## Basic properties of the reverse of a polynomial

Let $F$ be a field. Given a polynomial $p(x) \in F[x]$ of degree $n$, the reverse of $p$ is defined to be $x^np(\frac{1}{x})$.

1. Describe the coefficients of the reverse of $p(x)$ in terms of those of $p(x)$.
2. If $p$ has a nonzero constant coefficient, prove that $p(x)$ is irreducible if and only if its reverse is irreducible.

[Thanks to Marc van Leeuwen for pointing out a flaw in the original solution.]

1. Note that we can imagine computations involving $p(\frac{1}{x})$ as taking place in the ring of all Laurent series over $F$, so that arithmetic with negative powers of $x$ behaves as expected. Now if $p(x) = \sum_{i=0}^n a_i x^i$, then $x^np(x^{-1}) = \sum_{i=0}^n a_i x^{n-i}$; thus $x^np(x^{-1}) = \sum_{j=0}^n a_{n-j}x^j$. That is, the coefficients of the reverse of $p(x)$ are those of $p(x)$ in reverse order. (Hence the name.)
2. Let $\rho(p)$ denote the reverse of $p$, and let $p,q \in F[x]$ have degree $n$ and $m$ respectively such that both $a$ and $b$ have nonzero constant coefficients. Then $\rho(pq)(x) = x^{m+n}(pq)(x^{-1})$ $= x^mp(x^{-1}) x^n q(x^{-1})$ $= \rho(p)(x) \rho(q)(x)$. That is, the reverse of a product is the product of reverses.

Note that $\rho$ cannot increase degrees; that $\mathsf{deg}\ \rho(p) \leq \mathsf{deg}\ p$. Moreover, this inequality is strict if and only if the constant coefficient of $p$ is zero. If The constant coefficient of $p$ is nonzero, then $\rho(p)$ and $p$ have the same degree, so that $\rho(\rho(p)) = p$.

(If $R$ is an integral domain, then the subset of $R[x]$ consisting of those polynomials having nonzero constant coefficient is closed under multiplication (thus is a semigroup). The previous two paragraphs show that $\rho$ is an involution on this semigroup.)

Let $p$ be a polynomial having a nonzero constant coefficient. Suppose $p(x)$ is irreducible. If $\rho(p)$ is reducible, say $\rho(p) = a(x)b(x)$ where $a$ and $b$ are nonconstant, then (since the constant coefficient of $\rho(p)$ is nonzero and $R$ is a domain) the constant coefficients of $a$ and $b$ are nonzero. So $\rho(a)$ and $\rho(b)$ are nonconstant polynomials with nonzero constant coefficient with $p = \rho(a)\rho(b)$, a contradiction. So if $p$ is irreducible, then $\rho(p)$ is irreducible. Since $\rho(\rho(p)) = p$, we have that (for all $p(x)$ with nonzero constant term) $p(x)$ is irreducible if and only if $\rho(p)(x)$ is irreducible.

(Note to self: this probably generalizes to any subsemigroup that contains the units and has an involution.)

• Marc van Leeuwen  On March 21, 2011 at 7:41 am

The second statement is actually false, as exemplified by the reducible polynomial X^2+X, whose reverse X+1 is irreducible. The statement should mention that constant coefficients are assumed to be non-zero. People always forget about invertibles which are neither reducible nor irreducible. Also, there is no reason to restrict to polynomials over a field; with the indicated correction an integral domain (just so that divisibility is not to unusual a notion) will do. By the way, the proofs would work quite well in F[x,x^{-1}], where substitution of x^{-1} for x is an involutive automorphism.

• nbloomf  On March 21, 2011 at 7:49 am

Thank you!

• nbloomf  On March 21, 2011 at 8:14 am

I was going to take a few minutes to fix this, but on further reflection the details are more intricate than I realized. I will mark this solution “incomplete” and come back to it later today.

Thanks again!

• nbloomf  On March 22, 2011 at 10:21 am

Okay; I think it is fixed now. The problem was that I (erroneously) assumed that a polynomial and its reverse always have the same degree, so that the square of the reversion map is the identity. This is only true on polynomials whose constant term is nonzero.