Question

Consider the ring Q[x]/(x^3 − 2x).

(a) Prove that the set Ω := {a + bx + cx^2 : a, b, c ∈ Q} contains exactly one element of each coset of (x^3 − 2x) in Q[x].

(b) Show how to add and multiply elements in Ω. Make sure to reduce the answers so that the answers are elements in Ω.

(c) Find a zero divisor in Q[x]/(x^3 − 2x).

(d) Describe all the zero divisors in Q[x]/(x^3 − 2x). So for f(x) ∈ Q[x], f(x) + (x^3 − 2x) is a zero divisor if and only if . . . (something about f(x))

Answer 1

a) R = BBA/ <a?_2x) b Any element of R is of the form for) + <0x3_ 2017 row By divison algorithm we get fix) = 8 (W) (03-2x) +(«) with deg 360) < deg (2²-23) ise, deg 860) <3 . so, fra) + <223- 2x) = Q() (83-23) +2() + <213-220) = Y(u) + <903-22 ) with deg 003 < 3 Therefore, any element of R looks like a+bx+cx2 + <x²–2x) i aboce a Therefore, 2 =şa+bx+cx2. a,b,ctat contains exactly one element of each Coset of (x² – 2x ) in a (G] ut for), g(x) ER fon) + g(x) defined by f(x) = atbrtch? g(x) = e+ fx+ gut fr) + gen) (are) + (b+f) 2 + (C+9)#4 row f(x) g(x) = (atbrtc22) (e+fr+gne 2) - ae + (be taflx + (ce tbf tag) x² + (cf +69) x3 + cg21 now, in R. x3 = 2 x² – 2x =0 = x3 = 2x = x 4 = 2x2 - fon) gox) = ae + (be+af) x + cce +bf +ag)x? + Ccf +bg) (2x) + cg .222 =ae + cbe +af +206 +2bg) +(ce+bftag+ 20 g) x2 now at <x3_22) to in R. & 2. 2+423_2x) toine but (x + <23 an>) (x²_2 + (x² – 2x>) = x (x² – 20)+2x322) = (x3 - 2x)+(x²-227 00 (c)

therefore, at 4x² - 22 d) let, fear ore, x+<x3_za) is a zero divisor inR. let, fox) be a zero divisor inR.. Therefore, F9x) +<*3-22) such that (for) + <23 - 2x>) () +<33_201>) = OR 7 fen) f(x) E <3-20) . f(x) is a divisor of x²–2x. so all zero divisors are ... 2 + 3323-2), 22-2 + (x32), and powers of and scaler multiples of these two. so, fox) is a zero divisor iff foxt is power or scales multiple of these two.