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:
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:

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  data-recalc-dims= 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

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
r_{i}=q_{i+2}r_{i+1}+r_{i+2}\partial r_{i+2}\lt \partial r_{i+1}
\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}).


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^2-1=\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.
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:

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:

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


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