Let \(f(x)=x^2+1\in\mathbb{F}_3[x]\), where \(\mathbb{F}_3=\{0,1,2\}\) is the field of three elements. Let \(I=\langle f\rangle\) be the principal ideal in \(\mathbb{F}_3[x]\) generated by \(f\). (a) Prove that \(f\) is irreducible in \(\mathbb{F}_3[x]\). (b) Prove that the quotient ring \(\mathbb{F}_3[x]/I\) is a field. (c) Find the multiplicative inverse of \(I+x\) in the quotient ring \(\mathbb{F}_3[x]/I\).
Solution
Proof. (a) Since deg f = 2, f is reducible if and only if it has no roots in F3. Checking shows: f(0) = 1⁄= 0, f (1) = 2⁄= 0, f (2) = 2⁄= 0. Since f(a)⁄= 0 for all a∈ F3, it follows that f is irreducible. (b) Since f is irreducible, the ideal I =⟨f⟩ is maximal in F3[x], by a Math 350 theorem. So by another Math 350 theorem, the quotient F3[x]/I is a field. (c) We need a polynomial g∈ F3[x] such that (I +x)(I +g) =I + 1. However, (I +x)(I +g) =I + 1 ⇐⇒ I +xg =I + 1 ⇐⇒ xg− 1∈I ⇐⇒ xg + 2∈I, where the last equivalence uses −1 = 2 in F3. Choosing g(x) = 2x gives xg + 2 =x(2x) + 2 = 2(x2 + 1)∈I. Thus, by the above equivalences, and by commutativity, we have (I + 2x)(I +x) = (I +x)(I + 2x) =I + 1 = 1F3[x]/I, and hence I + 2x is the multiplicative inverse of I +x in F3[x]/I. QED