(January 2009) (a) Prove that \(f(X)=X^4+X^2+1\) is reducible in the polynomial ring \(\mathbb{F}_2[X]\), where \(\mathbb{F}_2=\{0,1\}\) is the field of two elements. (b) Prove that \(g(X)=X^3+2X^2+2X+3\) is irreducible in the polynomial ring \(\mathbb{F}_5[X]\), where \(\mathbb{F}_5=\{0,1,2,3,4\}\) is the field of five elements.
Solution
Proof. (a) To be reducible,f must either have a degree one factor (and hence have a root in F2), orf must be a product of two irreducible factors of degree two. But f(0) =f(1) = 1 so f has no roots in F2. Hence f is the product of two irreducible quadratic factors. The possible quadratic polynomials are X 2, X 2 + 1, X 2 +X and X 2 +X + 1. The first and third are clearly reducible, and over F2, we have X 2 + 1 = (X + 1)2, so X 2 +X + 1 is the only quadratic irreducible polynomial. Then an explicit calculation (made easier by using (a +b)2 =a2 +b2 since we are working over F2) shows that f(X) = (X 2 +X + 1)(X 2 +X + 1). QED (b) Since degg = 3, we know that g is reducible if and only if it has no roots. Checking shows: g(0) = 03 + 2(02) + 2(0) + 3 = 0 + 0 + 0 + 3 = 3⁄= 0 g(1) = 13 + 2(12) + 2(1) + 3 = 1 + 2 + 2 + 3 = 3⁄= 0 g(2) = 23 + 2(22) + 2(2) + 3 = 3 + 3 + 4 + 3 = 3⁄= 0 g(3) = 33 + 2(32) + 2(3) + 3 = 2 + 3 + 1 + 3 = 4⁄= 0 g(4) = 43 + 2(42) + 2(4) + 3 = 4 + 2 + 3 + 3 = 2⁄= 0 Since g(a)⁄= 0 for all a∈ F5, it follows that g is irreducible. QED