Cyclotomic Polynomial

In mathematics, the ''n''th cyclotomic polynomial, for any positive integer ''n'', is the unique irreducible polynomial with integer coefficients that is a divisor of $x^n-1$ and is not a divisor of $x^k-1$ for any Its roots are all ''n''th primitive roots of unity
$e^$, where ''k'' runs over the positive integers not greater than ''n'' and coprime to ''n'' (and ''i'' is the imaginary unit). In other words, the ''n''th cyclotomic polynomial is equal to
:$\Phi_n(x) = \prod_\stackrel \left(x-e^\right).$

It may also be defined as the monic polynomial with integer coefficients that is the minimal polynomial over the field of the rational numbers of any primitive ''n''th-root of unity ($e^$ is an example of such a root).

An important relation linking cyclotomic polynomials and primitive roots of unity is
:$\prod_\Phi_d(x) = x^n - 1,$
showing that is a root of $x^n - 1$ if and only if it is a ''d''th primitive root of unity for some ''d'' that divides ''n''.

Examples

If ''n'' is a prime number, then
:$\Phi_n(x) = 1+x+x^2+\cdots+x^=\sum_^ x^k.$
If ''n'' = 2''p'' where ''p'' is an odd prime number, then
:$\Phi_(x) = 1-x+x^2-\cdots+x^=\sum_^ (-x)^k.$

For ''n'' up to 30, the cyclotomic polynomials are:

:$\begin \Phi_1(x) &= x - 1 \\ \Phi_2(x) &= x + 1 \\ \Phi_3(x) &= x^2 + x + 1 \\ \Phi_4(x) &= x^2 + 1 \\ \Phi_5(x) &= x^4 + x^3 + x^2 + x +1 \\ \Phi_6(x) &= x^2 - x + 1 \\ \Phi_7(x) &= x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \\ \Phi_8(x) &= x^4 + 1 \\ \Phi_9(x) &= x^6 + x^3 + 1 \\ \Phi_(x) &= x^4 - x^3 + x^2 - x + 1 \\ \Phi_(x) &= x^ + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \\ \Phi_(x) &= x^4 - x^2 + 1 \\ \Phi_(x) &= x^ + x^ + x^ + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \\ \Phi_(x) &= x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 \\ \Phi_(x) &= x^8 - x^7 + x^5 - x^4 + x^3 - x + 1 \\ \Phi_(x) &= x^8 + 1 \\ \Phi_(x) &= x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1\\ \Phi_(x) &= x^6 - x^3 + 1 \\ \Phi_(x) &= x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1\\ \Phi_(x) &= x^8 - x^6 + x^4 - x^2 + 1 \\ \Phi_(x) &= x^ - x^ + x^9 - x^8 + x^6 - x^4 + x^3 - x + 1 \\ \Phi_(x) &= x^ - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 \\ \Phi_(x) &= x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ \\ & \qquad\quad + x^ + x^ + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \\ \Phi_(x) &= x^8 - x^4 + 1 \\ \Phi_(x) &= x^ + x^ + x^ + x^5 + 1 \\ \Phi_(x) &= x^ - x^ + x^ - x^9 + x^8 - x^7 + x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 \\ \Phi_(x) &= x^ + x^9 + 1 \\ \Phi_(x) &= x^ - x^ + x^8 - x^6 + x^4 - x^2 + 1 \\ \Phi_(x) &= x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ + x^ \\ & \qquad\quad + x^ + x^ + x^ + x^ + x^ + x^9 + x^8 + x^7 + x^6 + x^5 + x^4 + x^3 + x^2 + x + 1 \\ \Phi_(x) &= x^8 + x^7 - x^5 - x^4 - x^3 + x + 1. \end$

The case of the 105th cyclotomic polynomial is interesting because 105 is the least positive integer that is the product of three distinct odd prime numbers (3*5*7) and this polynomial is the first one that has a coefficient other than 1, 0, or −1:

:$\begin \Phi_(x) &= x^ + x^ + x^ - x^ - x^ - 2 x^ - x^ - x^ + x^ + x^ + x^ + x^ + x^ + x^ - x^ - x^ \\ &\qquad\quad - x^ - x^ - x^ + x^ + x^ + x^ + x^ + x^ + x^ - x^9 - x^8 - 2 x^7 - x^6 - x^5 + x^2 + x + 1. \end$

Properties

Fundamental tools

The cyclotomic polynomials are monic polynomials with integer coefficients that are irreducible over the field of the rational numbers. Except for ''n'' equal to 1 or 2, they are palindromics of even degree.

