# Some notes on Polynomials

(This is a work in progess...)

(Note: I am doing some exploration on Galois Theory and I have written these notes on polynomials here for my own convenience. Please write in the comments section if any clarification is required. Also, I will not provide proofs here for the time being.)

## Some notations:

The symbol $\mathbb{C}$ is the domain of complex numbers; i.e. set of all possible complex numbers.
The symbol $\mathbb{R}$ is the domain of real numbers; i.e. set of all possible real numbers.
The symbol $\mathbb{Q}$ is the domain of rational numbers; i.e set of all possible rational numbers.
The symbol $\mathbb{Z}$ is the domain of integers; i.e set of all possible Integers.
Generally, the symbol $K$ is used to denote a subfield of $\mathbb{C}$.
Generally, the symbol $R$ is used to denote a subring of $\mathbb{C}$.
In this blogpost, all subring are assumed to be subring with unity.
$a$ divides $b$ is denoted as $a|b$.
$a$ does not divide $b$ is denoted as $a\not|b$.
$\partial f$ denotes the degree of polynomial $f$.
$i$ denotes $\sqrt{-1}$.
The symbol $\in$ is used to denote element of. For example $a\in A$ means $a$ is an element of set $A$.
Complex number are of the form $x+iy$, where $x,y \in \mathbb{R}$.
Set of $n$ equivalence classes, over integers, is denoted by $\mathbb{Z}_n$.
$\mathbb{Z}/n\mathbb{Z}$ denotes the integers modulo $n$ or cyclic group of order $n$.
Under addition, $\mathbb{Z}_n=\mathbb{Z}/n\mathbb{Z}$.

## Polynomial in general context

A polynomial over $\mathbb{C}$ in the indeterminate $t$ is an expression:
$r_0+r_1t+\dots+r_nt^n$
where $r_0,\dots,r_n\in\mathbb{C}$,   $0\le n\in \mathbb{Z}$  and $t$ is undefined.
$r_i$ are the coefficients of the polynomial.

### Example 1:

$f(t)= (2+3i)t^2+5t+23+5i$
is a polynomial over $\mathbb{C}$

### Example 2:

$f(t)= 2t^3+5t+23$
is a polynomial over $\mathbb{Z}\subset \mathbb{C}$

## Ring of polynomials over $\mathbb{C}$

The set of all polynomials over $\mathbb{C}$ in the $t$ obeys all of the usual algebraic laws.

We denote this set by $\mathbb{C}[t]$, and call it the ring of polynomials over $\mathbb{C}$ in the indeterminate $t$.

We can also define polynomials in several indeterminates $t_1,t_2,\dots,t_n$, obtaining the ring of $n$-variable polynomials $\mathbb{C}[t_1,t_2,\dots,t_n]$ in an analogous way.

An element of $\mathbb{C[t]}$ will usually be denoted by a single letter, such as $f$, whenever it is clear which indeterminate is involved.

If there is ambiguity, we write $f(t)$ to emphasize the role played by $t$

## The $f(t)$ notation:

The $f(t)$ notation makes $f$ appear to be a function, with $t$ as its independent variable, and in fact we can identify each polynomial $f$ over $\mathbb{C}$ with the corresponding function.

Each polynomial $f \in \mathbb{C}[t]$ can be considered as a function from $\mathbb{C}$ to $\mathbb{C}$, defined as follows:
if $f=\sum r_it^i$ and $\alpha \in \mathbb{C}$, then $\alpha$ is mapped to $\sum r_i\alpha^i$

## Proposition: equality over a function

Two polynomials $f, g$ over $\mathbb{C}$ define the same function if & only if they have the same coefficients.

Proposition implies that we can safely consider a polynomial over a subfield of $\mathbb{C}$ as either a formal algebraic expression or a function.

Moreover, the same notational flexibility allows us to change the variable in a polynomial.

