Question

Let R be a ring and fix b∈R. (a) Show that S={n1 R​∈R:n∈Z} is a subring of R.

Answer 1

To show that S is a subring of R, we need to demonstrate closure under addition, subtraction, and multiplication, as well as the presence of additive and multiplicative identities.

Step-by-step explanation:

To show that S={n1 R∈R:n∈Z} is a subring of R, we need to demonstrate that S is closed under addition, subtraction, and multiplication, and that it contains the additive and multiplicative identities of R.

Let x, y be any two elements of S. Then x = n1r1 and y = n2r2, where n1, n2 are integers and r1, r2 are elements of R.

Now, let's consider the sum x+y and the product xy. For x+y, we have:

x + y = (n1r1) + (n2r2) = (n1 + n2)r1 + r2, which is an element of S since n1 + n2 is an integer.

For xy, we have:

xy = (n1r1)(n2r2) = (n1n2)(r1r2), which is also an element of S since n1n2 is an integer.