The degree of $\Phi_n$, or in other words the number of ''n''th primitive roots of unity, is $\varphi (n)$, where $\varphi$ is Euler's totient function.

The fact that $\Phi_n$ is an irreducible polynomial of degree $\varphi (n)$ in the ring \Z /math> is a nontrivial result due to Gauss. Depending on the chosen definition, it is either the value of the degree or the irreducibility which is a nontrivial result. The case of prime ''n'' is easier to prove than the general case, thanks to Eisenstein's criterion.

A fundamental relation involving cyclotomic polynomials is

:$\begin x^n - 1 &=\prod_ \left(x- e^ \right) \\ &= \prod_ \prod_ \left(x- e^ \right) \\ &=\prod_ \Phi_(x) = \prod_ \Phi_d(x).\end$

which means that each ''n''-th root of unity is a primitive ''d''-th root of unity for a unique ''d'' dividing ''n''.

The Möbius inversion formula allows the expression of $\Phi_n(x)$ as an explicit rational fraction:

:$\Phi_n(x)=\prod_(x^d-1)^,$

where $\mu$ is the Möbius function.

The cyclotomic polynomial $\Phi_(x)$ may be computed by (exactly) dividing $x^n-1$ by the cyclotomic polynomials of the proper divisors of ''n'' previously computed recursively by the same method:

:$\Phi_n(x)=\frac$

(Recall that $\Phi_(x)=x-1$.)

This formula defines an algorithm for computing $\Phi_n(x)$ for any ''n'', provided integer factorization and division of polynomials are available. Many computer algebra systems, such as SageMath, Maple, Mathematica, and PARI/GP, have a built-in function to compute the cyclotomic polynomials.

Easy cases for computation

As noted above, if is a prime number, then

:$\Phi_n(x) = 1+x+x^2+\cdots+x^=\sum_^x^k.$

If ''n'' is an odd integer greater than one, then

:$\Phi_(x) = \Phi_n(-x).$

In particular, if is twice an odd prime, then (as noted above)

:$\Phi_n(x) = 1-x+x^2-\cdots+x^=\sum_^(-x)^k.$

If is a prime power (where ''p'' is prime), then

:$\Phi_n(x) = \Phi_p(x^) =\sum_^x^.$

More generally, if with relatively prime to , then

:$\Phi_n(x) = \Phi_(x^).$

These formulas may be applied repeatedly to get a simple expression for any cyclotomic polynomial $\Phi_n(x)$ in term of a cyclotomic polynomial of square free index: If is the product of the prime divisors of (its radical), then

:$\Phi_n(x) = \Phi_q(x^).$

This allows to give formulas for the th cyclotomic polynomial when has at most one odd prime factor: If is an odd prime number, and and are positive integers, then:
:$\Phi_(x) = x^+1$
:$\Phi_(x) = \sum_^x^$
:$\Phi_(x) = \sum_^(-1)^jx^$

For the other values of , the computation of the th cyclotomic polynomial is similarly reduced to that of $\Phi_q(x),$ where is the product of the distinct odd prime divisors of . To deal with this case, one has that, for prime and not dividing ,

:$\Phi_(x)=\Phi_(x^p)/\Phi_n(x).$

Integers appearing as coefficients

The problem of bounding the magnitude of the coefficients of the cyclotomic polynomials has been the object of a number of research papers.

If ''n'' has at most two distinct odd prime factors, then Migotti showed that the coefficients of $\Phi_n$ are all in the set .

The first cyclotomic polynomial for a product of three different odd prime factors is $\Phi_(x);$ it has a coefficient −2 (see its expression above). The converse is not true: $\Phi_(x)=\Phi_(x)$ only has coefficients in .

