(March 2015) Let \(R=\mathbb{Z}[x]\) be the ring of polynomials (in one variable) with integer coefficients. Note that the constant polynomial \(6\) and the degree one polynomial \(x+1\) are both elements of \(R\). Define \[ I=\{6f+(x+1)g\mid f,g\in R\}. \] Prove that \(I\) is an ideal of \(R\).
Solution
Proof. (Nonempty) Let f =g = 0∈R. Then 6f + (x + 1)g∈I, so I⁄= ∅. (Closed under−) Givena,b∈I, writea = 6f1+(x+1)g1 andb = 6f2+(x+1)g2 withf1,f 2,g 1,g 2∈R. Then a−b = 6f1 + (x + 1)g1− ( 6f2 + (x + 1)g2 ) = 6(f1−f2) + (x + 1)(g1−g2)∈I. (Sticky) Given a∈I and h∈R, write a = 6f + (x + 1)g with f,g ∈R. Then ah =ha =h ( 6f + (x + 1)g ) = 6(hf) + (x + 1)(hg)∈I QED
10 Quotient Rings Recall that an ideal I⊆R is a subgroup under addition, so the cosets of I are usually written I +r ={s +r|s∈I}, although sometimes you see r +I since addition is commutative. The definition of ideal guarantees that the set of cosets R/I ={I +r|r∈R} becomes a ring, called the quotient ring, under the following operations: (I +a) + (I +b) =I + (a +b) and ( I +a)(I +b) =I +ab.