For example, if $t, u$ are two indeterminates and $f(t)=\sum r_it^i$, then we may define $f(u) =\sum r_iu^i$

## degree of polynomial $f$

If $f$ is a polynomial over $\mathbb{C}$ and $f\ne 0$, then the degree of $f$ is the highest power of $t$ occurring in $f$ with nonzero coefficient.

We write $\partial f$ for the degree of $f$

## Proposition: on degrees of polynomials

If $f, g$ are polynomials over $\mathbb{C}$, then
$\partial (f+g) \le \max(\partial f, \partial g)$ and $\partial(fg)=\partial f + \partial g$

## Theorem: Fundamental Theorem of Algebra

Let $p(t)$ be a polynomial over $\mathbb{C}$, with $\partial p \ge 1$ . Then there exists at least one $z\in\mathbb{C}$ such that $p(z) = 0$.

In simpler terms: Every complex polynomial has a root.

## Proof of Fundamental Theorem of Algebra

Read a proof I found on the internet here: https://www.math.brown.edu/~res/INF/handout5.pdf

Device 1:
Device 2:

## Theorem: Remainder Theorem

Let $p(t) \in \mathbb{C}[t]$ with $\partial p > 1$, and let $\alpha \in \mathbb{C}$
1. There exist $q(t) \in \mathbb{C}[t]$ and $r \in \mathbb{C}$ such that, $p(t) = (t-\alpha)q(t) + r$
2. The constant $r$ satisfies $r=p(\alpha)$

## Corollary: from Remainder Theorem:

The complex number $\alpha$ is a zero of $p(t)$ if and only if $t-\alpha$ divides $p(t)$ in $\mathbb{C}[t]$

## Proposition about factorization

Let $p(t) \in \mathbb{C}[t]$ with $\partial p = n \ge 1$. Then there exist $\alpha_1,\dots,\alpha_n \in \mathbb{C}$, & $0\ne k \in \mathbb{C}$, such that
$p(t) = k(t-\alpha_1)\dots(t-\alpha_n)$

## Multiplicity of zero

$\alpha_j$ are the only complex zeros of $p(t)$. The zeros $\alpha_j$ need not be distinct.

Collecting together those that are equal, we can rewrite Equation in the form
$p(t)=k(t-\beta_1)^{m_1}\dots(t-\beta_l)^{m_l}$

where $\beta_j$ are distinct, & $m_j$ are integers $\ge 1$ and $m_1+\dots+m_l=n$

We call $m_j$ the multiplicity of the zero $\beta_j$ of $p(t)$

Every complex polynomial of degree $n$ has precisely $n$ complex zeros, counted according to multiplicity.

## Proposition: Division Algorithm

Let $f$ & $g$ be polynomials over subfield $K\subset \mathbb{C}$, & suppose that $f$ is nonzero. Then there exist unique polynomials $q$ and $r$ over subfield $K\subset \mathbb{C}$ , such that $g=fq+r$ and $r$ has a strictly smaller degree than $f$.

With the above notation, $q$ is called the quotient & $r$ is called the remainder on dividing $g$ by $f$.

The inductive process we employed to find $q$ and $r$ is called the Division Algorithm.

## divisibility of polynomials

Let $f$ and $g$ be polynomials over subfield $K\subset \mathbb{C}$. We say that $f$ divides $g$ (or $f$ is a factor of $g$, or $g$ is a multiple of $f$) if there exists some polynomial $h$ over $K$ such that $g = fh$
The notation $f|g$ will mean that $f$ divides $g$, while $f\not |g$ will mean that $f$ does not divide $g$.

## Highest common factor (HCF) of polynomials

A polynomial $d$ over field $K\subset \mathbb{C}$ is a highest common factor (HCF) of polynomials $f$ and $g$ over $K$ if $d|f$ and $d|g$ and further, whenever $e|f$ and $e|g$, we have $e|d$.

HCF's need not be unique. The next lemma shows that they are unique apart from constant factors.

