Question

"Let m and n be positive integers with n dividing m. Prove that

the natural surjective ring projection Z/mᶻ ≥ Z/nᶻ is also
surjective on the units"

Answer 1

To prove the given statement, we show that for any element a in Z/mᶻ, there exists an element b in Z/nᶻ such that the projection of b onto Z/mᶻ is equal to a.

Step-by-step explanation:

The given statement can be restated as:

For positive integers m and n, if n divides m, then the natural surjective ring projection Z/mᶻ ≥ Z/nᶻ is surjective on the units.

To prove this, we need to show that for any element a in Z/mᶻ, there exists an element b in Z/nᶻ such that the projection of b onto Z/mᶻ is equal to a.

Let's consider a = [k], where k is a unit of Z/mᶻ (i.e., (k,m)=1). Since k is a unit, there exists an integer l such that kl ≡ 1 (mod m). We can write this as kl = 1 + km for some integer m.

Now, consider the element b = [1 + km] in Z/nᶻ. Since 1 + km is congruent to 1 (mod m) and n divides m, we have 1 + km ≡ 1 (mod n). Therefore, b is an element of Z/nᶻ and its projection onto Z/mᶻ is [1 + km] = [1] = a.

This proves that the natural surjective ring projection Z/mᶻ ≥ Z/nᶻ is surjective on the units.