## Basic properties of the reverse of a polynomial

Let be a field. Given a polynomial of degree , the *reverse* of is defined to be .

- Describe the coefficients of the reverse of in terms of those of .
- If has a nonzero constant coefficient, prove that is irreducible if and only if its reverse is irreducible.

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

- Note that we can imagine computations involving as taking place in the ring of all Laurent series over , so that arithmetic with negative powers of behaves as expected. Now if , then ; thus . That is, the coefficients of the reverse of are those of in reverse order. (Hence the name.)
- Let denote the reverse of , and let have degree and respectively such that both and have nonzero constant coefficients. Then . That is, the reverse of a product is the product of reverses.
Note that cannot increase degrees; that . Moreover, this inequality is strict if and only if the constant coefficient of is zero. If The constant coefficient of is nonzero, then and have the same degree, so that .

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

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

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

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## Comments

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.

Thank you!

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!

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.