10 SolutionLet J 0 5 in Z10 Verify that J is an ideal To ve

Solution

Let J = 0, 5 in Z10. Verify that J is an ideal

To verify that J is an ideal we note that it is a subring because it is closed under subtraction and multiplication (the four possible subtractions are 0 0 = 0, 5 5 = 0, 5 0 = 5 and 0 5 = 5 = 5; the four possible multiplicationsare0·0=0,5·5=25=5,5·0=0,and0·5=0). To show further that J is an ideal we need to show that it is closed under outside multiplication, namely, that 1a, 2a, 3a, 4a, 6a, 7a, 8a, and 9a are allinJ foranyaJ. Thisisclearifa=0,sosupposethata=5,whence we compute that 1 · 5 = 5, 2 · 5 = 10 = 0, 3 · 5 = 15 = 5, 4 · 5 = 20 = 0, 6·5=30=0,7·5=35=5,8·5=40=0,and9·5=45=5. ThusJ is an ideal. To show that Z10/I = Z5, it is enough to show that there is a surjective homomorphism from Z10 Z5, and that J is the kernel of this map.

Let f : Z10 Z5 be the map which sends [a]10 to [a]5. To show that f is well defined, we need to show that whenever [a]10 = [b]10 in Z10, then f([a]10) = f([b]10) in Z5. If [a]10 = [b]10, then a b = 10t for some t Z. Thus a b = 5(2t) and hence f([a]10) = [a]5 = [b]5 = f([b]10) as required.

To show that f is a homomorphism, note that for any [a]10,[b]10 Z10, f([a]10 +[b]10) = f([a+b]10) = [a+b]5 = [a]5 +[b]5 = f([a]10)+f([b]10) and f([a]10[b]10) = f([ab]10) = [ab]5 = [a]5[b]5 = f([a]10)f([b]10) as required.

To show that f is surjective note that f([0]10) = [0]5, f([1]10) = [1]5, f([2]10) = [2]5, f([3]10) = [3]5, and f([4]10) = [4]5. Because we can hit everything in Z5, f is a surjection.