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.)