If ''n'' is a product of more different odd prime factors, the coefficients may increase to very high values. E.g., $\Phi_(x) =\Phi_(x)$ has coefficients running from −22 to 23, $\Phi_(x)=\Phi_(x)$, the smallest ''n'' with 6 different odd primes, has coefficients of magnitude up to 532.

Let ''A''(''n'') denote the maximum absolute value of the coefficients of Φ_''n''. It is known that for any positive ''k'', the number of ''n'' up to ''x'' with ''A''(''n'') > ''n''^''k'' is at least ''c''(''k'')⋅''x'' for a positive ''c''(''k'') depending on ''k'' and ''x'' sufficiently large. In the opposite direction, for any function ψ(''n'') tending to infinity with ''n'' we have ''A''(''n'') bounded above by ''n''^ψ(''n'') for almost all ''n''.Maier (2008)

Gauss's formula

Let ''n'' be odd, square-free, and greater than 3. Then:

:$4\Phi_n(z) = A_n^2(z) - (-1)^nz^2B_n^2(z)$

where both ''A_n''(''z'') and ''B_n''(''z'') have integer coefficients, ''A_n''(''z'') has degree ''φ''(''n'')/2, and ''B_n''(''z'') has degree ''φ''(''n'')/2 − 2. Furthermore, ''A_n''(''z'') is palindromic when its degree is even; if its degree is odd it is antipalindromic. Similarly, ''B_n''(''z'') is palindromic unless ''n'' is composite and ≡ 3 (mod 4), in which case it is antipalindromic.

The first few cases are

:\begin
4\Phi_5(z) &=4(z^4+z^3+z^2+z+1)\\
&= (2z^2+z+2)^2 - 5z^2 \\ pt4\Phi_7(z) &=4(z^6+z^5+z^4+z^3+z^2+z+1)\\
&= (2z^3+z^2-z-2)^2+7z^2(z+1)^2 \\ pt4\Phi_(z)
&=4(z^+z^9+z^8+z^7+z^6+z^5+z^4+z^3+z^2+z+1)\\
&= (2z^5+z^4-2z^3+2z^2-z-2)^2+11z^2(z^3+1)^2
\end

Lucas's formula

Let ''n'' be odd, square-free and greater than 3. Then

:$\Phi_n(z) = U_n^2(z) - (-1)^nzV_n^2(z)$

where both ''U_n''(''z'') and ''V_n''(''z'') have integer coefficients, ''U_n''(''z'') has degree ''φ''(''n'')/2, and ''V_n''(''z'') has degree ''φ''(''n'')/2 − 1. This can also be written

:$\Phi_n \left ((-1)^z \right ) = C_n^2(z) - nzD_n^2(z).$

If ''n'' is even, square-free and greater than 2 (this forces ''n''/2 to be odd),

:$\Phi_ \left (-z^2 \right ) = \Phi_(z)= C_n^2(z) - nzD_n^2(z)$

where both ''C_n''(''z'') and ''D_n''(''z'') have integer coefficients, ''C_n''(''z'') has degree ''φ''(''n''), and ''D_n''(''z'') has degree ''φ''(''n'') − 1. ''C_n''(''z'') and ''D_n''(''z'') are both palindromic.

The first few cases are:

:\begin
\Phi_3(-z) &=\Phi_6(z) =z^2-z+1 \\
&= (z+1)^2 - 3z \\ pt\Phi_5(z) &=z^4+z^3+z^2+z+1 \\
&= (z^2+3z+1)^2 - 5z(z+1)^2 \\ pt\Phi_(-z^2) &=\Phi_(z)=z^4-z^2+1 \\
&= (z^2+3z+1)^2 - 6z(z+1)^2
\end

Cyclotomic polynomials over a finite field and over the -adic integers

Over a finite field with a prime number of elements, for any integer that is not a multiple of , the cyclotomic polynomial $\Phi_n$ factorizes into $\frac$ irreducible polynomials of degree , where $\varphi (n)$ is Euler's totient function and is the multiplicative order of modulo . In particular, $\Phi_n$ is irreducible if and only if is a primitive root modulo , that is, does not divide , and its multiplicative order modulo is $\varphi(n)$, the degree of $\Phi_n$.

