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Unit roots/Divisibility of polynomials

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This chapter demonstrates the use of unit roots in problems of divisibility of polynomials.

Divisibility of polynomials

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Let and be two polynomials.

Assume that there is another polynomial , such that:

.

If , and are polynomials with integer coefficients, then, is said to be divisible by over the integers.

Similarly, if , and are polynomials with rational coefficients, then, is said to be divisible by over the rational numbers.

If , and are polynomials with real coefficients or complex coefficients, then, is said to be divisible by over the real numbers or the complex numbers.

Example 1 Let and . The coefficients of these two polynomials are integers. However:

.

The polynomial in the second bracket of the right hand side does not have integer coefficients. Therefore, is not divisible by over the integers.

However, if we consider that and are polynomials with rational, real, or complex coefficients, then is also a polynomial with rational, real, or complex coefficients. So, is divisible by over the rational numbers, the real numbers, or the complex numbers. QED

Therefore, the divisibility of polynomials is highly associated with the set of numbers that we allow the coefficients to take. However, as both and are known polynomials, the coefficients of the quotient is key to the divisibility.

Now, let , where and have coefficients in the set of numbers we have specified. How can we determine whether is also a polynomial with coefficients in the same set? The answer of this question is given in the following rule:

Rule 1 Let .

  • If and have complex coefficients, then has complex coefficients. is divisible by over the complex numbers.
  • If and have real coefficients, then has real coefficients. is divisible by over the real numbers.
  • If and have rational coefficients, then has rational coefficients. is divisible by over the rational numbers.
  • If and have integer coefficients, and (sufficiently by not necessarily) then has integer coefficients. is divisible by over the integers.

This rule is proved in the appendix.

When we consider only integer coefficients, the condition is not necessary. For example, let and , then but .

Factor theorem and unit roots

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In elementary algebra, we have the remainder theorem. It is stated here without an accompanying proof:

Remainder theorem The remainder of the division of a polynomial by equals .

From this theorem we can have the following result, commonly called the "factor theorem":
Factor theorem A polynomial is divisible by , if and only if .
(Note that and are known, and we automatically assume that it falls in the set of numbers over which divisibility is considered.)

Proof Suppose is divisible by , then the remainder of the division of a polynomial by equals zero. So, .
On the other hand, if , then the remainder of the division of a polynomial by equals zero. So, . Moreover, the leading coefficient of the divisor is . Therefore, by rule 1, is divisible by over the integers, the rational numbers, the real numbers and the complex numbers.

Repeated applications of the factor theorem results in more complicated divisors:
Rule 2 If a polynomial satisfies:

,

then is divisible by .
Proof By factor theorem, since , there exists a polynomial such that:

.

Put :

.

Since and , we conclude that . By factor theorem, there exists a polynomial such that:

.

Therefore:

,
.

Note that the leading coefficient of the divisor is 1.

Corollary 1 If a polynomial with real coefficient satisfies , where and are real numbers and , then is divisible by .
Proof In algebra, we have already known that if a polynomial with real coefficients vanishes at some complex number , then it also vanishes at . So, we can apply rule 2 to obtain the required result, since implies .

The above corollary has a useful special case as shown below:
Corollary 2 If a polynomial with integer coefficient satisfies , where is a cube root of unity, then is divisible by .

Corollary 1 is an example of the application of the complex numbers in dealing with divisibility of polynomials with real coefficients. Corollary 2 further restricts to divisibility over the integers by the aid of the cube root of unity.

We can also apply other roots of unity in problems of polynomial divisibility, but we need some results similar to those shown above.

Rule 3 If a polynomial satisfies:

for distinct numbers , then is divisible by the product .

Rule 4 is analogous to rule 3: they are repeated applications of the factor theorem.

Corollary 3 If a polynomial with integer coefficient satisfies:

where

then is divisible by .

To prove this corollary, we have to use the result that conjugate of non-real roots of unity are also non-real roots of unity.

Using unit roots in problems of divisibility of polynomials

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Example 2 Let , where and are integers. Prove that is divisible by .
Proof Put , then:

,

By corollary 2, is divisible by .

Example 3 Let be a natural number, and:

.

Prove that for any integer , is a multiple of .
Proof It is easy to show that and is a polynomial with integer coefficient. Therefore, is divisible by . As both and are polynomials with integer coefficients, so is the quotient, say . Therefore, when is an integer, is an integral multiple of .

Example 4 Find the remainder from the division of by .
Solution Write and . Let the remainder be , then:

,

where represents the quotient. Since the degree of the divisor is 4, the degree of the remainder is at most three. Therefore, we let:

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Now we are going to determine the coefficients of .
From:

,

we know that:

,

So:

Compare the real parts and imaginary parts of both sides of these equations:

Solving, . So, the remainder is .

Example 5 Let , , and be polynomials satisfying the identity:

.

Prove that is a factor of .
Proof Let:

,
.

Note that only has -terms, -terms and -terms for non-negative integers . Therefore, coefficients of -terms are all zero. Then:

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On the other hand, by comparing the coefficients of -terms:

,

Therefore, .
Now, we will prove . First, note that:

So,

for any .

In other words, . Observe that is a polynomial and therefore it has a finite degree. will result in a contradiction to this. Therefore, .
So:

. QED

Alternative proof Let be a complex fifth root of unity, then:

, , .

Put into the given identity:

.

We can obtain the following by first putting and then comparing the real parts and the imaginary parts respectively:

,
,
,
,

We can solve these equation easily:

.

From factor theorem, is a factor of . QED

The alternative proof makes use of the properties of unit roots. By doing so, we can also conclude that is a factor of and , which implies that also contains the factor .

In the first proof, we only requires that the coefficients of -terms be zero. So, we can add one more term and the statement still holds:

Example 6 Let , , , and be polynomials satisfying the identity:

.

Prove that is a factor of .
Proof Repeat the first proof of the previous example.
Alternative proof Repeat the alternative proof of the previous example: putting and comparing the real parts and imaginary parts. By this method, we can also conclude that all the polynomials , , , and have the factor .