $f,g \in R[x]$ are nonzero then $\text{deg}\big( f(x) + g(x) \big) \leq \max\big( \text{deg}\big(f(x)\big), \text{deg}\big( g(x) \big) \big)$
>- $R[x]$ is an [[integral domain]] iff $R$ is an [[integral domain]].
^properties
> [!proof] Proof of Properties.
> **1.** Write $d_{f}$ and $d_{g}$ for the respective degree and recall the definition of addition:
$f(x)+g(x)=\sum_{i \geq 0} (a_{i}+b_{i})x^{i}.$
If $d> \max(d_{f},d_{g})$ then $a_{d}=0=b_{d}$ and therefore $a_{d}+b_{d}=0$. This is the result?
**2.** One direction follows from the natural identification of $R$ with a [[subring]] of $R[x]$: if $R[x]$ is an [[integral domain]] and $a,b \in R$ with $ab=0$, then we consider the two constant polynomials $a+0x+0x^{2}+\dots$ and $b+0x+0x^{2}+\dots$. The product of these polynomials is zero if and only if one of them is the zero polynomial, if and only $a=0$ or $b=0$. Hence $R$ is an [[integral domain]].
>
Conversely, let $R$ be an [[integral domain]] and take two polynomials $f(x),g(x) \in R[x]$ satisfying $f(x) \cdot g(x)=0+0x+0x^{2}+\dots$. That is, every inner sum in $\sum_{k \geq 0} \sum_{i+j=k}a_{i}b_{j}x^{i+j}$
is zero, meaning $\sum_{i+j=k}a_{i}b_{j}=0$ for all $k \in \mathbb{N} \cup \{ 0 \}$. If one of $f(x)$ or $g(x)$ is zero, we are done, so assume neither are zero; fix positive degrees $d_{f}$ and $d_{g}$. Then the product has degree $d_{f}+d_{g}$; consider the term when $k=d_{f}+d_{g}$, for it we have $\sum_{i+j=d_{f}+d_{g}}a_{i}b_{j}=0.$
Of course, this sum actually only has one term, that being $a_{d_{i}}b_{d_{j}}$. So we've shown $a_{d_{i}}b_{d_{j}}=0$ with $a_{d_{i}} \neq 0$ and $b_{d_{j}} \neq 0$, contradicting that $R$ is an [[integral domain]].
^proof
> [!justification]
> First we clarify the definition of multiplication. It agrees with how one learns to multiply polynomials in high school: $\begin{align}
\left( \sum_{i \geq 0} a_{i}x^{i} \right)\left( \sum_{j \geq 0} b_{j}x^{j} \right)= & \sum_{i \geq 0} \big( a_{i}x_{i}\sum_{j \geq 0 } b_{j}x^{j} \big) & \text{(distribute)} \\
= & \sum_{i \geq 0} \sum_{j \geq 0} a_{i}b_{j}x^{i+j} & \text{(distribute)} \\
= & \sum_{k \geq 0} \sum_{i+j=k} a_{i}b_{j}x^{i+j} & \text{(order by power)}
\end{align}$
Next we check $R[x]$ endowed with these [[binary operation|operations]] is a [[ring]]. The [[abelian group]] structure is clear: $0=0x^{0}+0x^{1}+\dots$ is the additive identity, $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\dots$ has additive inverse $-a_{0} + (-a_{1})x+(-a_{2})x^{2}$ inherited from the [[abelian group]] structure on $R$; likewise associativity is inherited: if $f(x)$ has coefficients $a_{i}$, $g(x)$ has coefficients $b_{i}$, $h(x)$ has coefficients $c_{i}$, then $\begin{align}
f(x)+\big( g(x) + h(x) \big)= & \sum_{i \geq 0} \big( f(x) + (g(x)+h(x)) \big)x^{i} \\
= & \sum_{i \geq 0} \big( (f(x) + g(x)) + h(x) \big)x^{i} \\
= & \big( f(x) + g(x) \big) + h(x).
\end{align}$The multiplicative identity is the polynomial $1_{R[x]}=1_{R}+0x+0x^{2}+\dots$. Multiplication is associative because, noting that the polynomial $g(x)h(x)$ has as its $k^{th}$ coefficient $\sum_{j+\ell=k}b_{j}c_{\ell}$ , $\begin{align}
f(x)\big( g(x)h(x) \big) = & \sum_{k \geq 0} \sum_{i+j=k} ( a_{i} \sum_{j+\ell=k}b_{j}c_{\ell})x^{i+j}\\ = & \sum_{k \geq 0} \sum_{i+j=k} \sum_{j+\ell=k}a_{i} (b_{j}c_{\ell})x^{i+j} \\
=& \sum_{k \geq 0} \sum_{i+j=k} \sum_{j+\ell=k}(a_{i} b_{j})c_{\ell}x^{i+j} \\
= & \sum_{k \geq 0} \sum_{i+j=k} (a_{i} b_{j} \sum_{j+\ell=k}c_{\ell})x^{i+j} \\
= & \big( f(x) g(x) \big)h(x).
\end{align}$
That distributivity holds is a similarly straightforward check.
^justification
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####
# Legacy
# Definition
Let $\ff$ denote $\mathbb{R}$ or $\cc$. A function $p: \ff \to \ff$ is called a **polynomial** with coefficients in $\ff$ if there exist $a_0, \dots, a_m \in \ff$ such that $p(z) = a_0 + a_1z + a_2z^2 + \dots + a_mz^m$
for all $z \in \ff$.
*It can be shown that the coefficients are unique.*
We call the set of all polynomials with coefficients in $\ff$ [[vector space of all polynomials with coefficients in F]].
# On Degree
A **polynomial** $p \in \PP(\ff)$ is said to have **degree** $m$ if there exist scalars $a_0, a_1, \dots ,a_m \in \ff$ with $a_m \neq 0$ such that $p(z) = a_0 + a_1z + \dots + a_mz^m$
for all $z \in \ff$. If $p$ has degree $m$, we write $\deg p=m$. The **polynomial** that is identically $0$ is said to have degree $-\ii$.
For $m \in \zz^+$, we call the set of all **polynomials** with coefficients in $\ff$ and **degree** at most $m$ $\PP_m(\ff)$
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#### References
> [!backlink]
> ```dataview
TABLE rows.file.link as "Further Reading"
FROM [[]]
FLATTEN file.tags
GROUP BY file.tags as Tag
> [!frontlink]
> ```dataview
TABLE rows.file.link as "Further Reading"
FROM outgoing([[]])
FLATTEN file.tags
GROUP BY file.tags as Tag