## Lemma about HCF of Polynomials

If $d$ is an hcf of the polynomials $f$ and $g$ over subfield $K\subset \mathbb{C}$, and if $0\ne k \in K$, then $kd$ is also hcf for $f$ & $g$.

If $d$ and $e$ are two hcf's for $f$ and $g$, then there exists a nonzero element $k \in K\subset\mathbb{C}$ such that $e = kd$.

## Euclidean Algorithm  for calculating HCF of polynomials

Ingredients: Two polynomials $f$ and $g$ over subfield $K \subset \mathbb{C}$, both nonzero.

Recipe: For notational convenience let $f = r_{-1},g=r_0$.
Use the Division Algorithm to find successively polynomials $q_j$ and $r_i$ such that

$r_{-1}=q_1r_0+r_1$$\partial r_1\lt \partial r_0$
$r_{0}=q_2r_1+r_2$$\partial r_2\lt \partial r_1$
$r_{1}=q_3r_2+r_3$$\partial r_3\lt \partial r_2$
$\vdots$
$r_{i}=q_{i+2}r_{i+1}+r_{i+2}$$\partial r_{i+2}\lt \partial r_{i+1}$
$\vdots$
$\huge r_{s}=q_{s+2}r_{s+1}$ and  $\huge r_{s+2}=0$

Because the degrees of the $r_i$ decrease, we must eventually reach a point where the process stops; this can happen only if some $r_{s+2} = 0$

The last equation in the list then reads
$r_s = q_{s+2}r_{s+1}$
and it provides the answer we seek.

With the above notation, $r_{s+1}$ is an hcf for $f$ and $g$.

The above algorithm is the same as long division algorithm that we have always used to find HCF over integers ($\mathbb{Z}$).

### Example:

Let $f = t^4 + 2t^3 + 2t^2 + 2t + 1$, $g = t^2 - 1$ over $\mathbb{Q}$

We compute an hcf as follows:

$f = t^4 + 2t^3 + 2t^2 + 2t + 1$
$=(t^2+2t+3)(t^2-1)+4t+4$

Now,
$t^2-1=\left(4t+4\right)\left({1\over 4}t-{1\over 4}\right)$

Hence $4t + 4$ is an hcf. So is any rational multiple of it; e.g $t+1$

## Theorem: w.r.t hcf of two polynomials

Let $f$ and $g$ be nonzero polynomials over subfield $K\subset \mathbb{C}$, & let $d$ be an hcf for $f$ and $g$. Then there exist polynomials $a$ and $b$ over $K$ such that:
$d = af + bg$

## polynomial analogue of a prime number

A polynomial of positive degree over a subring $R$ of $\mathbb{C}$ is reducible if it is a product of two polynomials over $R$ of smaller degree. Otherwise, it is irreducible

## Theorem: w.r.t irreducible polynomials

Any nonzero polynomial over a subring $R$ of $\mathbb{C}$ is a product of irreducible polynomials over $R$.

## coprime polynomials

definition: If $f$ and $g$ are polynomials over a subfield $K$ of $\mathbb{C}$ with hcf equal to $1$, we say that $f$ and $g$ are coprime, or $f$ is prime to $g$.

The key to unique factorization is a statement analogous to an important property of primes in $\mathbb{Z}$, and is used in the same way.

## Lemma w.r.t irreducibility and divisibility

Let $K$ be a subfield of $\mathbb{C}$, $f$ an irreducible polynomial over $K$, & $g$, $h$ polynomials over $K$. If $f$ divides $gh$, then either $f$ divides $g$ or $f$ divides $h$.

## Theorem: Uniqueness Theorem

For any subfield $K$ of $\mathbb{C}$, factorization of polynomials of positive degree over $K$ into irreducible polynomials is unique up to constant factors & the order in which the factors are written.

