(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 \(a ot|b\).
\(\partial f\) 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 \(f e 0\), 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

Applet devices to understand proof of the Fundamental Theorem of Algebra

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 \(\alpha_1,\dots,\alpha_n \in \mathbb{C}\), & \(0 e k \in \mathbb{C}\), 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 \(f ot |g\) 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 \(0 e k \in K\), 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) \(q ot | a_n\)
(2) \(q|a_i\)  \((i=0,\dots,n-1)\)
(3) \(q^2 ot | a_0\)
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.

Example:

Algebra