These results are also true over the -adic integers, since Hensel's lemma allows lifting a factorization over the field with elements to a factorization over the -adic integers.

Polynomial values

If takes any real value, then $\Phi_n(x)>0$ for every (this follows from the fact that the roots of a cyclotomic polynomial are all non-real, for ).

For studying the values that a cyclotomic polynomial may take when is given an integer value, it suffices to consider only the case , as the cases and are trivial (one has $\Phi_1(x)=x-1$ and $\Phi_2(x)=x+1$).

For , one has

:$\Phi_n(0) =1,$
:$\Phi_n(1) =1$ if is not a prime power,
:$\Phi_n(1) =p$ if $n=p^k$ is a prime power with .

The values that a cyclotomic polynomial $\Phi_n(x)$ may take for other integer values of is strongly related with the multiplicative order modulo a prime number.

More precisely, given a prime number and an integer coprime with , the multiplicative order of modulo , is the smallest positive integer such that is a divisor of $b^n-1.$ For , the multiplicative order of modulo is also the shortest period of the representation of in the numeral base (see Unique prime; this explains the notation choice).

The definition of the multiplicative order implies that, if is the multiplicative order of modulo , then is a divisor of $\Phi_n(b).$ The converse is not true, but one has the following.

If is a positive integer and is an integer, then (see below for a proof)
:$\Phi_n(b)=2^kgh,$
where
* is a non-negative integer, always equal to 0 when is even. (In fact, if is neither 1 nor 2, then is either 0 or 1. Besides, if is not a power of 2, then is always equal to 0)
* is 1 or the largest odd prime factor of .
* is odd, coprime with , and its prime factors are exactly the odd primes such that is the multiplicative order of modulo .

This implies that, if is an odd prime divisor of $\Phi_n(b),$ then either is a divisor of or is a divisor of . In the latter case, $p^2$ does not divide $\Phi_n(b).$

Zsigmondy's theorem implies that the only cases where and are

:$\begin \Phi_1(2) &=1 \\ \Phi_2 \left (2^k-1 \right ) & =2^k && k >0 \\ \Phi_6(2) &=3 \end$

It follows from above factorization that the odd prime factors of

:$\frac$

are exactly the odd primes such that is the multiplicative order of modulo . This fraction may be even only when is odd. In this case, the multiplicative order of modulo is always .

There are many pairs with such that $\Phi_n(b)$ is prime. In fact, Bunyakovsky conjecture implies that, for every , there are infinitely many such that $\Phi_n(b)$ is prime. See for the list of the smallest such that $\Phi_n(b)$ is prime (the smallest such that $\Phi_n(b)$ is prime is about $\lambda \cdot \varphi(n)$, where $\lambda$ is Euler–Mascheroni constant, and $\varphi$ is Euler's totient function). See also for the list of the smallest primes of the form $\Phi_n(b)$ with and , and, more generally, , for the smallest positive integers of this form.

* ''Values of'' $\Phi_n(1).$ If $n=p^$ is a prime power, then
::$\Phi_n(x)=1+x^+x^+\cdots+x^ \qquad \text \qquad \Phi_n(1)=p.$
:If is not a prime power, let $P(x)=1+x+\cdots+x^,$ we have $P(1)=n,$ and is the product of the $\Phi_k(x)$ for dividing and different of . If is a prime divisor of multiplicity in , then $\Phi_p(x), \Phi_(x), \cdots, \Phi_(x)$ divide , and their values at are factors equal to of $n=P(1).$ As is the multiplicity of in , cannot divide the value at of the other factors of $P(x).$ Thus there is no prime that divides $\Phi_n(1).$

*''If'' ''is the multiplicative order of'' ''modulo'' , ''then'' $p \mid \Phi_n(b).$ By definition, $p \mid b^n-1.$ If $p \nmid \Phi_n(b),$ then would divide another factor $\Phi_k(b)$ of $b^n-1,$ and would thus divide $b^k-1,$ showing that, if there would be the case, would not be the multiplicative order of modulo .

