Reason/proof for a particular method of factoring cyclic polynomials

Question

My book says that if a-b is a factor of a cyclic polynomial P(a,b,c), then b-c and c-a will also be factors. I have found this statement on the internet, but not it's proof or intuition. Also, I know that if I find that (a-b)(b-c)(c-a) is in fact a factor of the polynomial I'm trying to factorize, and the polynomial has a degree of 5, then the remaining factor must be k(a^2 + b^2 + c^2 ) + m(ab+bc+ca). But again, why is it so?

Answer 1

If $Q(x,y,z)$ is a factor of $P(x,y,z)$, then by substitution, $Q(y,z,x)$ is clearly a factor of $P(y,z,x)$. If $P$ is cyclic, we have $P(x,y,z)=P(y,z,x)$. Hence if $x-y$ is a factor of cyclic $P(x,y,z)$, then $y-z$ is also a factor, and so is $z-x$. consequently, the least common multiple of $x-y$, $y-z$, and $z-y$ must be a factor, and for such simple polynomials, the least common multiple is of course their product $(x-y)(y-z)(z-x)$.

As $(x-y)(y-z)(z-x)$ has degree $3$, factoring it out from a degree $5$ polynomial leaves you with a degree $2$ polynomial. There cannot be another factor $(x-y)(y-z)(z-x)$ in it as that would exceed the degree. Nor can there be any other irreducible factor $Q(x,y,z)$ of degree $\ge1$ in it that would lead to a factor $Q(x,y,z)Q(y,z,x)Q(z,x,y)$ of degree $\ge 3$ -- unless $Q(x,y,z)$ is itself cyclic in whic case the least common multiple of $Q(x,y,z)$, $Q(y,z,x)$, $Q(z,x,y)$ is not the procduct but already $Q(x,y,z)$ itself So $Q$ is cyclic of degree $\le 2$, which means that as monomials we can have

• a constant term $\alpha$ (which you left out)
• If we have $\beta x$, we need also $\beta y$ and $\beta z$, hence the degree one part of $Q$ must be of the form $\beta(x+y+z)$
• If we have $\gamma x^2$, we need $\gamma y^2$ and $\gamma z^2$ ba the same argument
• If we have $\delta xy$, we need $\delta yz$ and $\delta zx$ by the same argument.

(Alternatively, write down the general polynomial of degree $\le 2$ in three variables, $$\alpha+\beta_1x+\beta_2y+\beta_3z+\gamma_1x^2+\gamma_2y^2+\gamma_3z^2+\delta_1yz+\delta_2xz+\delta_3xy,$$ and evaluate the equations in the coefficients that cyclic symmetry imposes, namely $\beta_1=\beta_2=\beta_3$ etc.)

Hence the only cyclic polynomials of degree $<3$ are those of the form $$\alpha+\beta(x+y+z)+\gamma(x^2+y^2+z^2)+\delta(xy+yz+zx). $$ Contrary to what your question suggests, $\alpha$ and $\beta$ need not be zero.