(January 1999) Let \(k\) be a field. Show that a cubic polynomial \(f(x)\in k[x]\) is irreducible in \(k[x]\) if and only if \(f(x)\) has no roots in \(k\).
Solution
Proof. We’ll prove f reducible iff f has a root in k. (=⇒): Since f is reducible, there exist g,h∈k[x] with f =gh and with degg, degh< 3. Since deg g and degh are nonnegative integers summing to deg f = 3, one of g and h must have degree 1 and the other must have degree 2. Without loss, deg g = 1. That is, g(x) =ax +b for some a,b∈k with a⁄= 0. Then −b/a∈k, and f ( − b a ) =g ( − b a ) h ( − b a ) = [ a ( − b a ) +b] h ( − b a ) = (−b +b)h ( − b a ) = 0· ( − b a ) = 0, and hence−b/a is a root of of f in k. (⇐=): Let c∈k be a root of f, and let g(x) = x−c∈k[x]. By the division algorithm for f and g, there exist polynomials q,r ∈k[x] with f =qg +r and with either r = 0 or degr< degg = 1. If r⁄= 0 but degr< 1, then degr = 0, which means that r =b is a nonzero constant. Hence, 0 =f(c) =q(c)g(c) +r(c) =q(c)· (c−c) +b = 0 +b =b⁄= 0, a contradiction. Thus, we must in fact have r = 0, and hence f =qg. Since deg f = 3 and degg = 1, we must have deg q = 3− 1 = 2, and hence we have factored f as a product of two lower-degree polynomials g and q, proving that f is reducible. QED