*''The other prime divisors of'' $\Phi_n(b)$ ''are divisors of'' . Let be a prime divisor of $\Phi_n(b)$ such that is not be the multiplicative order of modulo . If is the multiplicative order of modulo , then divides both $\Phi_n(b)$ and $\Phi_k(b).$ The resultant of $\Phi_n(x)$ and $\Phi_k(x)$ may be written $P\Phi_k+Q\Phi_n,$ where and are polynomials. Thus divides this resultant. As divides , and the resultant of two polynomials divides the discriminant of any common multiple of these polynomials, divides also the discriminant $n^n$ of $x^n-1.$ Thus divides .

* ''and'' ''are coprime''. In other words, if is a prime common divisor of and $\Phi_n(b),$ then is not the multiplicative order of modulo . By Fermat's little theorem, the multiplicative order of is a divisor of , and thus smaller than .

* ''is square-free''. In other words, if is a prime common divisor of and $\Phi_n(b),$ then $p^2$ does not divide $\Phi_n(b).$ Let . It suffices to prove that $p^2$ does not divide for some polynomial , which is a multiple of $\Phi_n(x).$ We take
::$S(x)=\frac = 1 + x^m + x^ + \cdots + x^.$
:The multiplicative order of modulo divides , which is a divisor of . Thus is a multiple of . Now,
::$S(b) = \frac = p+ \binomc + \cdots + \binomc^.$
:As is prime and greater than 2, all the terms but the first one are multiples of $p^2.$ This proves that $p^2 \nmid \Phi_n(b).$

Applications

Using $\Phi_n$, one can give an elementary proof for the infinitude of primes congruent to 1 modulo ''n'',S. Shirali. ''Number Theory''. Orient Blackswan, 2004. p. 67. which is a special case of Dirichlet's theorem on arithmetic progressions.

Suppose $p_1, p_2, \ldots, p_k$ is a finite list of primes congruent to $1$ modulo $n.$ Let $N = np_1p_2\cdots p_k$ and consider $\Phi_n(N)$. Let $q$ be a prime factor of $\Phi_n(N)$ (to see that $\Phi_n(N) \neq \pm 1$ decompose it into linear factors and note that 1 is the closest root of unity to $N$). Since $\Phi_n(x) \equiv \pm 1 \pmod x,$ we know that $q$ is a new prime not in the list. We will show that $q \equiv 1 \pmod n.$

Let $m$ be the order of $N$ modulo $q.$ Since $\Phi_n(N) \mid N^n - 1$ we have $N^n -1 \equiv 0 \pmod$. Thus $m \mid n$. We will show that $m = n$.

Assume for contradiction that $m < n$. Since

:$\prod_ \Phi_d(N) = N^m - 1 \equiv 0 \pmod q$

we have

:$\Phi_d(N) \equiv 0 \pmod q,$

for some $d < n$. Then $N$ is a double root of

:$\prod_ \Phi_d(x) \equiv x^n -1 \pmod q.$

Thus $N$ must be a root of the derivative so

:$\left.\frac\_N \equiv nN^ \equiv 0 \pmod q.$

But $q \nmid N$ and therefore $q \nmid n.$ This is a contradiction so $m = n$. The order of $N \pmod q,$ which is $n$, must divide $q-1$. Thus $q \equiv 1 \pmod n.$

See also

* Cyclotomic field
* Aurifeuillean factorization
* Root of unity

Notes

References

Gauss's book ''Disquisitiones Arithmeticae'' has been translated from Latin into English and German. The German edition includes all of his papers on number theory: all the proofs of quadratic reciprocity, the determination of the sign of the Gauss sum, the investigations into biquadratic reciprocity, and unpublished notes.
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External links

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*{{OEIS el, sequencenumber=A013594, name=Smallest order of cyclotomic polynomial containing n or −n as a coefficient

Polynomials
Algebra
Number theory