Question

If U, V are ideals of R, let U+V = {u+v | u in U, v in V }. Prove that UV is an ideal of R​

Answer 1

Answer:

10. If U is an ideal of R let [R : U] = {x ∈ R | rx ∈ U for every r ∈ R}. Prove

that [R : U] is an ideal of R and that it contains U.

Solution: Suppose some x, y ∈ [R : U], therefore rx ∈ U

∀ r ∈ R and

ry ∈ U

∀ r ∈ R. But since U being an ideal, we have rx xry = r(x

y) ∈

U

∀ r ∈ R, showing x x y ∈ [R : U]. Thus [R : U] is a subgroup of R under

addition. Next suppose x ∈ [R : U] and r1 ∈ R. Therefore rx ∈ U

∀ r ∈ R.

Also r(xr1) = (rx)r1 = u1r1 for some u1 ∈ U. But U being an ideal, there-fore u1r1 ∈ U. So r(xr1) ∈ U

∀ r ∈ R, implying xr1 ∈ [R : U] ∀ r1 ∈ R.

Again, r(r1x) = (rr1)x = r2x for some r2 ∈ R. So r(r1x) = r2x ∈ U, implying

r1x ∈ [R : U]. So r1x ∈ [R : U] ∀ r1 ∈ R. Hence [R : U] is an ideal of R.

Also if x ∈ U, therefore rx ∈ U

∀ r ∈ R as U is an ideal of R. But

rx ∈ U

∀ r ∈ R implies x ∈ [R : U]. Thus U ⊂ [R : U]