## Lemma: Gauss's Lemma

Let $f$ be a polynomial over $\mathbb{Z}$ that is irreducible over $\mathbb{Z}$. Then $f$, considered as a polynomial over $\mathbb{Q}$, is also irreducible over $\mathbb{Q}$

## Corollary over Gauss's Lemma

Let $f \in \mathbb{Z}[t]$ and suppose that over $\mathbb{Q}[t]$ there is a factorization into irreducibles:
$f = g_1\dots g_s$
Then there exist $a_i\in\mathbb{Q}$ such that $a_ig_i \in\mathbb{Z}[t]$ and $a_1\dots a_s = 1$.
Furthermore,
$f = (a_1g_1)\dots(a_sg_s)$
is a factorization of $f$ into irreducibles in $\mathbb{Z}[t]$

## Theorem: Eisenstein's Criterion

Let, $f(t)=a_0+a_1t+\dots+a_nt^n$ be a polynomial over $\mathbb{Z}$. Suppose that there is a prime $q$ such that
(1) $q\not | a_n$
(2) $q|a_i$  $(i=0,\dots,n-1)$
(3) $q^2\not | a_0$
Then $f$ is irreducible over $\mathbb{Q}$

### Example 1:

Consider
$f(t) ={2\over 9}t^5 + {5\over 3}t^4 + t^3+ {1\over 3}$ over $\mathbb{Q}$.

This is irreducible over $\mathbb{Q}$ if and only if
$9 f(t) = 2t^5 + 15t^4 + 9t^3 + 3$
is irreducible over $\mathbb{Q}$.

Eisenstein's criterion now applies with $q = 3$, showing that $f$ is irreducible.

### Example 2:

Consider
$f(t) = t^{16} + t^{15}+ \dots + 1$ over $\mathbb{Q}$

As it stands, $f$ is not susceptible to the Eisenstein treatment.

But  $f(t)$ is obviously irreducible if and only if $f(t + 1)$ is irreducible.
Hint: $f(t) = g(t)h(t)$ if and only if $f(t + 1) = g(t + 1)h(t + 1)$

If we expand this monster, we obtain

$f(t + 1) = t^{16} + 17t^{15} + 136t^{14} + 680t^{13} + 2380t^{12}+ 6188t^{11}$
$+ 12376t^{10} + 19448t^{9} + 24310t^{8} + 24310t^{7}+ 19448t^{6}$
$+ 12376t^{5} + 6188t^{4} + 2380t^{3} + 680t^{2} + 136{t} + 17$

$= t^{16} + 17(t^{15} + 8t^{14} + 40t^{13} + 140t^{12}$ $+ 364t^{11}$
$+ 728t^{10} + 1144t^{9} + 1430t^{8} + 1430t^{7} + 1144t^{6}$
$+ 728t^{5} + 364t^{4} + 140t^{3} + 40t^{2} + 8{t} + 1)$

Therefore, Eisenstein's criterion applies with $q = 17$, so $f$ is irreducible over $\mathbb{Q}$.

## Reduction Modulo p

The idea is this. There is a natural map $\mathbb{Z}\to \mathbb{Z}_n$ in which each $m \in \mathbb{Z}$ maps to its congruence class modulo $n$.

The natural map extends in an obvious way to a map $\mathbb{Z}[t]\to\mathbb{Z}_n[t]$.

Now a reducible polynomial over $\mathbb{Z}$ is a product $gh$ of polynomials of lower degree, & this factorization is preserved by the map. Provided $n$ does not divide the highest coefficient of the given polynomial, the image is reducible over $\mathbb{Z}_n$.

So if the image of a polynomial is irreducible over $\mathbb{Z}_n$, then the original polynomial must be irreducible over $\mathbb{Z}$.

But unfortunately, in practice, the trick is to choose the right value for $n$ which shows irreducibility.

Also, Since $\mathbb{Z}_n$ is finite, there are only finitely many possibilities to check when deciding
\irreducibility.